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ELECTRO CHEMISTRY
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ELECTROCHEMISTRY

1.
INTRODUCTION
Chemical reactions can be used to produce electrical energy, conversely, electrical energy can
be used to carry out chemical reactions that do not proceed spontaneously.
Electrochemistry is the study of production of electricity from energy released during
spontaneous chemical reactions and the use of electrical energy to bring about non -spontaneous
chemical transformations. The subject is of importance
importance both for theoretical and practical
considerations. A large number of metals, sodium hydroxide, chlorine, fluorine and many other
chemicals are produced by electrochemical methods. Batteries and fuel cells convert chemical
energy into electrical energy and are used on a large scale in various instruments and devices.
The reactions carried out electrochemically can be energy efficient and less polluting.
Therefore, study of electrochemistry is important for creating new technologies that are eco
friendly. The transmission of sensory signals through cells to brain and vice versa and
communication between the cells are known to have electrochemical origin. Electrochemistry,
is therefore, a very vast and interdisciplinary subject. In this Unit, we will cover only some of
its important elementary aspects.
ELECTROCHEMICAL CELL :
If a metal electrode is immersed in an aqueous solution containing cations of that metal, an
equilibrium that leads to negative charge formation on the electrode is established. This
configuration of electrode and solution is called a half-cell. Two half-cells can be combined to
form an electrochemical cell. The equilibrium condition in an electrochemical cell is that the
electrochemical potential, rather than the chemical potential, of
of a species is the same in all parts
of the cell. The electrochemical potential can be changed through the application of an
electrical potential external to the cell. This allows the direction of spontaneous change in the
cell reaction to be reversed. Electrochemical
Electrochemical cells can be used to determine the equilibrium
constant for the cell reaction. Electrochemical cells can also be used to provide power, in which
case they are called batteries. Electrochemical cells in which the reactants can be supplied
continuously are called fuel cells.
Electrochemical cell are of two types
Galvanic cells or Voltaic cell
A spontaneous chemical reaction
generates an electric current.
Electrolytic cell
An electric current drives a
nonspontaneous reaction.
Thus the two types of cells are reverse of each other.
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2.
GALVANIC CELL OR VOLTAIC
VO
CELL :
A galvanic cell is an electrochemical cell that converts the chemical energy of a spontaneous
redox reaction into electrical energy. In this device the Gibbs energy of the spontaneous redox
reaction is converted into electrical work which may be used for running
a motor or other
r
electrical gadgets like heater, fan, geyser, etc.
2.1
Constriction
onstriction of galvanic cell :
Anode : Some metals (which are reactive) are found to have tendency to go into the solution
ns or their salt solutions.
phase when
en these are placed in contact with their ions
solution For example : Zn
rod is placed in ZnSO4 solution.
If a Zn electrode is partially immersed in an aqueous solution of ZnSO4, an equilibrium is
established between Zn(ss) and Zn2+(aq)) as a small amount of the Zn goes into solution as
2+
Zn2+(aq)) ions as depicted in above figure. Zn(s)  Zn +2e–
However, the electrons remain on the Zn electrode. Therefore, a negative charge builds up on
the Zn electrode, and a corresponding positive charge builds up in the surrounding solution.
This charging leads to a difference in the electrical potential between the electrode and the
solution, which we call the half-cell
half
potential or electrode potential. This particular electrode is
known as anode.
*
On anodee oxidation will take place. (release of electrons).
electron
*
To act as source of electrons.
*
It is of negative polarity.
*
The electrode potential is represented by EZn(s) / Zn2+ (aq.)
Cathode :
Some metals (Cu, Ag, Au etc.,) are found to have the opposite tendency i.e., when placed in
contact with their aqueous ions, the ions from the solution will get deposited on the metal rod.
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2+
The following equilibrium
librium will be established :
Cu +2e–  Cu(s).
So rod will have deficiency of electron (positive charge). Extra negative charge will surround
this positively charged rod and form double layer. An electrical double layer is developed in the
system and hence a potential difference is created between the rod and the solution which is
is will be known as cathode.
known as half cell potential or electrode potential. This
*
At cathode reduction will take place. (gain of electrons will take place)
*
To act as sink of electron.
*
Positive polarity will be developed.
*
Their electrode potential can be represented by : E Cu2+(aq.)/Cu(s)
Is where oxidation occurs
Is where reduction occurs


Anode : Is where electrons are produced
Cathode : Is where electrons are consumed
Has a negativesign
Has a positive
positivesign
sign


*
cell potentials cannot be measured directly. They are measured relative to one
Half-cell
another rather than absolutely. To understand how this is done, it is useful to consider
cells, such as the one shown figure.
an electrochemical
rochemical cell, which consists of two half-cells,
This particular cell is known as the Daniell cell, after its inventor. On the left, a Zn
). The solute is completely dissociated
electrode is immersed in a solution of ZnSO4(aq).
2+
2
to form Zn (aq)) and SO 4 (aq).. On the right, a Cu electrode is immersed in a solution
of CuSO4, which is completely dissociated to form Cu2+(aq)) and SO 24  (aq) . The two
cells are connected by an ionic conductor known as a salt bridge. The salt bridge
half-cells
consists of an electrolyte such as KCl suspended in a gel. A salt bridge allows current to
solutions. A metal wire
flow between the half-cells
half cells while preventing the mixing of the solu
fastened to each electrode allows the electron current to flow through the external part
of the circuit.
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2.2
Salt bridge :
A salt bridge is a U–shaped
shaped inverted tube that contains a gel permeated with an inert
agar powder with a natural
electrolyte. Generally tube is filled with a paste of agar
agar-agar
electrolyte/generally not common to anodic/cathodic compartment with porous plugs at each
mouth of tube. The ions of the inert electrolyte do not react with other ion in the solution and
the ions are not oxidised or reduced at the electrodes.
The electrolyte in salt bridge should be such that speed of it's cation equals speed
of it's anion in electrical field. For that charge and sign of the ions should be
almost equal i.e. Mobility of cation = Mobility of anion
KCl is generally preferred but KNO3 or NH4NO3 can also be used.
*
If Ag+, Hg22+, Pb2+, Tl+ ions are present in a cell then in salt bridge KCl is not used
because there can be formation of precipitate of
o AgCl, Hg2Cl2, PbCl2 or TlCl at mouth
of tube which will prevent the migration of ions and its functioning will stop.
Functions of Salt Bridge :
(i)
It connects the solution of two half cell to complete the circuit.
(ii)
It minimize the liquid junction
junction potential. The potential difference between the junction
of two liquids.( Liquid-Liquid Junction Potential :The
The potential difference which
arises between two solutions (during the progress of reaction) when in contact with
each other.)
(iii)
It maintains the electrical neutrality of the solution in order to give continuous flow or
generation of current.
"The simultaneous electrical neutrality of the anodic oxidation chamber and cathodic
reduction chamber is due to same mobility or velocity of K+ and NO3– ions taken into
salt bridge.
(iv)
If the salt bridge is removed then voltage drops to zero.
(v)
It prevents mechanical mixing of two electrolytic solution.
2.3
Shorthand Notation for Galvanic Cells
We require two half cells to produce an electrochemical cell, which can be represented by
following few rules;
*
The anode half-cell
cell is always written on the left followed on the right by cathode half
cell.
*
The separation of two phases (state of matter) is shown by a vertical line.
*
The various materials present in the same phase are shown together using commas.
*
The salt bridge is represented by a double slash (||).
*
The significant features of the substance viz. pressure of a gas, concentration of ions
etc. are indicated in brackets immediately after writing the substance.
*
For a gas electrode, the gas is indicated after the electrode for anode and before the
electrode in case of cathode. (i.e Pt H2 / H+ or H+ /H2 Pt)
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Ex.1
Sol.
Ex.2
Sol.
2.4
Write short hand notation for the following reaction,
Sn2+ (aq) + 2Ag+ (aq) 
 Sn4+ (aq) + 2Ag(s).
The cell consists of a platinum wire anode dipping into an Sn+2 solution and a silver cathode
dipping into an Ag+ solution ttherefore
Pt(s) | Sn2+(aq), Sn4+ (aq) || Ag+ (aq) | Ag(s).
Write the electrode reaction and the net cell reaction for the following cells. Which electrode
would be the positive terminal in each cell ?
(a) Zn | Zn2+ || Br–, Br2 | Pt
(b) Cr| Cr3+ || I– , I2 | Pt
(c) Pt | H2, H+ || Cu2+ | Cu
(d) Cd | Cd2+ || Cl– , AgCl | Ag
(a)
Oxidation half cell reaction, Zn 
 Zn2+ + 2e–
reduction half cell reaction, Br2 + 2e– 
 2Br–
Net cell reaction
Zn + Br2 
 Zn2+ + 2Br–
(Positive terminal : cathode Pt)
(b)
Oxidation half reaction,
[Cr 
 Cr3+ + 3e–] x 2
reduction half reaction,
[I2 + 2e– 
 2I–] x 3
Net cell reaction
2Cr + 3I2 2Cr3+ + 6I–
(Positive terminal : cathode Pt)
(c)
Oxidation half reaction,
H2 
 2H+ + 2e–
reduction half reaction,
Cu2+ + 2e– 
 Cu
2+
Net cell reaction
H2 + Cu 
 Cu + 2H+
(Positive terminal : cathode Cu)
(d)
Oxidation half reaction,
Cd 
 Cd2+ + 2e–
reduction half reaction,
[AgCl + e– Ag + Cl–] x 2
Net cell reaction
Cd + 2AgCl Cd2+ + 2Ag + 2Cl–
(Positive terminal : cathode Ag)
Half cell potential/ Electrode Potential :
The
he potential difference developed between metal electrode and its ions in solution in known as
electrode potential.
The potential difference developed between metal electrodes and the solution in
standard conditions (i.e. 1M
1 concentration of its ions at 1 bar pressure and at a particular
temperature) is known as standard electrode potential. According to IUPAC convention,
standard reduction potentials are
are now called standard electrode potentials
Two types of oxidation potentials :
Standard Oxidation Potential(SOP) :
The electrode potential for oxidation half
reaction
Greater is the SOP, greater will be tendency
to get oxidized in standard conditions.
conditions
Reduction Potential (R.P.)
The electrode potential for half reduction half
reaction
Greater is the R.P. greater will be tendency to
get reduced in standard conditions.
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Type of Electrode
Representation
Half reaction
2+
–
1. Metal electrode
Reduction : Zn + 2e  Zn(s)
(Zn
electrode,Cu Oxidation : Zn(s)  Zn2+ + 2e–
electrode etc.)
2
E 0Zn 2 /Zn(s) (SRP)
E 0Zn(s)/ Zn2 (SOP)
Hydrogen peroxide Reduction : 2e– + 2H+ + H2O2  2H2O
electrode
Oxidation : H2O2  O2 + 2H+ + 2e–
3. Redox electrode
Reduction : MnO 4– + 8H+ + 5e–  Mn
4H2O
E 0H2O2 /H2O (SRP)
2+
E 0H2O2 /O2 (SOP)
+
E 0MnO /Mn 2 (SRP)
4
2.3
Free energy changes for cell reaction :
The free energy change  G (a thermochemical quantity) and the cell potential
E(an electrochemical quantity) both measure the driving force of a chemical
reaction.
The values of  G and E are directly proportional and are related by the equation,
 G = –nFE
nFE
where n = Number of moles of electron transferred in the
t reaction.
–
F = Faraday constant = 96485 C/mole e
2.4
Cell potential :
The cell potential is the difference between the electrode potentials (reduction potentials) of the
cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is
drawn through the cell.
ed from the values of electrode
The emf of the cell or cell potential can be calculat
calculated
potential of the two half cell constituting the cell. The following three method are
in use :
(i)
When oxidation potential of anode and reduction potential of cathode are taken
into account :
Ecell = oxidation potential of anode + reduction potential of cathode
= E ox (anode) + E red (cathode)
 In standard conditions
E 0cell = SOP of anode + SRP of cathode
E 0cell = E 0ox (anode) + E 0red (cathode)
(ii) When reduction potential of both electrodes are taken into account :
Ecell = Reduction potential of cathode – Reduction potential of anode
= E red (cathode) – E red (anode)
 In standard conditions
E 0cell = SRP of cathode + SRP of anode
E 0cell = E 0red (cathode)– E 0red (anode)
E°cell is intensive property so on multiplying/Dividing
multiplying/Dividing cell reaction by any
number, the E°cell value would not change.
2.4.1 Measurement of Electrode Potential :
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The potential of individual half-cell
half cell cannot be measured. We can measure only the difference
cell potentials that gives the emf of the cell. If we arbitrarily choose the
between the two half-cell
potential of one electrode (half-cell)
(half cell) then that of the other can be determined with respect to
this. According to convention, a half-cell
half cell called standard hydrogen electrode (Reference
(
electrode) is assigned a zero potential at all temperatures
temperatures corresponding to the reaction
+
–
E 0H /H = 0 V
2H (aq, 1M) + 2e  H2 (g, 1 bar)
2
H2(g, 1bar)  2H+ (aq, 1M) + 2e–
E
0
H 2 /H 
=0V
*
 f G 0 [H  (aq .)]  0
The standard hydrogen electrode consists of a platinum electrode coated with platinum black.
+
To achieve equilibrium on a short timescale, this reaction 2H + 2e–  H2 is carried out
over a platinum black catalyst electrode.
ydrogen gas is bubbled through it. The
The electrode is dipped in an acidic solution and pure hhydrogen
concentration of both the reduced and oxidised forms of hydrogen is maintained at unity. This
implies that the pressure of hydrogen gas is one bar and the concentration of hydrogen ion in
the solution is one molar.
half-cell constructed by
Att 298 K the emf of the cell, standard hydrogen electrode with second half
cell) and the other half-cell
as
taking standard hydrogen electrode as anode (reference half
half-cell)
half
cathode, gives the reduction potential of the other half-cell.
half
If the concentrations of the oxidised
and the reduced forms of the species in the right hand half-cell
half cell are unity, then the cell potential
is equal to standard electrode potential,
E0cell  E 0ox
 E0red
  
 0  E 
SHE
E
0
cell
0
red Electrode

Electrode
  E 0red 
Electrode
The measured emf of the cell:
Pt(s) | H2(g, 1 bar) | H+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu
is 0.34 V and it is also the value for the standard electrode potential of the half – cell
corresponding
ing to the reaction:
reaction
Cu2+ (aq, 1 M) + 2 e–  Cu(s)
Similarly, the measured emf of the cell:
Pt(s) | H2(g, 1 bar) | H+ (aq, 1 M) || Zn2+ (aq, 1 M) | Zn
half-cell reaction:
is – 0.76 V corresponding to the standard electrode potential of the half
2+
–
Zn (aq, 1 M) + 2e  Zn(s)
The positive value of the standard electrode potential in the first
first case indicates that Cu2+ ions
get reduced more easily than H+ ions. The reverse process cannot occur, that is, hydrogen ions
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cannot oxidize Cu (or alternatively we can say that hydrogen gas can reduce copper ion) under
the standard conditions described above. Thus, Cu does not dissolve in HCl. In nitric acid it is
oxidised by nitrate ion and not by hydrogen ion. The negative value of the standard electrode
potential in the second case indicates that hydrogen ions can oxidise zinc (or zinc can reduce
hydrogen ions).
In view of this convention, the half reaction for the Daniell cell in Fig. 3.1 can be written as:
Left electrode : Zn(s)  Zn2+ (aq, 1 M) + 2e–
Right electrode : Cu2+ (aq, 1 M) + 2e–  Cu(s)
The overall reaction of the cell is the sum of above two reactions and we obtain the equation:
Zn(s) + Cu2+ (aq)  Zn2+ (aq) + Cu(s)
emf of the cell = E 0cell = E 0R  E 0L
= 0.34V – (– 0.76) V = 1.10 V
Sometimes metals like platinum or gold are used as inert electrodes. They do not participate in
the reaction but provide their surface for oxidation or reduction reactions and for the
conduction of electrons.
For example, Pt is used in the following half – cells:
Hydrogen electrode : Pt(s) | H2(g) | H+(aq)
1
With half-cell
cell reaction : H+(aq) + e–  H2(g)
2
–
Bromine electrode : Pt(s) | Br2(aq) | Br (aq)
1
With half –cell
cell reaction : Br2(aq) + e–  Br– (aq)
2
The standard electrode potentials are very important and we can extract a lot of useful
information from them. The values of standard electrode potentials for some selected half-cell
half
reduction reactions are given in Table. If the standard electrode potential
potenti of an electrode is
greater than zero then its reduced form is more stable compared to hydrogen gas. Similarly, if
the standard electrode potential is negative then hydrogen gas is more stable than the reduced
form of the species. It can be seen that the standard electrode potential for fluorine is the
highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced
to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion
is the weakest
st reducing agent. Lithium has the lowest electrode potential indicating that lithium
ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in
an aqueous solution. It may be seen that as we go from top to bottom in Tabl
Table the standard
electrode potential decreases and with this, decreases the oxidizing power of the species on the
left and increases the reducing power of the species on the right hand side of the reaction.
Electrochemical cells are extensively used for determining
determining the pH of solutions, solubility
product, equilibrium constant and other thermodynamic properties and for potentiometric
titrations.
Electrode
Reaction
SRP (at 298 K)
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* Li
Li+ + e–  Li(s)
– 3.05 V
K
Ba
Ca
K+ + e–  K (s)
– 2.93 V
Ca+2 + 2e–  Ca(s)
– 2.87 V
Na
–2.71 V
Na+ + e–  Na(s)
+2
–
Mg
–2.71 V
Mg + 2e  Mg(s)
Al
* Electrolytes (H2O) H2O(l) + e–  H2 + OH– – 0.828 V
Zn
Cr
* Fe
Cd
Co
Ni
Sn
Pb
*H2
Cu
I2
Fe
Hg
Ag
Hg
Br2
Zn+2 + 2e–  Zn(s)
Cr+3 + 3e–  Cr(s)
Fe2+ + 2e-  Fe
Cd+2 +2e–  Cd(s)
–0.76 V
–0.74 V
–0.44 V
–0.41 V
Ni+2 + 2e–  Ni(s)
Sn+2 + 2e–  Sn(s)
Pb+2 + 2e–  Pb(s)
2H+ + 2e–  H2(g)
Cu2+ + 2e–  Cu(s)
–0.24 V
–0.14 V
–0.13 V
0.00V
0.34 V
Fe3+ + e–  Fe2+
Hg22+ + 2e-  Hg(l)
Ag+ + e–  Ag
Hg2+  Hg(l)
Br2 + 2e–  2Br–
0.77 V
0.79 V
0.80 V
0.85 V
1.06 V
Electrochemical Series :
Ex.3
*
Electrolytes
O2 + 2H+ + 2e–  H2O()
*
*
*
*
Cr2O72–
+
–
+3
+ 14H + 6e  2Cr + 7H2O
Cl2 Cl2 +2e–  2 Cl–
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
F2 + 2e–  2F–
1.23 V
1.33 V
1.36 V
1.51 V
2.87 V
Calculate E0cell of (at 298 K), Zn(s) / ZnSO4(aq) || CuSO4(aq) / Cu(s)
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given that
Sol.
E0Zn/Zn2+(aq) = 0.76 V, E0Cu(s) / Cu2+(aq) = – 0.34 V
E0cell = (S.R.P)cathode – (S.R.P)anode
= 0.34 – (– 0.76) = 1.1 V
Ex.4
Given the cell Ag AgCl(s) | NaCl (0.05 M) || Ag NO3 (0.30 M) | Ag
(a) Write half reaction occurring at the anode.
(b) Write half reaction occurring at the cathode.
(c) Write the net ionic equation of the reaction.
(d) calculate E°cell at 25°C.
(e) Does the cell reaction go spontaneous as written ?
(Given E°AgCl,Cl = + 0.22 volt) ; E 0Ag  /Ag = + 0.80 volt)
Sol.
(a)
LHS electrode is anode and half reaction is oxidation.
Ag+ + Cl–  AgCl(s) + e–
(b)
... (i)
RHS electrode is cathode and half reaction is reduction.
Ag + e  Ag(s)
... (ii)
(c)
From equation (i) and (ii) cell reaction is : Cl– (0.05 M) + Ag+ (0.30 M) AgCl(s)
(d)
E°cell = E°right – E°left
= (0.80 – 0.22 volt = 0.58 volt
(e)
Yes, the e.m.f. value is positive, the reaction will be spontaneous as written in the cell
eaction.
2.4.2Calculation of Electrode Potential of unknown electrode with the help of given (two)
electrode.
(i)
Obtain the reaction of the 3rd electrode with the help of some algebraic operations on
reactions of the given electrodes.
(ii)
Then calculate Gº of the 3rd reaction with the help of some algebaric operations of G0
of 1st and 2nd reactions.

(ii)
Use G0 = –nF E0elec. to calculate unknown E.P.
ivide electrode reaction by any number the E 0cell
E 0cell is intensive property so if we multiply/
multiply/divide
value would not changed
i.e.
Zn2+ + 2e–  Zn(s)
E° = – 0.76 V
2Zn2+ + 4e–  2Zn(s)
E° = – 0.76 V (remain same)
Multiply by 2
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Ex.5
Given that E0Cu2+/Cu = 0.337 V and E0Cu+
/Cu2+
= – 0.153 V. Then calculate
E 0Cu  /Cu .
Sol.
(i)
(ii)
Cu2+ + 2e– Cu
G1
+
2+
–
Cu Cu + e
G2
____________________
after adding
Cu++ e–  Cu
G1
+ G2 = G3
–2F
2F E10 – F E 02 = – F E30
E3 = 2 E10 + E 02
= 2 x 0.337 – 0.153
= 0.674 – 0.153 = 0.521 V
Ex.6
E0
E0
E0
= –1.51 V
Mn 2 /MnO4
= + 1.23 V
MnO2 /Mn 2
MnO4 /MnO2
=?
(All in acidic medium)
4H2O + Mn2+  Mn O4 + 8H+ + 5e–
Sol.
G1
(i) Mn O 4 + 8H+ + 5e– 4H
 2O + Mn2+
–G1
2e– + MnO2 + 4H+  Mn2+ + 2H2O
G2
(ii) 2H2O + Mn2+  MnO2 + 4H+ + 2e–
–G2
(iii) 4H+ + Mn O4 + 3e– MnO

2 + 2H2O

G3
(i) + (ii) = (iii)
G3 = – G1 – G2
–3E3F = 5 E10 F + 2 E 02 F
E=
Ex.7
[5E1  2E 2 ] [5(1.51)  2(1.23)] [7.55  2.46] 5.09
=
= 1.69 V
=
=
3
3
3
3
Will Fe2+ disproportionate or not
anode
cathode
0.77 V
 0.44 V
Fe  

  Fe2+  
  Fe
3+
-0.036
Sol.
This is known as Latimer
atimer diagram.
S.R.P to right of the species greater than SRP of its left species will undergo disproportionation.
C.O.: NAIVEDHYAM, Plot No. SP-11,
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2.5

NERNST EQUATION :
Cell potentials depend on temperature and on the composition of the reaction mixtures.
It depends upon the concentration of the solute
solute and the partial pressure of the gas, if any.
The dependence upon the concentration can be derived from thermodynamics.
From thermodynamics
G = G° + RT ln Q
– nFE = – nFE° + 2.303 R T log Q
2.303RT
E = E° –
log Q
nF
Take T = 298 K , R = 8.314 J/mol K, F = 96500 C
0.059
Now we get, E = E° –
log Q
n
Where n = number of transfered electron, Q = reaction quotient
Nernst equation can be used to calculate cell potentials for non standard conditions also.
Nernst equations can be applied to half cell reactions also.
Applications of Nernst equation
2.5.1Nernst Equation for Electrode Potential
Mn+(aq) + ne–  M(s)
RT
 M(s) 
ERed = E 0red –
n
 M n  
nF
ERed = E 0red –
at 298K,
ERed = E 0
Re d n
2.303RT
M(s)
log  n  
nF
M 
0.059
1
–
log  n  
n
M 
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Hydrogen
Electrode
H2(g)  2H+(aq) + 2e–
E = E0 –

 (H )2 
0.0591
log 

2
 PH2 
Metal–metal soluble salt electrode.
Zn2+ + 2e– Zn(s)
Zn(s)

ERed = E 0red 
0.059
1
log  2 
2
 Zn 
Re d n

2.303RT
1
log  2  at 298K
nF
 Zn 
ERed = E 0
Gas – electrode Hydrogen electrode.
2H+ + 2e– H2(g)
ERedn= E 0
Re d n


 PH 
0.059
log  2 2 
2
 [H ] 
Redox electrode
4H2O + Mn2+ Mn
Mn O4 + 8H+ + 5e–
EOX = E 0ox 
Ex.8
0.059
[MnO 4 ][H  ]8
log
5
[Mn 2 ]
Calculate R.P. of hydrogen electrode at 298K which is prepared with
with the help of aq. solution of
acetic acid with 0.1 M concentration at 1 atm pressure Ka = 1.8 x 10–55.
Sol.
[H+] =
Ka  c = 1.8  105 101 = 1.8 106
2H+ + 2e– H2
ERedn= E 0red 
ERedn= 
PH2
0.059
log
2
[H  ]2
0.059
1

log 
6 
2
 1.8  10 
ERedn = 
(E0Redn = 0)
=
0.059
[6 – log (1.8)]
2
0.059
x 5.74 = –0.169 V
2
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Ex.9
Which is stronger oxidizing agent
(i) K2Cr2O7 in solution in which
[Cr2O72–] = 0.1 M, [Cr3+] = 10–2 M and [H+] = 10–1 M
(ii) KMnO4 in a solution in which
[MnO4–] = 10–1M, [Mn2+] = 10–2M, [H+] = 10–2 M
E 0Cr O 2– /Cr 3 = 1.33 V E 0 MnO /Mn 2 = 1.51 V
Sol.
(i) 14H+ + Cr2O72– 2Cr+3 + 7H2O + 6e–
2
7
4
104 10 
0.059
0.059
log 
= 1.33 –
x 11

14
6
6
 10

0.649
ERedn = 1.33 –
= 1.330 – 0.108 = 1.222 V
6
(ii) 5e– + 8H+ + Mn O 4  Mn2+ + 4H2O
ERedn = 1.33 –
ERedn = 1.51 –
 102

0.059
log  16
1  = 1.51 – 0.059 x 3 = 1.51 – 0.18 = 1.33 V
5
10 10 
ERedn is more so, good oxidising agent
2.5.2Nernst Equation for cell Potential :
 cC + dD
aA + bB 
RT
lnQ
nF
n – no. of electrons which gets cancelled out while making cell reaction.
Ecell = E 0cell –
Equilibrium in electrochemical cell
0
G = – nF Eºcell
G = – nF Ecell
From thermodynamics
0
G = G + RTnQ
at chemical equilibrium G = 0
Ecell = 0 cell will be of no use
so, G0 = – RTn Keq
at equilibrium – nF Eºcell = –2.303 RT log (Keq)
nF
log Keq =
Eºcell
2.303 RT
at 298 K and R = 8.314 J/mol K
n
log Keq =
Eºcell
0.059
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Cl2 (g) Cl- (aq)
0.1 atm 10-2 M
Ex.10 Calculate Ecell of Pt(s)
Cr2 O72- ,Cr
, Cr +3  in H2SO4  = 0.05M
Pt
0.01 M
0.1 M
given that E 0Cr O2 //Cr
= 1.33 V
Cr 3
2
0
Sol.
7
–
E Cl | Cl2 = –1.36
1.36 V
6e– + 14H+ + Cr2O72–  2Cr+3 + 7H2O
[2Cl–  Cl2 + 2e–] x 3
14H+ + 6Cl– + Cr2O72–  3Cl2 + 2Cr+3 + 7H2O
E 0cell = 1.33 – (+1.36) = – 0.03
Ecell = – 0.03 –
[Cr 3 ]2 [PCl2 ]3
0.059  23
0.059
log
= – 0.03 –
 14
 6
2
[H ] [Cl ] [Cr2O7 ]
6
6
Ecell = – 0.26 V
Example problem 11.5
For the Daniell cell E° = 1.10 V. calculate K for the reaction at 298.15 K
Zn(s) + Cu2+ (aq)  Zn2+ (aq) + Cu(s).
Solution.
In K =
nF
2  96485C mol –1  1.10V
E =
= 85.63
RT
8.314J K –1 mol 1  298.15K
K = 1.55 × 1037
Note that the equilibrium constant calculated in Example Problem 11.5 is so large that it could
not have been measured by determining the activities of a 2 and a 2 by spectroscopic
Zn
Cu
methods. This would require a measurement technique that is accurate over more than 30 orders
of magnitude in the activity. By contrast, the equilibrium constant in an electrochemical cell can
be determined with high accuracy using only a voltmeter.
A further example of the use of electrochemical measurements to determine equilibrium
constants is the solubility constant for a weakly soluble salt. If the overall reaction corresponding
otentials, then the solubility constant can
to dissolution can be generated by combining half-cell
half
potentials,
cell reaction can be combined
be calculated from the potentials. For example, the following half
half-cell
to calculate the solubility product of AgBr.
and
AgBr(s) + e–  Ag(s) + Br–(aq)
E° = 0.07133 V
+
–
Ag(s)  Ag (aq) + e
E° = –0.7996 V
___________________________________________
AgBr(s)  Ag+ (aq) + Br– (aq)
E° = –0.7283 V
In
KSP =
nF
1 96485C mol1  (–0.7283V)
E° =
= – 28.35
RT
8.314J K –1 mol1  298.15K
The value of the solubility constant is KSP = 4.88 × 10–13.
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 Zn + Fe2+, is – 0.32 volt at 25°C. What will be the
Ex.11 The E°cell for the reaction Fe + Zn2+ 
equilibrium concentration of Fe2+, when a piece of iron is placed in a 1 M Zn2+ solution ?
Sol.
We have the Nernst equation at equilibrium at 25°C
E° =
0.0591
log K
n
... (i)
Since E°Cell for the given reaction is negative, therefore, the reverse reaction is feasible
for which E°cell will be + 0.32 V, Thus for
Zn
–
x
Now , E° =
log
 Fe

Fe2+
+
[Zn2  ]
[Fe2 ]
+
–
0.0591
[Zn 2  ]
log
n
[Fe 2  ]
= – 10.829
Zn2+ ;
E°Cell + 0.32 V
(1–x)
or
0.32 =
0.0591
[Zn 2+ ]
log
2
[Fe2+ ]
Taking antilog,
[Fe2+] = 1.483 × 10–11 M
Work done by a cell :
(i)
Let 'n' faraday charge be taken out of a cell of EMF 'E' ; then work done by the cell will
be calculated as :
(ii)
work = Charge × Potential = nFE
Work done by cell = Decrease in free energy
so
– G = nFE
Ex.12 Calculate the maximum work that can be obtained from the Daniel cell given below Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s). Given that Eº Zn
Sol.
Cell reaction is :
2
/ Zn
= – 0.76 V and Eº Cu
Zn(s) + Cu2+ (aq)  Cu(s) + Zn2+ (aq)
2
/Cu
= + 0.34 V.
Here n = 2
Eºcell = Eºcathode – Eºanode (On the basis of reduction potential)
= + 0.34 – (0.76) = 1.10 V
We know that :
Wmax = Gº = – nFEº
= – (2 mol) × (96500 C mol) × (1.10 V) = – 212300
C.V. = – 212300 J
or
Wmax = – 212300 J
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2.6
CONCENTRATION CELLS :
A concentration cell consists of two electrodes of the same material, each electrode dipping in a
solution of its own ions and the solution being at different concentrations.
The two solutions are separated by a salt bridge.
e.g. Ag(s) | Ag+ (a1) || Ag+ (a2) | Ag(s)
(a1 < a2) a1 , a2 are concentrations of each half cell
At LHS electrode Anode : Ag (s)  Ag+(a1) + e–
At RHS electrode Cathode : Ag+(a2) + e–  Ag(s)
The net cell reaction is : Ag+ (a2)  Ag+ (a1)
The nernst eq. is
0.059
a
Ecell = –
log 1
(Here n = 1, Temp, 298 K)
n
a2
Likewise,
the e.m.f. of the cell consisting of two hydrogen electrodes operating at different
pressure P1 and P2 (P1 > P2 ) and dipping into a solution HCl is :
0.059
P
Ecell =
log 1 (at 298 K)
2
P2
Q.
If for the concentration cell.
 n  1 
n  1
X(s) X(aq)
 M  X(aq)  M  X(s)
 20 
2 
Ecell = 0.029 V at 298 K calculate n.
Sol. X(s)
 Xn+(aq) + ne– anodic reaction
cathodic reaction
X n  (aq)  ne 
 X (s)
 X n  (aq)anode
X n  (aq)cathod 
0.029 = 0 –
[X n ]
0.059
log  n anode
n
[X ]cathode
 1 
 
0.059
20
 0.029 = 0 –
log  
n
1
 
2
0.059
1
log
n
10
0.059
log10 –1
0.029 = –
n
0.059
0.029 =
n
 n=2
0.029 = –
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Q.
A hydrogen electrode is immersed in a solution pH = 0 (HCl). By how much will the reuction
potential change if an equivalent amount of NaOH is added to this solution, so that solution, so


that solution becomes neutral PH2  1atm T =298 K
Sol. 2H+(aq) + 2e–

ERP = E°SRP –
ERP = 0 –
H2(g)
 
0.0591  PH2 

log  
  H (aq)  
n


0.059
1
log  2 
1
1 
E RP1 = 0 –
0.059
1
log  2  = 0
1
1 
E RP1 = 0
E RPfinal = 0 –
0.059
0.059
1
=
× 14V = 0.4137 V
log
7 2
2
2
(10 )
Change in reduction potential [0.4137 – 0] = 0.4137
Q.
Calculate Ecell of :
Pt
H2
HCOOH (aq.) CH 3COOH H 2
1atm C1 K a1
Ecell = 
Prove that
Sol. HCOOH
C2 Ka 2


1atm
Pt(S)
0.059
C Ka
log 1 1
2
C 2 Ka 2
HCOO–
+
H+
C1(1–1)
c11
c11
Anode
H2(g) 
2Ha+ + 2e–
Cathode
2H+C (aq) + 2e–  H2(g)
(H2(g))a + 2H+c (aq)



2Ha+(aq) + (H2(g))c

[H  ]a2 [PH2 ]c
0.059
log  2
Ecell = 0 –
2
[H ]C [PH2 ]a

[H+]a =

Ecell = 
[H+]c =
c1Ka1
P   P 
H2 c
H2 a
 1atm
c 2 Ka 2
c Ka
0.059
log 1 1
2
c 2 Ka 2
C.O.: NAIVEDHYAM, Plot No. SP-11,
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2.6
METAL -METAL
METAL INSOLUBLE - ION ELECTRODE :
In this half cell, a metal coated with its insoluble salt is in contact with a solution containing the
anion of the insoluble salt.
(i) Construction :
(ii) Half cell reaction :
(a) If it act as Anode : At anode oxidation take place.
If it is normal silver electrode than.
Ag(s) 
 Ag+(aq) + e– ..... (i)
But solution have Cl.(aq) ion soAg+(aq) & Cl.(aq) Form PPT ofAgCl(s)
Ag+(aq) + Cl– (aq) 
 AgCl(s) ..... (ii)
Overall half cell reaction at anode [from eq. (i) & (ii)].
Ag(s) + Cl– (aq) 
 AgCl(s) + e– ..... (iii)
(b) If it act as cathode : At cathode reduction take place.
If it is normal silver electrode than.
Ag+(aq) + e– 
 Ag(s) ..... (i)
AgCl(s) 
 Ag+(aq) + Cl– (aq) ..... (ii)
Overall half cell reaction at cathode [from eq. (i) & (ii)].
AgCl(s) + e. 
 Ag(s) + Cl– (aq) ..... (iii)
(iii) Cell representation :
(a)
(b)
(c)
Abode :
Cathode :
(aq)
||
(aq)
Metal will be at the end.
Metal insoluble salt will be in middle.
Ag(s) | AgCl(s) | Cl– (aq) ||
EAg s
ECl  aq 
AgCl Cl –  aq 
—————
||Cl (aq) | Ag Cl(s) | Ag(s)
AgCl(s) Ag  s 
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Relation between ECl (aq)|AgCl(s)|Ag(s) & Ksp of AgCl
(iv)
unsoluble salt
If a cell is constructed using metal–metal
metal
soluble salt and metal–metal
metal
electrode.
0
 0.22V
Given E 0Ag |Ag  0.80V, ECl

|AgCl|Ag
Anode
Ag|AgCl|Cl–(aq)
||
Cathode
Ag+(aq) | Ag
Anode :
Ag(s) + Cl– (aq)  AgCl(s) + e–
Cathode:
Ag+ + e– Ag
Overall cell reaction:
Ag+ (aq)c + Cl–(aq)a AgCl(s)
At equilibrium the concentratuion of [Ag+]a = [Ag+]c
So that at equilibrium Ecell = 0
0.059
[AgCl(s)]
But
Ecell = E 0cell 
log
1
[Ag  ]c [Cl  ]a
(Metal – metal insoluble salt)
(Metal–metal
metal soluble salt)
At equilibrium
0 = E 0cell 
0.059
1
log

1
[Ag ]c [Cl  ]a
Now for AgCl
AgCl(s)  Ag+ (aq)a + Cl–(aq)a
Ksp = [Ag+]a [Cl–]a
So,
Ksp (AgCl)
[Cl–]a =
0= E
[Ag  ]a
0
cell
E 0cell =
[Ag  ]a
0.059
–
log
[Ag  ]a  Ksp
1
[At equilibrium the concentratuion of [Ag+]a = [Ag+]c]
0.059
1
log
1
Ksp
E 0cell = –
0.059
1
log
1
Ksp
E 0cell = E 0cathode  E 0Anode
= E 0Ag /Ag  E 0Cl /agCl/Ag
So,
0
E 0Ag  |Ag  E Cl
=

|AgCl|Ag
0.059
1
log
1
Ksp
E 0Cl /AgCl/Ag = E 0Ag /Ag 
0.059
log K sp
1
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Some other example :
Pb | PbSO4 | SO42–– || Pb2+ | Pb
White ppt.
0
ESO
= E0Pb2 |Pb 
2
|PbSO |Pb
4
4
0.059
log Ksp
2
Ex. Calculate Ksp of AgI with the help of following cell.
Ag | Agl(s) | 0.05 M KI (aq) || 0.05 M AgNO3 | Ag(s)
If E 0cell = 0
Sol. First methode :
Ag+a + e–

Anode :
Ag
Cathode :
Ag+c + e–  Ag


 Aga+
Agc+ 
According to nernst equation.
Ecell = E0cell 
[Ag  ]a
0.059
log
1
[Ag  ]c
0.7884 = 0 
[Ag  ]a
0.059
log
1
0.05

0.7884
= log [Ag+]a – log (5 × 10–2)
0.059
log [Ag+]a = – 1.3 –
0.7884
0.059
[Ag+]a = 2.17 × 10–15
Ksp = [Ag+]a [I–]a
Ksp = 2.17 × 10–15 × 0.05 = 1 × 10–16.
Second Methode :
Ecell = E 0cell –
if E 0cell  0
0.059
1
log

1
[Ag ]c [I  ]a
Ecell = 
0.059
0.059
1
log K sp 
log

1
1
[Ag ]c [I  ]a
So
[Ag  ]a [I ]a
0.059
log
Ecell = 
1
[Ag  ]c [I  ]a
But
E0cell  
0
Ecell

0.059
log Ksp
1
0.059
log [Ag  ]c [I ]a
1
[Ag  ]a
0.059
log
Ecell = 
1
[Ag  ]c
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(4)
Calomel Electrode :

A calomel cell consists of a platinum electrode dipping into mercury in contact
conta with
calomel (dimercury) (l) chloride, Hg2Cl2) and potassium chloride solution.

Usually the solution is saturated with potassium chloride.

The cell has an e.m.f of 0.246 V at 28° C
Standrad (normal) calomel electrode when [Cl–] = 1 M = 1 N
At Anode
2Hg(l)  Hg 22  (aq) + 2e–
At Cathode
–
Hg(
Hg 22  2e  Hg(l)
–
Hg 22  (aq) + 2Cl  Hg2Cl2(s)
Hg2Cl2  Hg 22  (aq) + 2Cl–(aq)
2Hg(l) + 2Cl–(aq)  Hg2Cl2(s) + 2e–
Hg2Cl2(s) + 2e–  Hg(l) + 2Cl–

Cell representation
Pt(s) | Hg(l) | Hg2Cl2(s) | Cl–(aq) || cathode
E 0Pt Hg|Hg Cl  Cl– = SOP
2
2
E Cl– /Hg Cl
2

Cell representation
(Anode) || Cl–(aq) | Hg2Cl2 | Hg(l) | Pt(s)
E 0Cl |Hg Cl |Hg(l ) = SRP
2 /Hg
0
 ECl

/Hg Cl
2
2 /Hg

2
2
RT
l n[Cl ]
F
DO YOURSELF–1
1.
Colour of KI solution containing starch turns blue when Cl2 water is added. Explain.
2.
E° of some oxidants are given as :
I2 + 2e– 2I–
E° = + 0.54 V
MnO4– + 8H + + 5e–  Mn2+ + 4H2O
E° = + 1.52 V
Fe3+ + e–  Fe2+
E° = + 0.77 V
4+
–
2+
Sn + 2e  Sn
E° = + 0.1 V
(a) Select the strongest reductant and oxidant in these.
(b) Select the weakest reductant and oxidant in these.
(c) Select the spontaneous reaction from the changes given below.
(i)
Sn4+ + 2Fe2+  Sn2+ + 2Fe3+
(ii)
2Fe2+ + I2  2Fe3+ + 2I–
(iii)
3.
Sn4+ + 2I–  Sn2+ + I2
(iv)
Sn2+ + I2  Sn4+ + 2I–
Given the standard electrode potentials ;
K+/K = –2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = – 2.37 V, Cr3+/Cr = –0.74V.
Arrange these metals in their increasing order of reducing power.
C.O.: NAIVEDHYAM, Plot No. SP-11,
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0744
ONLINE PARTNER UNACADEMY
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4.
Answer of following :
ve sign and cathode by +ve sign. What is the type
(i)
In a cell, anode is represented by –ve
of the cell?
(ii)
Write the reaction taking place at anode in the following cell :
Pt, H2 (1atm) | HCl (1M) Cl2 , Pt.
(iii)
(iv)
(v)
5.
With the help of the following possible reactions, arrange Mg, Zn, Cu and Ag in
decreasing order of their reduction potentials.
Cu + 2Ag+  Cu2+ + 2Ag
Mg + Zn2+  Mg2+ + Zn
Zn + Cu2+  Zn2+ + Cu
Cl2 can displace I2 from KI solution but I2 does not displace.
The E° (standard oxidation potentials) values of the elements
elements A, B, C and D are +
85 volt respectively. Arrange these in decreasing order of
0.76, – 0.34, + 0.25 and – 0.85
reactivity.
A cell is prepared by dipping a copper rod in 1 M CuSO4 solution and a nickel rod in 1M
NiSO4. The standard reduction potentials of copper and nickel electrodes are + 0.34 V and –
0.25 V respectively.
(i)
Which electrode will work as anode and which as cathode ?
(ii)
What will be the cell reaction ?
(iii) How is the cell represented ?
(iv)
Calculate the EMF of the cell.
6.
Predict whether the following reaction can occur under standard conditions or not.
Sn2+ (aq) + Br2 () Sn4+ (aq) + 2Br–(aq)
Given E°Sn4+/Sn2+ = + 0.15, E°Br2/Br– = 1.06V.
7.
Co3+ + e–  Co2+ E° = + 1.82V
2H2O  O2 + 4H+ + 4e– ; E° = –1.23V.
Given that,
Explain why Co3+ is not stable in aqueous solutions.
8.
The measured e.m.f. at 25°C for the cell reaction,
Zn (s) + Cu2+ (1.0M)  Cu (s) + Zn2+ (0.1 M)
is 1.3 volt Calculate E° for the cell reaction.
9.
 Fe2+ (aq) + Cu (s) . Given that
Calculate G° for the reaction : Cu2+ (aq) + Fe (s) 
E°Cu2+/Cu= + 0.34 V, E°
Fe+2 Fe
= – 0.44 V
C.O.: NAIVEDHYAM, Plot No. SP-11,
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0744
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10.
Calculate the equilibrium constant for the reaction at 298 K
 Zn2+(aq) + Cu (s)
Zn (s) + Cu2+ (aq) 
Given E °Zn2+ /Zn =-0.76 V and E° 2+ = +0.34 V
Cu
11.
/Cu
Calculate the cell e.m.f. and G for the cell reaction at 298 K for the cell.
Zn (s) |Zn2+ (0.0004M) ||Cd2+ (0.2M) | Cd(s)
o
Given E oZn2+ /Zn = – 0.763 V ; E Cd
= – 0.403 V at 298 K.
+2
Cd
F = 96500 C mol–1.
2.6
CALCULATION OF THERMODYNAMICS FUNCTION OF CELL REACTION :
(1)
G°
= – n F E 0cell
G°
= – n F Ecell
(2)
G
= H – TS
(But: H = E + PV)
G = (E + PV) – TS
dG
= dE + PdV + VdP – TdS – SdT
st
(a)
From 1 law of thermodynamics
dE
= dq + dW
= dq – pdV
dq
= dE + pdV
nd
(b)
From 2 law of thermodynmics
dq
dS =
T
dq = TdS
Using 1st law and 2nd law of thermodynamics
dG = (dE + PdV) + Vdp – TdS – SdT

dG = Vdp – SdT
If a cell reaction is taking place at constant pressure (which is generally the case).
dG = Vdp – SdT
at constant pressure
So,
Vdp = 0
And dG = – SdT


 dG 
S=– 
 dT  p
(At constant pressure)
 d( G) 
S = – 
 dT  p
(At constant pressure)
G = – n R Ecell
 d

(Ecell )  (–nF)
S = – 
nF)
 dT
p
C.O.: NAIVEDHYAM, Plot No. SP-11,
SP 11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
0744
ONLINE PARTNER UNACADEMY
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 d

(Ecell ) 
S = nF 
 dT
p
dE
= temp. cofficient of the cell
dT
E = a + bT + CT2 + ........
3.
G = H – TS
H = G + TS
H = – n F Ecell + n F T
4.
dEcell
dT
Cp of cell reaction
dH
Cp =
dT
d
( H)
Cp =
dT
dE cell 
d 
=
  nFE cell  nFT

dT 
dT 
dE cell
dE
d 2 E cell
d 2 E cell
So,
C
=
 nF cell  nFT
nFT
p
dT
dT
dT 2
dT 2
The standard potential of the following cell is 0.23 V at 15° C & 0.21 at 35°C.
Pt | H2(g) | HCl (aq) | AgCl (s) | Ag (s)
(i)
Write the cell reaction.
(ii)
Calculate H°, S°
S° for the cell reaction by assuming that these
thes quantities remain
unchanged in the range 15° C to 35° C.
(iii) Calculate the solubility of AgCl in water at 25°C. Given standard reduction potential of
[JEE 2001]
the Ag+/Ag couple is 0.80 V at 25° C.
0.21  0.23
 dE 
Sol.
= 1 × 10–3 volt deg–1.

 
20
 dT  P
=  nF
Q.
S° = – 96500 × 10–3 J/K/mol. = – 96.5 J/K/mol.
(i)
Cell reaction:
H2 + 2AgCl(s)  2H+ + 2Ag(s) + 2Cl–.
(ii)
S° = – 96.5 J/K/mol.
– 22195 = H° – 288 × S°

– 20265 = H° – 288 × S°

S° = – 96.5 J/K/mol.
H° = – 49987 J/K/mol.
C.O.: NAIVEDHYAM, Plot No. SP-11,
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0744
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(iii)
Cl–/AgCl/Ag.
Ag+ (C) + e– Ag,
C=
KSP
1
0.0591
1
log
1
C
–10
C = KSP = 1.53 × 10 .
0.22 = 0.80 –
Solubility =
3.
3.1





K SP = 1.24 × 10–5 M. Ans.
ELECTROLYSIS & ELECT
ELECTROLYTIC CELL :
Electrolysis :
Electrolyte is a combination of cations and anions which in fused
fused state or in aqueous solution
can conduct electricity.
This is possible due to the movement of ions from which it is made of.
The process of using an electric current to bring about chemical change is called electrolysis.
Electrolysis is a process of oxidation and reduction due to current in the electrolytic solution.
The product obtained during electrolysis depends on following factors.

 The nature of the electrolyte 

 The concentration of electrolyte

The charge density flowing during electrolysis.

 The nature of the electrode
Active vs Inactive (Inert) electrodes :

The metal electrodes in the cell that are active, because the metals themselves are
components of the half reactions.

As the Daniel cell operates, the mass of the zinc electrode gradually decreases, and the
[Zn2+] in the anode half – cell increases. At the same time, the mass of the copper
electrode increases and the [Cu2+] in the cathode half – cell decreases; we say that the
Cu2+ "plates out" on the electrode.

For many redox reactions, however, there are no reactants or products capable of
serving as electrodes. Inactive electrodes are used, most commonly rods of graphite or
platinum, materials that conduct electrons into or out of the cell but cannot take part in
the half -reactions.
reactions.

In a voltaic cell based on the following half reactions, for instance, the species cannot
act as electrodes :
2I–(aq)  I2(s) +2e– [anode ; oxidation]
MnO4– (aq) + 8H+ (aq) + 5e–  Mn2+ (aq) + 4H2O() [cathode ; reduction]
Therefore, each half – cell consists of inactive electrodes immersed in an electrolyte solution
reaction. In the anode half
half-cell, I– ions are
that contains all the species involved in that half -reaction.
oxidized to solid I2. The electrons released flow into the graphite anode, through the wire, and
into the graphite cathode. From there, the electrons are consumed by MnO4– ions as they are
reduced to Mn2+ ions.
C.O.: NAIVEDHYAM, Plot No. SP-11,
SP 11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
0744
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3.1
Electrolysis using inert electrodes :
(1) Electrolysis of CuSO4 molten
Cathode (red) : Pb2+ + 2e–  Pb(s) E0 = 0.126V
Anode :
2Br-  Br2 + 2e- E0 = – 1.08 V
Ecell = – 0.126 – (0.108) x 10 = – 1.206 V
Eext > 1.206 V
(2) Electrolysis of CuSO4 molten
Cathode :
Cu2+ + 2e– Cu
E0 = +0.34 V
Anode :
2 SO 24  S2 O 82  + 2e–
E0 = – 2.05 V
H2S2O8 – Marchall's acid peroxy disulphuric acid.
Ecell = 0.34 – (2.05) = – 1.71 V (negative  not feasible)
(3) Electrolysis of aq CuSO4
Cathode :
Cu2+ + 2e– Cu(s)
2e + 2H2O()  H2(g) + 2OH–(aq)
Anode :
2 SO 24  S2 O 82  + 2e–
2.05 V
E0 = –2.05
2H2O()  O2 + 4H+ + 4e–
1.23 V
E0 = –1.23
(4) Electrolysis of aq NaBr solution (initially PH = 7)
Cathode :
Na+(aq) + e–  Na(s)
2e– + 2H2O() H2 + 2OH–
Anode :
(6)
E0 = – 2 V
E0 = – 0.83 V
2Br-  Br2 + 2e–
E 0OX = – 1.08 V
2H2O () O2 + 4H+ + 4e–
E 0OX = – 1.23 V
(5) Electrolysis of aq NaCl
Cathode :
Na+ + e– Na
2e– + 2H2O() H2(g) + 2OHAnode :
E0 = 0.34 V
0.83V
E0 = –0.83V
E0 = – 2V
0.83 V
E0 = –0.83
2Cl–  Cl2 + 2e–
1.30 V
E 0OX = –1.30
2H2O() O2 + 4H+ + 4e–
1.23 V
E 0OX = –1.23
Rate of production of Cl2 is more than rate of production of O2 gas.
Electrolysis of CH3COONa
Cathode :
2H2O + 2e–  H2(g) + 2OH–
Anode :
2CH3COO–  2CO2 + C2H6 + 2e–
Note : According to thermodynamics, oxidation of H2O to produce O2 should take place on anode but
experimentally (experiment from chemical kinetics) the rate of oxidation of water is found to
be very slow. To increasee it's rate, the greater potential difference is applied called over voltage
or over potential but because of this oxidation of Cl– ions also become feasible and this takes
place on anode.
C.O.: NAIVEDHYAM, Plot No. SP-11,
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0744
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3.2
Electrolysis using attackable (reactive) electrodes :
(2) Electrolysis of aq. CuSO4 using Cu electrode.
Cathode (reduction) :
Cu2+ + 2e– Cu
2H2O() + 2e– H2(g) + 2OH–
Anode (oxidation) : SO 24  S2 O 82  + 2e–
2H2O()  O2 + 2H+ + 4e–
Cu(s) Cu2+ + 2e–
electrolytic refining
(2) AgNO3(aq) using Cu cathode & Ag anode.
Cathode :
Ag+ + e–  Ag(s)
2H2O () + 2e– H2(g) + 2OH–
Anode :
NO3–  X (No reaction)
2H2O() O2 + 4H+ + 4e–
Ag(s)  Ag+(aq) + e–
Eº = + 0.34 V
Eº = –0.83 V
E 0OX = – 2.05 V
Eº = 1.23 V
Eº = –0.34 V
Eº = 0.8 V
Eº = – 0.83 V
Eº = – 1.23 V
Eº = – 0.80 V
3.3 Faraday's Law of Electrolysis :
3.3.1 1st Law : The mass deposited/released/produced of any substance during electrolysis is
proportional to the amount of charge passed into the electrolyte.
W Q
W = ZQ
Z – electrochemical equivalent of the substance.
mass
Unit of Z =
= Kg/C or g/C
coulomb
Z = Mass deposited when 1 C of charge is passed into the solution.

Equivalent mass (E) : mass of any substance produced when 1 mole of e– are passed through
the solution during electrolysis.
Molar mass
E=

no.of e involved in oxidation / reduction
M
M
E=
e.g.
Ag+ + e– Ag
E=
Cu2+ + 2e-  Cu(s)
1
2
M
Al3+ + 3e–  Al(s)
E=
3
1 mole of e– = 1 Faraday of charge.
 96500 C – Charge deposit E gram metal charge
E
E 

1C  
Z=
 g
96500
 96500 
EQ
Q
Molar mass
W=
=
x

96500 (no.of e involved) 96500
ixt
Molar mass
Q = it
W=
x
 dQ = I  dt
96500 (no. of e  involved)
C.O.: NAIVEDHYAM, Plot No. SP-11,
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3.3.2 2nd Law : When equal charge is passed through 2 electrolytic cells and this cells are connected
in series then mass deposited at electrode will be in the ratio of their electrochemical
equivalents or in the ratio of their equivalent masses.
EQ
W = ZQ =
96500
W1
z
E
= 1 = 1 ( Q = same)
W2
z2
E2
Current Efficiency :
Current efficiency =
charge actually used in electricity
× 100
ch arg e passed
Current efficiency =
mass actually produced
× 100
mass that should have been produced
Ex.13 Calculate volume of the gases liberated at STP if 1 L of 0.2 molar solution of CuSO4 is
electrolysed by 5.79 A current for 10000 seconds.
5.79 10000 579
Sol. No. of moles of e– =
=
= 0.6
96500
965
Cathode :
Cu2+ + 2e– Cu(s)
0.2 mole 0.4 mole
2H2O() + 2e–  H2 + 2OH–
0.2 mole of e– 0.1
 mole of H2 at S.T.P.
Anode :
2H2O()  O2 + 4H+ + 4e–
4 mole of e–  1 mole of O2
0.6 mole of e– 0.15 mole of O2
so, total moles = 0.25 mole
Total volume = 5.6 Ltr.
Ex.14 The electrochemical equivalent of copper is 0.0003296 g coulomb–1. Calculate the amount of
copper deposited by a current of 0.5 ampere flowing through copper sulphate solution for 50
minutes.
Sol. According to Faraday's first law, W = Zit
W = 0.5 × 50 × 60 × 0.003296 = 0.4944 g
Ex. 15 An electric current is passed through three cells connected in series containing ZnSO4,
acidulated water and CuSO4 respectively. What amount of Zn and H2 are liberated when
6.25 g of Cu is deposited? Eq. wt. of Cu and Zn are 31.70 and 32.6 respectively.
Sol. 
Eq. of Cu = Eq. of Zn = Eq. of H2
WH2
6.25
W
= Zn =
31.70 32.6
1
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Ex. 16 The cell consists of three compartments
separated by porous barriers. The first
contains a cobalt electrode in 5.00 L of
0.100 M cobalt (II) nitrate; the second
0.20 M NO
Co
contains 5.00 L of 0.100 M KNO3 and
0.10 M Co
third contains 0.1 M AgNO3. Assuming
that the current within the cell is carried
5.0 L
equally by the positive and negative ions,
tabulate the concentrations of ions of each
type in each compartment of the cell after
the passage of 0.100 mole electrons.
Given
Co2+ + 2e–  Co
Ag+ + e–  Ag
–
3
2+
Sol.
0.10 M NO3
–
0.10 M NO 3
–
Ag
0.10 M K
+
0.10 M Ag
5.0 L
+
5.0 L
Eº = – .28 V
Eº = 0.80 V
Spontaneous reaction is :
2Ag+ + Co Co2+ + 2Ag
Eº = 1.08
In the left compartment. cobalt will be oxidized to cobalt (II) ion. In the right compartment,
silver ion will be reduced to silver. The passage of 0.100 mol electrons will cause the following
quantities of change.
Compartment 3
Compartment 1
Compartment 2
3+
– 0.100 mol Ag+
Effect of electrode
+ 0.0500 mol Co
Positive ion movement
–0.0250 mol Co2+
+ 0.0250 mol Co2+ + 0.0500 mol K+
– 0.0500 mol K+
Negative ion movement
+ 0.0500 mol NO3–
– 0.0500 mol NO3–
Changing the numbers of mol to concentrators in 5.0 L compartments and adding or subtracting
yields the following results :
Ag+ 0.0800
Final concentrations (M) Co2+
0.015
Co2+ 0.00500
NO3– 0.0900
NO3– 0.210
K+
0.090
NO3– 0.100
K+
0.0100
DO YOURSELF–2
1.
How many moles of electrons are needed for the reduction of 20 mL of 0.5M solution of
KMnO4 in acid medium ?
2.
An aqueous solution of NaCl is electrolysed with inert electrodes. Write the equations for the
reactions taking place at cathode and anode. What happens if NaNO3(aq.) is used instead of
NaCl ?
3.
tion becomes colourless, blue
On electrolysing CuSO4 solution in presence of Pt, the solution
colour of solution disappears. Why ?
4.
How will show that Faraday's second law of electrolysis is simply corollary of the first law.
5.
A Solution of copper (II) sulphate is electrolysed between copper electrodes by a current
curre of
10.0 amperes passing for one hour. What changes occur at the electrodes and in the solution ?
C.O.: NAIVEDHYAM, Plot No. SP-11,
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4.
ELECTROLYTIC CONDUCTANCE
CONDUCT
:
4.1
Factors Affecting Conductance & Resistance :
(i)
Solute – Solute interactions (Inter – Ionic force of attraction) Greater the force of
attraction, greater will be the resistance.
Force Charge
(ii)
Solute – Solvent Interaction (Hydration/Solvation of Ions)
Greater the solvation
1
Solvation Charge
arge 
greater will be resistance
size
Li+ (Hydrated largest) Cs+ (Hydrated smallest)
resistance of LiCl > resistance of CsCl
(iii)
Solvent-solvent
solvent interaction (Viscosity) : greater the viscosity greater will be resistance
(iv)
Temperature
T
R
(v)
Nature of electrolyte
Strong electrolyte – Low resistance
Weak electrolyte – High resistance
Some definations :
Resistance (R) :
V
R=
(Ohm's law ())


R=
A
 – resistivity/specific resistance
– resistance of unit length wire of unit area of cross section = constant = (
( m)
RA
=

Resistivity of a solution is defined as the resistance of the solution between two electrodes of 1
cm2 area of cross section and 1 cm apart.
or
Resistance of 1 cm3 of solution will be it's resistivity.
4.2
C.O.: NAIVEDHYAM, Plot No. SP-11,
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Conductance () :
1
C=
= mho = –1
R
= S (Siemens)
Conductivity/specific conductance


1
=
=
= 
RA
A

unit –1 cm–1
= conductivity of 1 cm3 of solution
 concentration of ions
1
1
=
C=
R

 (( no. of ions) no. of charge carriers

Since conductivity or resistivity of the solution is dependent on it's concentration, so
two more type of conductivities are defined for the solution.
Molar conductivity/molar conductance (m) :
Conductance of a solution containing 1 mole of an electrolyte between 2 electrodes which are
unit length apart.
Let the molarity of the solution 'C'
 C
C moles of electrolyte are present in 1 Lt. of solution.
so molar conductance = m
κ × 1000
κ ×1000

m = V
m =

m =
C
molarity
–1
–
2
–1

Its units are Ohm cm mol




Equivalent conductance (eq): Conductivity of a solution containing 1 g equivalent of the
electrolyte. 
κ ×1000
eq =
Normality
–

Its units are Ohm–1
cm2 eq–1
Ionic Mobility

Ionic Mobility = speed of the ion per unit electrical field
speed
speed
=
=
electrical field
potential gradient

Its units are V–1 cm2 sec–1
Ionic mobility = u =
Λ 0M
Λ0
= M
96500
F


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Transport Number
Transport Number of any ion is fraction of total current carried by that ion.

Λ0M
Transport Number of cation = 0
Λ M electrolyte
Ex.17 If resistivity of 0.8 M KCl solution is 2.5 x 103  cm calculate m of the solution.
Sol. = 2.5 x 10–3 cm
K=
4  102  1000  10
= 5 x 105 –1 cm2 mole–1
0.8
Variation of conductivity and molar conductivity with concentration

Conductivity always decreases with the decrease in concentration both for weak and
strong electrolytes.

The number of ions per unit volume that carry the current in a solution decreases on
dilution.

Molar conductivity increases with decreases in concentration. This is because the total
volume, V of solution containing one mole of electrolyte also increases.

Molar conductivity is the conductance of solution.

When concentration approaches zero, the molar conductivity is known as limiting molar
mo
conductivity and is represented by the symbol °.

4.3





4.3.1

103
= 4 x 102
2.5
m =
Strong Electrolytes :
increases slowly with dilution and can be represented by the

For strong electrolytes. increases
equation
 =  – A C1/2

The value of the constant 'A' for a given solvent and temperature depends on the type of
electrolyte i.e. the charges on the cations and anion produced on the dissociation of the
electrolyte in the solution.
2 electrolyte respectively.
Example : Thus NaCl, CaCl2, MgSO4 are known as 1-1 , 2-1 and 2-2
All electrolytes of a particular type have the same value for 'A'.

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4.3.2



Weak electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at higher concentration and
with dilution is due to increases in the number of
hence for such electrolytes, the change in with
ions in total volume of solution that contains 1 mol of electrolyte.
el
At infinite dilution (i.e. concentration c zero) electrolyte
dissociates completely ( = 1), but at such low concentration the
conductivity of
the solution is so low that it cannot be measured accurately.
Molar conductivity versus c1/2 for acetic acid
(weak electrolyte) and potassium chloride
(strong electrolyte in aqueous solutions.
Kohlarausch's Law :
"At infinite dilution, when dissociation is complete, each ion makes a definite contribution
towards equivalent conductance of the electrolyte irrespective of the nature of the ion with
which it is associated and the value of equivalent conductance at infi
infinite dilution for any
electrolyte is the sum of contribution of its constituent ions."
i.e.,
  = + + –
At infinite dilution or near zero concentration when dissociation is 100%, each ion makes a

definite contribution towards molar conductivity of electrolyte irrespective of the nature of the
other ion. (because interionic forces of attraction are zero)
0
 m electrolyte = + + Λ 0m +  – Λ 0m+ = no. of cation in one formula unit of electrolyte
– = no. of anions
in one formula unit of electrolyte
For NaCl, + = 1 – = 1
For Al2(SO4)3, + = 2 – = 3
 0eq  electrolyte =  eq  +
 0eq 
=
Λ 0eq =
 eq 
Λ 0m
charge on the cation
Λ 0eq .Al3+ =
Λ 0m
charge on the anion
Λ 0m Al3+
3
Λ 0eq , electrolyte =
Λ 0m electrolyte
total + ve charge on cation
or
total - ve charge on anion
Λ
0
eq
Λ
0
eq
0
eq
3+
0
eq
Al2(SO4)3 = Λ Al + Λ SO
Al2(SO4)3 =
2–
4
0
m
Λ Al
=
3
3+
0
m
Λ SO 42–
+
2
2Λ 0mAl3+ + 3Λ 0mSO24
6
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Ex.18 Λ 0m Na+ = 150–1 cm2 mole–1
;
–
Λ 0eq Ba2+ = 100–1
cm2 eq–1
-1
2
–1
Λ0eqSO2–
4 = 125 cm eq
;
–1
2
–1
3+
–
Λ 0m Al = 300 cm mole
–1
2
–1
Λ 0m NH 4+ = 200 cm mole
;
1
2
–1
–
–1
Λ 0m , Cl = 150 cm mole
Then calculate
Sol.
(a) Λ 0eq , Al3+
(b) Λ 0eq Al2(SO4)3
(c) Λ 0m (NH4)2SO4
(d) Λ 0m NaCl, BaCl2. 6H2O
(e) Λ 0m , (NH4)2 SO4 Al2(SO4)3 . 24H2O
(f) Λ 0eq NaCl
(a) Λ 0eq Al3+ =
300
(b) Λ 0eq Al2(SO4)3 = 100 + 125 = 225
= 100
3
0
m
(c) Λ (NH4)2SO4 = 2 x 200 + 2 x 125 = 650
(d) Λ 0m NaCl.BaCl2.6H2O = 150 + 200 + 3 x 150 = 800 r–1
(e) Λ 0m (NH4)2 SO4 Al2(SO4)3.24H2O = 400 + 600 + 4 x 250 = 2000
(f) Λ 0eq NaCl = 300–1 cm2 eq–1
0
Ex.19 To calculate Λ 0m or Λ eq
of weak electrolyte
Sol.
Λ0mCH3COOH = Λ0mCH COO- + Λ 0m H+
3
= (Λ
0
mCH3COO –
+ Λ 0mNa ) – Λ 0mNa + + Λ 0mH+ + Λ 0mCl+ – Λ 0mCl –
+
Λ0CH3COOH = Λ 0mCH3COONa + Λ 0mHCl – Λ 0mNaCl
Ex.20 Calculate Λ 0m of oxalic acid, given that
Λ 0eq Na2C2O4 = 400–1 cm2 eq–1
–1
2
–1
Λ 0m H2SO4 = 700 cm mole
Λ 0eq Na2SO4 = 450–1 cm2 eq–1
Sol.
–1
2
Λ 0m H2C2O4 = 700 + 800 – 900 = 600 cm mole
Λ 0eq = 400 +
700
– 450
2
Λm
= 350 – 50 = 300
2
m = 600
Applications of Kohlaraushch's law
 Calculate ° for any electrolyte from the ° of individual ions.
 Determine the value of its dissociation constant once we known the ° and ° at a given
concentration c.
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
Degree of dissociation : At greater dilution the ionization become 100%, therefore called
infinite dilution.
ution the ionization (dissociation into ions) is less than 100% and equivalent
At lower dilution
conductance become lower,
i.e., eq < °eq
degree of dissociation
Λeq
equivalent conductance at a given concentration
concentrati
=
0 =
Λeq
equivalent conductance at at infinite dilution
dilutio


Dissociation constant of weak electrolyte :
KC =
C 2
;  = degree of dissociation
1 
C = concentration


The degree of dissociation then it can be approximated to the ratio of molar conductivity c at
the concentration c to limiting molar conductivity, º, Thus we have :
 = /º
But we known that for a weak electrolyte like acetic acid.
Ka =
c 2
c 2
C 2
=
=
 1   / 
    
(1   )


Solubility(s) and KSP of any sparingly soluble salt.
Sparingly soluble salt = Very small solubility
Solubility = molarity = 0
so, solution can be considered to be of zero conc or infinite dilution.
K ×1000
K ×1000
m, saturated =  M =
S =
KSP = S2 (for AB type salt)
Solubility
Λ 0M
Example-21 If conductivity of water used to make saturated solution of AgCl is found to be 3.1 x 10–
5–1 cm–1 and conductance of the solution of AgCl = 4.5 x 10–55–1 cm–1
If Λ 0M AgNO3 = 200
200 –1 cm2 mole–1, Λ 0M NaNO3 = 310–1 cm2 mole–1
Solution
calculate KSP of AgCl
Λ 0M AgCl = 140
Total conductance = 10–5
140× 4×10 -5 ×1000 1.4×10 -4
S=
=
140
14
–4
S = 5.4 x 10
S2 = 1 x 10–8
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Example-22 To calculate KW of water
H2O() + H2O() H2O+(aq) + OH–(aq)
m = Λ0M,H2O = Λ 0M H+ + Λ 0M OH–
K ×1000
- Concentration of water molecules 100% dissociated Ask
molarity
K  1000
molarity = [H+] = [OH–] =
 M
=
2
 K 1000 
[H ][OH ]
KW = [H ][OH ] = 
K
or
K
=
a
b

0
H2O
 M 
Variation of , m &eq of solutions with Dilution
+
–
conc.
of ions in the solution. In case of both strong and weak electrolytes on dilution the
concentration of ions will decrease hence  will decrease.
m or eq
( C) strong electrolyte
1000 × κ
m =
(  Ka C ) weak electrolyte.
molarity
eq =
5.
5.1
5.1
1000 × κ
normality
For strong electrolyte
m 
For weak electrolyte
m
κ
C

= constant
C
C

KaC
κ
1


C
C
C
TYPE OF BATTRIES :
Primary cells : These cells can not be recharge i.e., dry cell (lechlanche cells)
mercury cells (miniature cell used in the electronic devices)
Ecell = constant
as all substances used are either pure solids or pure liquids.
DRY CELLS and alkaline batteries :
•
Cell potential = 1.5 V
•
Anode : Zn (s)  Zn+2(aq) + 2e–
•
Cathode : MnO2 + NH4+ + e– MnO(OH) + NH3
Zn2+ + 4NH3  [Zn(NH3)4]2+
•
Alkaline batteries contain basic material inside it.
•
NaOH / KOH is used instead of the acidic salt NH4Cl
•
Cathode : 2MnO2(s) + H2O() + 2e– Mn2O3(s) + 2OH–(aq)
Anode : Zn(s) + 2OH–(aq)  ZnO(s) + H2O() + 2e–
•
•
Voltage produced by these cells = 1.54 V
The cell potential does not decline under high current loads because no gases are formed.
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5.2
Mercury cell :
•
Suitable for law current devices - Hearingaeds, watches
•
Anode : Zn(Hg) + 2OH–  ZnO(s) + H2O + 2e–
Cathode : HgO + H2O + 2e–  Hg() + 2OH–
___________________________________
Cell reaction : Zn(Hg) + HgO(s) ZnO(s) + Hg()
•
Cell potential = 1.35 V and remains constant.
5.2
Secondary cells :
5.2.1 Lead storage batteries used is automobiles (Cars/bikes)
Anode
: Pb(s)
Cathode : PbO2(s)
H2SO4(conc.) about 38% solution of H2SO4 is taken.
Anode : Pb(s)  Pb2+ (aq) + 2e–
Pb2+(aq) + SO 24  (aq) PbSO
PbSO4(s)
Pb(s) + SO 24  (aq)  PbSO4 + 2e–
Most of the PbSO4(s) ppt sticks to the lead rod.
Cathode : 2e– + 4H+ + PbO2(s)  Pb2+(aq) + 2H2O()
Pb2+(aq) + SO 24  (aq) + 4H+ + 2e– PbSO4(s) + 2H2O()
PbSO4(s) sticks to cathode rod.
Pb(s) + PbO2 + 4H+ + 2 SO 24  (aq) 2PbSO4(s) + 2H2O()
Ecell = 2.05 V
Note : During the working of the cell or discharge H2SO4 will be consumed so it's concentration in the
solution hence density of the solution will decrease, during charging of the cell PbSO4 will get
converted into Pb(s) and, PbO2(s) and H2SO4 will be produced.
5.2.2 Nickel – cadmium battery.
Ecell = constant as cell reaction has pure solid/liquids only.
Anode : Cd(s)
Cathode : NiO2(s)
Electrolyte : KOH
Cd + 2OH– Cd(OH)
Cd(OH)2 + 2e–
2e– + NiO2 + 2H2O Ni(OH)2(s) + 2OH–
Cd(s) + NiO2(s) + 2H2O() Cd(OH)2(s) + Ni(OH)2(s)
5.2.3
Fuel cells
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(i) H2 – O2 fuel cell :
Anode :
H2 2H+ + 2e– x 2
Cathode :
4e– + 4H+ + O2 2H2O
–––––––––––––––––––––––
2H2 + O2  2H2O()
–––––––––––––––––––––––
(ii) CH4 – O2 fuel cells :
Anode :
2H2O + CH4  CO2 + 8H+ +8 e–
Cathod :
4e– + 4H+ + O2 2H2O
–––––––––––––––––––––––
CH4 + 2O2 CO2 + 2H2O()
–––––––––––––––––––––––
6.
Mechanism of corrosion
Oxidation : Fe(s)  Fe2+ (aq) + 2e–
Reduction : 2O2–(g) + 4H+ (aq)  2H2O(I)
Atmospheric
Oxidation : 2Fe2+(aq) + 2H2O(l) + 1/2O2 Fe2O3(s) + 4H+(aq)
Ex.23 During the discharge of a lead storage battery the density of H2SO4 falls from 1 g/cc to 2 g/C,
H2SO4 of density of 1 g/C. C is X% by weight and that of density of 2 g/c.c is Y% by weight.
The battery holds V litre of acid before discharging. Calculate the total charge released at anode
of the battery. The reactions occurring during discharging are.
Solution
At anode :
Pb + SO 24   PbSO4 + 2e–
At cathode :
PbO2 + 4H+ + SO42– +2e–  PbSO4 + 2H2O
Mass of acid solution before discharge of lead storage battery
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(LSB) = (V x 103 x 1) g
= (1000 x V1)g
X 
Mass of H2SO4 before discharge of LSB =  1000  V1 
 g = (10 x V1 X)g
100 

Net reaction during discharging : Pb + PbO2 + 2H2SO4  PbSO4 + 2H2O
From the reaction, it is evident that the moles of electron exchanged (lost at anode and
gain at cathode) is equal to the moles of H2SO4 consumed or moles of H2O produced.
Let the
he moles of H2SO4 produced be x, then
Mass of H2O produced during discharge of LSB = (18x) g
Mass of H2SO4 consumed during discharge of LSB = (98x) g
Mass of H2SO4 after discharge of LSB = [(10V1 X)] – 98x] g
Mass of acid solution after discharge of LSB = [(1000 V1)– 98x + 18x] = [(1000 V1)
– 80x]g
Mass of H 2SO 4 after disch arg e
× 100
% of H2SO4 after discharge of LSB =
Mass of acid solution after disch arg e
Y=
[(1000  V1 )  98x]
x 100
[(1000  V1 )  80x]
x can be calculated as all other quantities
are known.
Total charge released at cathode,
cathod Q = nF = xF.
Example-24 A lead storage cell is discharged which causes the H2SO4 electrolyte to change from a
concentration of 34.6% by weight (density 1.261 g ml–1 at 25°C) to one of 27% by
Calculate the total charge
weight. The original volume of electrolyte is one litre. Calc
released at anode of the battery. Note that the water is produced by the cell reaction as
H2SO4 is used up. Over all reaction is.
Pb(s) + PbO2(s) + 2H2SO4()  2PbSO4(s) +2H2O()
Solution
Before the discharge of lead storage battery,
Mass of solution = 1000 x 1.261 = 1261 g
1261 34.6
Mass of H2SO4 =
= 436.3 g.
100
Mass of water = 1261 – 436.3 = 824.7 g
After the discharge of lead storage battery,
uring discharge
Let the mass of H2O produce as a result of net reaction during
(Pb + PbO2 + 2H2SO4  2PbSO4 + 2H2O) is x g
x
Moles of H2O produced =
= moles of H2SO4 consumed
18
x
Mass of H2SO4 consumed =
x 98
18
98x
Now, mass of solution after discharge = 1261 –
+x
18
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% by the mass of H2SO4 after discharge =
Mass of H 2SO 4 left
x 100 = 27
Mass of solution after disch arg e
98x
18 x 100 = 27
=
98x
1261 
x
18
436.3 
 x = 22.59 g
DO YOURSELF - 3
1.
The resistance of a 0.01 N solution of an electrolyte was found to 210 ohm at 298 K using a
conductivity cell with a cell constant of 0.88 cm–1. Calculate specific conductance and
equivalent conductance of solution.
2.
The conductivity of pure water in a conductivity cell with electrodes of cross
cross-sectional area 4
cm2 placed at a distance 2 cm apart is 8 × 10–7 S cm–1. Calculate :
(a) the resistance of water.
(b) the current that would flow through the cell under the applied potential difference of 1
volt.
3.
The specific conductivity of a solution containing 1.0 g of anhydrous BaCl2 in 200 cm3 of the
solution has been found to be 0.0058 S cm–1. Calculate the molar and equivalent conductivity
of the solution. Molecular wt. of BaCl2 = 208.
4.
The resistance of a solution 'A' is 50 ohm and that of solution 'B' iiss 100 ohm, both solution
being taken in the same conductivity cell. If equal volumes of solutions A and B are mixed,
what will be the resistance of the mixture using the same cell ?
(Assume that there is no increase in the degree of dissociation of A and B on mixing.)
5.
The value of  for NH4Cl, NaOH and NaCl are 129.8, 248.1 and 126.4 ohm–1 cm2 mol–1
respectively. Calculate  for NH4OH solution.
6.
Calculate molar conductance for NH4OH, given that molar conductances for Ba(OH)2, BaCl2
and NH4Cl are 523.28, 280.0 and 129.8 ohm–1 cm2 mol–1 respectively.
7.
The equivalent conductivities of acetic acid at 298 K at the concentrations of 0.1 M and
0.001 M are 5.20 and 49.2 S cm2 eq.–1 respectively. Calculate the degree of dissociation of
acetic acid at these
hese concentrations. Given that, (H+) and (CH3COO–) are 349.8 and 40.9
ohm–1 cm2 mol–1 respectively.
ANSWER KEY
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1.
DO YOURSELF - 1
Chlorine placed below iodine in electrochemical series having more reduction potential and
thus shows reduction whereas I– undergoes oxidation. The I2 so formed get absorbed in starch
to give blue colour.
2I–  I2 + 2e–
Cl2 + 2e–  2Cl–
—————————
Cl2 + 2I–  I2 + 2Cl–
2.
(a)
More the E°OP, more is the tendency for oxidation. Therefore, since maximum E°OP
stands for :
Sn2+  Sn4+ + 2e–


(b)
E°OP = – 0.1V
strongest reductant
:
Sn2+
and weakest oxidant :
Sn4+
More or +ve is E°RP ,more is the tendency for reduction. Therefore, since maximum E°RP

stands for :
MnO 4– + 8H + + 5e– Mn2+ + 4H2O

strongest oxidant

E°RP = + 1.52 V
MnO4–
:
and weakest reductant :
Mn2+
Note : Stronger is oxidant, weaker is its conjugate reducant and vice-versa.
vice versa.
(c)
For (i) E°Cell = E°OP
Fe2+ / Fe3+
+ E°RP
= –0.77 + 0.1
Sn 4+ /Sn 2+
Fe2+ oxidizes and Sn4+ reduces in change.
E °Cell = – 0.67 V

E °Cell is negative.

(i)
For (ii) E

°
Cell
(ii)
For (iii)
Is non-spontaneous
non
change.
= E°OP
Fe2+ / Fe3+
+ E°RP
Is non-spontaneous
non
change.
E°Cell =E°OP – +E°RP
I /I2

(iii)
For (iv)
(iv)
3.
= –0.77 + 0.54 = – 0.23 V
I2 / I-
Sn 4+ /Sn 2+
–
= –0.54+0.1 = – 0.44 V
Is non-spontaneous
non
change.
E°Cell =E°OP
Sn 2+ /Sn 4+
+E°RP
I 2 /I
–
= – 0.1+0.54 = + 0.44 V
Is spontaneous change.
More is E°RP, more is the tendency to get reduced or more is the oxidising power or lesser is
reducing power.
Ag < Hg < Cr < Mg < K
C.O.: NAIVEDHYAM, Plot No. SP-11,
SP 11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
0744
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4.
(i)
The cell is an electrochemical cell i.e. galvanic cell.
(ii)
Reaction taking place at anode is : H2  2H+ + 2e–
(iii)
The decreasing order of reduction potentials in : Ag > Cu > Zn > Mg
(iv)
I– is stronger reducing agents than Cl–. Therefore, I– can reduce Cl2 to Cl– ions whereas
Cl–cannot reduct I2 to I–. Thus, only the following reaction is feasible.
Cl2 + 2I– (2KI) 2KCl (2Cl–) + I2
(v)
The reactivity of an elements is directly porportional to its standard oxidation potential
(E°).The decreasing order of reactivity of elements is : A > C > B > D.
5.
(i)
The nickel electrode with smaller E° value ((–0.25
0.25 V) will work as anode while copper
electrode with more E° value (+0.34V) will work as cathode.
(ii)
The cell reaction may be written as :
At anode
:
Ni (s)  Ni2+ (aq.) + 2e–
At cathode
:
Cu2+ (aq) + 2e– Cu (s)
Cell reaction :
(iii)
Ni (s) + Cu2+ (aq)  Ni2+ (aq) + Cu (s)
The cell may be represented as :
Ni(s)/Ni2+ (aq) ||Cu2+ (aq)/Cu (s)
(iv)
6.
EMF of cell = E°cathode -E°anode =(+0.34) – ( – 0.25) = 0.59 V
E°cell =E°cathode -E°anode = 1.06 – 0.15 = 0.91V.
Since, E °cell comes out be positive, this means that the reaction can occur.
7.
The E°cell can be calculated as follows :
4 [Co+3 + e–  Co2+] ;
2H2O  O2 + 4H+ + 4e– ;
E° = + 1.82V
E° = – 1.23 V.
Add : 4Co3+ + 2H2O  4Co2+ + 4H+ + O2 ; E° = 1.82 – 1.23 = + 0.59V.
Since E °cell is positive, the cell reaction is spontaneous. This means that Co3+ ions will take part
in the reaction. Therefore, Co3+ is not stable.
8.
Using Nernst equation (at 298 K),
E cell =E°cell -
Here
0.0591V
[Zn 2+ (aq)]
log
2
[Cu 2+ (aq)]
Ecell = 1.3 V, [Cu2+ (aq)] = 1.0 M,
[Zn2+ (aq)] = 0.1 M, Ecell = ?
C.O.: NAIVEDHYAM, Plot No. SP-11,
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0744
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43
Substituting the values,
1.3 V = E °cell -
0.0591V
0.1
log
2
1.0
1.3 V = Ecell – 0.02955 V log 10–1
1.3 V = Ecell + 0.02955 V log 10
Ecell = 1.3 V – 0.02955 V = 1.27 V
9.
The cell reactions are :
Fe (s) Fe2+ (aq) + 2e–
............ At (anode)
Cu2+ (aq) + 2e–  Cu (s)
............. At (cathode)
We know that :
Gh°
= –nF ; n = 2 mol,
°

E cell =  E°(Cu2+ /Cu) –E (Fe
2+
/Fe) 

= (+ 0.34 V) – (–0.44 V) = + 0.78 V
F = 96500 C mol–11

G° =
– nF Ecell
= –(2mol)
(2mol) × (96500 C mol–1) × (+0.78 V)
= – 150540 CV = –150540 J
10.
We know that, log Kc =
(
1 CV = 1J)
nE ocell
0.0591
E°cell =  E°Cathode –E°Anode 
= [(+0.34 V) – (– 0.76 V)] = 1.10 V, n =2,


log K c =
2×(1.10 V)
=37.29
(0.0591 V)
Kc = Antilog 37.29 = 1.95 × 1037
11.
Step I. Calculation of cell e.m.f.
According to Nernst equation
E=E o -
0.0591
[Zn 2+ (aq)]
log
n
[Cd 2+ (aq)]
C.O.: NAIVEDHYAM, Plot No. SP-11,
SP 11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
0744
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o
o
o
E cell
= E (Cd
–E(Zn
= (–0.403) – (–0.763) = 0.36 V
2+
2+
/Cd)
/Zn)
[Zn+ (aq)] = 0.0004 M, [Cd2+ (aq)] = 0.2 M, n = 2
E=(0.36)-
= 0.36-
(0.0591 V)
0.0004
log
2
0.2
(0.0591 V)
×(–2.69990)
2
= 0.36 V + 0.08 = 0.44 V
Step II.
Calculation of G

G
= –nFEcell
Ecell
= 0.44 V, n = 2 mol, F = 96500 C mol–1
G
= – (2 mol) × (96500 C mol–1) × (0.44 V)
= – 84920 CV = – 84920 J
DO YOURSELF - 2
1.
2.
Moles of KMnO4 = M × V (L) = 0.5 × 20 × 10–3 = 10–2

1 mol KMnO4 required

10–2 mol KMnO4 required
For NaCl(aq.) anode
cathode
For NaNO3(aq.) anode
cathode
3.
[ Mn7+ + 5e–  Mn2+]
= 5 mol e–
= 5 × 10–2 mol e–
:
2Cl–  Cl2 + 2e–
:
2H+ + 2e–  H2
:
2OH–  H2O +
:
2H+ + 2e–  H2
At cathode
:
Cu2+ + 2e–  Cu
At anode
:
2OH–  H2O +
1
O2 + 2e–
2
1
O + 2e–
2 2
Cu2+ ions are reduced and blue colour is due to Cu2+.
4.
Accroding to Faraday's first law of electrolysis.
w=Z× Q
If same quantity of electricity is passed through two electolytes, i.e., Q1 = Q2 = Q, then
In case of first electrolyte,
w1 = Z1 × Q and
In case of second electrolyte, w2 = Z2 × Q
C.O.: NAIVEDHYAM, Plot No. SP-11,
SP 11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
0744
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On dividing ,
w1 Z1 E1 /96500 E1
= =
=
w 2 Z 2 E 2 /96500 E 2
where E1 and E2 are their equivalent masses.
5.
According to Faraday's first law of electrolysis :
The reaction at cathode : Cu2+ +
2e–   Cu
63.5
2 × 96500 C
The quantity of charge passed = I × t = (10 amp) × (60 × 60s) = 36000 C.
2 × 96500 C of charge deposit copper = 63.5 g
(63.5 g)
36000 C of charge deposit copper =
×(36000 C)=11.84 g
(2×96500 C)
Thus, 11.84 g of copper will dissolve from the anode and the same amount from the solution
will get deposited on the cathode. The concentration of the solution will remain unchanged.
DO YOURSELF - 3
1.
Given for 0.01 N solution.
R = 210 ohm

= 0.88 cm–1
A
Specific conductance
1 

 
R A
1
× 0.88 = 4.19 × 10–3 mho cm–1

210
k  1000
 eq 
N
 eq 
eq.=
2.
4.19  10 3  1000
0.01
419 mho cm2 eq–1.
 2 1
  cm–1
A 4 2
1 
Also   
R A
1 
1
1
=
R 
 = 6.25 × 105 ohm
7
 A
8  10
2
V
1
= 1.6 × 10–6 ampere
From Ohm's law,
=R

I=
I
6.25  105
Cell constant =
(a)
(b)
C.O.: NAIVEDHYAM, Plot No. SP-11,
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3.
1  1000
= 0.024 M
208  200
= 0.024 × 2 = 0.048 N
Molarity of BaCl2
Also
=
normality of BaCl2
(N = M × V. f)
Now
  k×
1000 0.0058  1000
= 241.67 S cm2 mol–1

CM
0.024
Also
  k×
1000 0.0058  1000
= 120.83 S cm2 equivalent–1

CN
0.048
4.
Let k1 and k2 be the specific conductance of the solution A and B respectively and the cell
constant of the cell be X.
Sp. conductance = Conductance × Cell constant
For solution A :
1
k1  × X ..................... (i)
50
1
For solution B :
Sp. conductance, k 2 
.................. (ii)
X
100
When equal volumes of A and B are mixed, both the solutions get doubly diluted, hence their
k
k
individual contribution towards the sp. conductance of the mixture will be 1 and 2
2
2
1
respectively and the sp. conductance of the mixture will be
(k + k2).
2 1
1
1

For the mixture :
(k1 + k2) = × X
..................... (iii)
2
R
(R is the resistance of mixture)
From equation (i), (ii) and (iii) ;
R = 66.67 ohm
5.
 NH 4OH   NH 4Cl   NaOH   NaCl

6.

NH 4OH
= 129.8 + 248.1 – 126.4
= 251.5 ohm–1 cm2 mol–1

 Ba(OH)2   Ba 2+  2  OH
–
= 523.28
......... (i)

 BaCl2   Ba 2+  2 Cl
-
= 280.00
......... (ii)

 NH 4Cl   NH +   Cl
–
= 129.80
........
4


(iii)

  NH 4OH   NH 4   OH

– 
Eq.(iii) +
Eq.(i) Eq.(ii)
2
2


 NH    OH
=
 = 
NH OH 
4
4
will gives
502.88
= 251.44 ohm–1 cm2 mol–1
2
C.O.: NAIVEDHYAM, Plot No. SP-11,
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0744
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7.
c
Degree of dissociation is given by   


(i)
Evaluation of  CH 3COOH


+  H +
=  CH
 CH
COO –
3COOH
3
(ii)
= 40.9 + 349.8 = 390.7 ohm–1 cm2 eq.–1
Evaluation of degree of dissociation
At
C = 0.1 M
or
c
5.20
=
= 0.013


390.7
 = 1.3%
At C = 0.001 M


c
49.2
=
= 0.125


390.7
i.e.
1.3%
i.e.
12.5%
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
48
EXERCISE # (S-I)
1.
2.
3
GALVANIC CELL, ITS REPRESENTATION & SALT BRIDGE
In the galvanic cell Cu | Cu2+ || Ag+ | Ag, the electrons flow from Cu-electrode to Ag-electrode.
Answer the following questions regarding this cell :
(a) Which is the anode ?
(b) Which is the cathode ?
(c) What happens at anode-reduction or oxidation ?
(d) What happens at cathode-oxidation or reduction ?
(e) Which electrode loses mass ?
(f) Which electrode gains mass ?
(g) Write the electrode reactions.
(h) Write the cell reaction
(i) Which metal has greater tendency to loss electron-Cu or Ag ?
(j) Which is the more reactive metal-Cu or Ag ?
(k) What is the function of salt bridge represented by the symbol || ?
Write cell reaction of the following cells :
(a) Ag | Ag+ (aq) | | Cu2+ (aq) | Cu
(c) Pt,Cl2 | Cl– (aq) | | Ag+ (aq) | Ag
(b) Pt | Fe2+ , Fe3+ | | MnO–4 , Mn2+, H+ | Pt
(d) Pt, H2 | H+ (aq) | | Cd2+ (aq) | Cd
Write cell representation for following cells.
(a) Cd2+ (aq) + Zn (s)  Zn2+ (aq) + Cd (s)
(b) 2Ag+ (aq) + H2 (g)  2H+ (aq) + 2Ag (s)
(c) Cr2O2–7 (a) + 14H+ (aq) + 6Fe2+ (aq)  6Fe3+ (aq) + 2Cr3+ (aq) + 7H2O ()
4.
STANDARD CELL POTINTIAL AND ELECTROCHEMICAL SERIES
For the cell reaction 2Ce4+ + Co  2Ce3+ + Co2+
o
E°cell is 1.89 V. If E CO
is – 0.28 V, what is the value of E oCe4 |Ce3 ?
2
Co
5.
Determine the standard reduction potential for the half reaction :
Cl2 + 2e–  2Cl–
6.
Given Pt2+ + 2Cl–  Pt + Cl2,
E oCell = – 0.15 V
Pt2+ + 2e–  Pt
E° = 1.20 V
Is 1.0 M H+ solution under H2SO4 at 1.0 atm capable of oxidising silver metal in the presence
of 1.0 M Ag+ ion?
E oH |H  Pt  = 0.0 V
E oAg  |Ag = 0.80 V,
2
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7.
Cu2+ + e–  Cu+
Cu2+ + 2e–  Cu
If for the half cell reactions
E° = 0.15 V
E° = 0.3
0.34 V
Calculate E° of the half cell reaction
Cu+ + e–  Cu
also predict whether Cu+ undergoes disproportionation or not.
8.
If s E oFe2 |Fe = –0.44 V, E oFe3 |Fe2 = 0.77 V. Calculate E oFe3 |Fe .
9.
Consider the standard reduction potentials (in volts) as shown in Fig. Find Eº.
1
0.936
0.576
SO 24 
 SO32 
 S2 O32
2
E  ?
|______________________________
______________________________

10.
What is ECell if :
2Cr + 3H2O + 3OCl–  2Cr3+ + 3Cl– + 6 OH–
Cr3+ + 3e–  Cr,
OCl– + H2O + 2e–  Cl– + 2OH–
E° = – 0.74 V
E° = 0.94 V
1.5V
11.
The standard oxidation potentials for Mn3+ ion acid solution are Mn2+  Mn3+
1.0 V

 MnO2 . Is the reaction 2 Mn3+ + 2 H2O  Mn2+ + MnO2 + 4H+ spontaneous
under conditions of unit activity ? What is the change in free energy ?
12.
The reduction potential values are given below
Al3+/Al = –1.67
1.67 volt, Mg2+/Mg = –2.34 volt, Cu2+/Cu = +0.34 volt
2 /¯¯ = +0.53 volt. Which one is the best reducing agent ?
13.
The standard reduction potential value of the three metallic cations X, Y and Z are 0.52, – 3.03
and – 1.18 V respectively. Write the decreasing order of reducing power of the corresponding
metals :
14.
(i)
(ii)
(iii)
15.
Which of the following oxides is reduced by hydrogen ?
MgO, CuO and Na2O
Which of the following oxides will decompose most easily on heating ?
ZnO, CuO, MgO, and Ag2O
The value of EºOX for electrode reactions,
Fe Fe2+ + 2e– Cu  Cu2+ + 2e– and
Zn  Zn2+ + 2e–
are 0.444, –0.337
0.337 and 0.763 volt respectively. State which of these metals can replace
the other two from the solution of their salts ?
Using the G0 for the reactions
C + O2  CO2
G0 = 395 kJ / mole,
2Al(l) + 3/2O2  Al2O3(s) G0 = 1269 kJ/mole
Al2O3(s)  Al2O3 (melt) G0 = 16 kJ/mole
Calculate the EMF for the cell reaction
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2Al2O3 (melt) + 3C  4Al() + 3CO2(g)
16.
For the reaction, 4Al(s) + 3O2(g) + 6H2O + 4 OH–  4 [Al(OH)4–] ; E°cell = 2.73 V. If G°f
(OH–) = –157 kJ mol–1 and G°f (H2O) = –237.2 kJ mol–1, determine G°f [Al (OH)4–].
17.
NERNST EQUATION & ITS APPLICATIONS
Calculate the EMF of a Daniel cell when the concentration of ZnSO4 and CuSO4 are 0.001 M
and 0.1M respectively. The standard potential of the cell is 1.1V.
18.
Calculate the oxidation potential of a hydrogen electrode at pH = 1 (T = 298 K).
19.
Calculate E0 and E for the cell Sn | Sn2+ (1M) || Pb2+(10–3M) | Pb, E0 (Sn2+| Sn) = -0.14V, E0
(Pb2+| Pb) = –0.13V.
0.13V. Is cell representation is correct?
20.
At what concentration of Cu2+ in a solution of CuSO4 will the electrode potential be zero at
25°C? Given : E0 (Cu | Cu2+) = –0.34 V.
21.
A zinc electrode is placed in a 0.1M solution at 25°C. Assuming that the salt is 20% dissociated
at this dilutions calculate the electrode reduction potential. E0 (Zn2+| Zn) = -0.76V.
22.
The standard reduction potential of Cu2+ / Cu couple is 0.34 V at 25°C. Calculate
Calcul the reduction
potential at pH = 14 for this couple.
(Given : Ksp, Cu (OH)2 = 1.0 × 10–19).
23.
Consider the following electrochemical cell :
(a) Write a balanced net ionic equation for the spontaneous reaction that take place in the cell.
(b) Calculte the standard cell potential Eº for the cell reaction.
(c) If the cell emf is 1.6 V, what is the concentration of Zn2+ ?
cell ?
(d) How will the cell potential be affected if KI is added to Ag+ half-cell
24.
The standard oxidation potential of Zn referred to SHE is 0.76V and that of Cu is –0.34V at
25ºC. When excess of Zn is added to CuSO4 , Zn displaces Cu2+ till equilibrium is reached.
What is the approx value of log
[Zn 2 ]
at equilibrium?
[Cu 2 ]
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25.
Calculate the equilibrium constant for the reaction
Fe2+ +Ce4+  Fe3+ + Ce3+, [given : E0 Ce4+ / Ce3+ = 1.44V;E0 Fe3+/Fe2+ = 0.68V]
26.
Calculate the equilibrium constant for the reaction Fe + CuSO4  FeSO4 + Cu at 25°C. Given
E0 (Fe/Fe2+) = 0.44V, E0 (Cu/Cu2+) = -0.337V.
27.
For a cell Mg(s) | Mg2+(aq) || Ag+ (aq) | Ag,
(i) Calculate the equilibrium constant at 25°C.
(ii) Also find the maximum work per mole Mg that can be obtained by operating the cell.
E0 (Mg2+/Mg) = –2.37V,
2.37V, E0 (Ag+/Ag) = 0.8 V.
28.
The standard reduction potential at 25°C for the reduction of water
0.8277 volt. Calculate the equilibrium constant for the reaction
2H2O + 2e–  H2 + 2OH– is -0.8277
2H2O  H3O+ + OH– at 25°C.
CONCENTRATION CELL & METAL, METAL INSOLUBLE SALTSALT ION ELECTRODE
29.
Equinormal Solutions of two weak acids, HA (pKa = 3) and HB (pKa = 5) are each placed in
contact with equal pressure of hydrogen electrode at 25°C. When a cell is constructed by
interconnecting them through a salt bridge, find the emf of the cell.
30.
In two vessels each containing 500ml water, 0.5m mol of aniline (Kb= 10–9) and 50mmol of
HCl are added separately. Two hydrogen electrodes are constructed using these solutions.
Calculate the emf of cell made by connecting them appropriately.
31.
Calculate the EMF of the following cell
Zn | Zn2+ (0.01M) || Zn2+ (0.1 M) | Zn
at 298 K.
32.
Calculate pH using the following cell :
Pt (H2) | H+ (x M) | | H+ (1 M) | Pt (H2)
1 atm
1 atm
33.
if Ecell = 0.2364 V.
Write cell reaction from given cell diagrams :
(A) Cu | Cu 2 || Cl – | Hg 2 Cl2 | Hg | Pt
(B) Ag (s) | AglO3 (s)|Ag+, HlO3 || Zn2+ | Zn (s)
(C) Mn (s) | Mn (OH)2 (s) Mn2+, OH– || Cu2+|Cu (s)
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34.
For the galvanic cell : Ag | AgCl (s)|KCl (0.2M) || KBr (0.001M) |AgBr (s) |Ag,
2.303RT
Calculate the EMF generated? (Take
= 0.06 )
F
[Ksp(AgCl)= 10–10 ; Ksp(AgBr)=10–13]
35.
Given, E° = –0.27
0.27 V for the Cl– | PbCl2 |Pb couple and – 0.12 V for the Pb2+ | Pb couple,
2.303RT
determine Ksp for PbCl2 at 25°C ? (Take
= 0.06 )
F
36.
The pKsp of Agl is 16. if the E° value for Ag+ | Ag is 0.8 V. Find the E° for the half cell
2.303RT
reaction AgI(s) + e–  Ag + I– ? (Take
= 0.06 )
F
ELECTROLYTIC CELL
37. 
ELECTROLYTE
Product at ANODE Product at CATHODE
1 NaCl (Molten) with Pt electrode
2 NaCl (aq) with Pt electrode
3 Na2SO4 (aq) with Pt electrode
4 NaNO3 (aq) with Pt electrode
5 AgNO3 (aq) with Pt electrode
6 CuSO4 (aq) with Inert electrode
7 CuSO4 (aq) with Copper electrode


38.
FARADAY’S
FARADAY LAW & ITS APPLICTIONS
Calculate the no. of electrons lost or gained during electrolysis of
(c) 2.7 gm of Al3+ ions
(a) 3.55 gm of Cl– ions
(b) 1 gm Cu2+ ions
39.
How many faradays of electricity are involved in each of the case
(a) 0.25 mole Al3+ is converted to Al.
(b) 27.6 gm of SO3 is convered to SO2–3
(c) The Cu2+ in 1100 ml of 0.5 M Cu2+ is converted to Cu.
40.
0.5 mole of electron is passed through two electrolytic cells in series. One contains silver ions,
and the other zinc ions. Assume that only cathode reaction
reaction in each cell is the reduction of the
ion to the metal. How many gm of each metals will be deposited.
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41.
The electrosynthesis of MnO2 is carried out from a solution of MnSO4 in H2SO4 (aq). If a
current of 25.5 ampere is used with a current efficiency
efficiency of 85%, how long would it take to
produce 1 kg of MnO2?
42.
A constant current of 30 A is passed through an aqueous solution of NaCl for a time of 1.0 hr.
How many grams of NaOH are produced? What is volume of Cl2 gas at 1atm,273K produced
(in litre) ?
43.
If 0.224 litre of H2 gas is formed at the cathode, how much O2 gas is formed at the anode under
identical conditions?
44.
Assume 96500 C as one unit of electricity. If cost of electricity of producing x gm Al
A is Rs x,
what is the cost of electricity of producing x gm Mg?
45.
Chromium metal can be plated out from an acidic solution containing CrO3 according to
following equation:
CrO3(aq) + 6H+ (aq) + 6e–  Cr(s) + 3H2O
Calculate :
(i) How many grams of chromium will be plated out by 24125 coulombs and
(ii) How long will it take to plate out 1.5 gm of chromium by using 12.5 ampere current
46.
A certain metal salt solution is electrolysed in series with a silver coulometer. The weights of
silver and the metal deposited are 0.5094 g and 0.2653g. Calculate the valency of the metal if
its atomic weight is nearly that of silver.
47.
3A current was passed through an aqueous solution of an unknown salt of Pd for 1Hr. 2.977g
ind n. (Given Atomic mass of Pd = 106.4)
of Pd+n was deposited at cathode. F
Find
48.
A metal is known to form fluoride MF2. When 10A of electricity is passed through a molten
salt for 330 sec., 1.95g of metal is deposited. Find the atomic weight of M. What will be the
quantity of electricity required to deposit the same mass of Cu from CuSO4?
49.
After electrolysis of NaCl solution with inert electrodes for a certain period of time. 600 mL of
the solution was left. Which was found to be 1N in NaOH. During the same time, 31.75 g of Cu
was deposited
posited in the copper voltameter in series with the electrolytic cell. Calculate the
percentage yield of NaOH obtained.
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50.
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 ampere
for 20 mintue. What mass of Ni is deposited
de
at the cathode?
51.
A current of 3.7A is passed for 6hrs. between Ni electrodes in 0.5L of 2M solution of
Ni(NO3)2. What will be the molarity of solution at the end of electrolysis?
52.
Electrolysis of a solution of HSO4– ions produces S2O82–. Assuming 75% current efficiency,
what current should be employed to achieve a production rate of 1 mole of S2O82– per
hour ?
CONDUCTIVITIES AND CELL CONSTANT
The resistance of a conductivity cell filled with 0.01N solution of NaCl is 200 ohm
at18oC.Calculate the equivalent conductivity of the solution. The cell constant of the
conductivity cell is 0.88 cm–1.
53.
54.
The molar conductivity of 0.1 M CH3COOH solution is 4 S cm2 mole–1 . What is the specific
conductivity and resistivity of the solution
solut
?
55.
The conductivity of pure water in a conductivity cell with electrodes of cross sectional area 4
cm2 and 2 cm apart is 8 × 10–7 S cm–1.
(i) What is resistance of conductivity cell ?
(ii) What current would flow through the cell under an applied potential difference of 1 volt?
For 0.01N KCl, the resistivity 800 ohm cm. Calculate the conductivity and equivalent
conductance.
56.
57.
58.
59.
A solution containing 2.08 g of anhydrous barium chloride is 500 CC of water has a specific
conductivity 0.005 ohm–11cm–1. What are molar and equivalent conductivities of this solution.
APPLICATION OF KOHLRAUSCH'S LAW
Equivalent conductance of 0.01 N Na2SO4 solution is 120 ohm–1 cm2 eq–1. The equivalent
conductance at infinite dilution is 150 ohm–1 cm2eq–1. What is the degree of dissociation in
0.01 N Na2SO4 solution?
Specific conductance of a saturated solution of AgBr is 8.075×10–7 ohm–1cm–1 at 250C.
Specific conductance of pure water at 25°C is 0.75 ×10–7 ohm–1 cm–2.  m for KBr , AgNO3
and KNO3 are 140 , 130 , 110 ( S cm2 mol–1) respectively. Calculate the solubility of AgBr in
gm/litre.
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60.
Saturated solution of AgCl at 25°C has specific conductance of 1.12×10–6 ohm–1 cm–1. The 
Ag+ and Cl– are 54 and 58 ohm–1 cm2 / equi. respectively. Calculate the solubility product of
AgCl at 25°C.
–
If your answer is x × 10–y
then fill ‘y’ in OMR as sum of digits (excluding decimal places) till
you get the single digit answer.
61.
Hydrofluoric acid is weak acid. At 25°C, the molar conductivity of 0.002M HF is 200 ohm–1
cm2
mole–1. If its m = 400 ohm–1 cm2 mole–1, calculate its degree of dissociation and equilibrium
constant at the given concentration.
62.
The value of m for HCl, NaCl and CH3CO2Na are 425, 125 and 100 S cm2 mol–1
respectively. Calculate the value of m for acetic acid. If the equivalent conductivity of the
given acetic acid is 48 at 25° C, calculate its degree of dissociation.
63.
at infinite
For the strong electroytes NaOH, NaCl and BaCl2 the molar ionic conductivities
conducti
–4
–4
–4
2
–1
dilution are 240 ×10 , 125 ×10 and 280.0 ×10 mho cm mol respectively. Calculate the
molar conductivity of Ba(OH)2 at infinite dilution.
64.
At 25°C, (H+) = 3.5 ×10–2 S m2 mol–1 and (OH–) = 2 ×10–2 S m2mol–1.
Given: Sp. conductance = 5.5 ×10–6 S m–1 for H2O, determine pH and Kw.
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EXERCISE # (S-II)
1.
Calculate the emf of the cell
Pt, H2(1.0 atm) | CH3COOH (0.1M) || NH3(aq, 0.01M) | H2 (1.0 atm),
Pt Ka(CH3COOH) = 1.8 × 10–5, Kb (NH3) = 1.8 × 10–5.
2.
The Edison storage cell is represented as Fe(s) | FeO(s) | KOH(aq) | Ni2O3 (s) | Ni(s) The halfcell reaction are
Ni2O3(s) + H2O(i) + 2e–  2NiO(s) + 2OH–,
E0 = + 0.40V
FeO(s) + H2O(l) + 2e–  Fe(s) + 2OH–,
E0 = – 0.87V
(i) What is the cell reaction?
(ii) What is the cell e.m.f.? How does it depend on the concentration of KOH?
First multiply your answer with 100 Then fill your answer as sum of digits (excluding
decimal places) till you get the single
sing digit answer.
(iii) What is the maximum amount of electrical energy that can be obtained from one mole of
Ni2O3?
3.
The standard reduction potential for Cu2+ / Cu is 0.34 V. Calculate the reduction potential at
pH = 14 for the above couple. Ksp of Cu(OH)2 is 1 × 10–19.
4.
The emf of the cell Ag|AgI|KI(0.05M) || AgNO3(0.05M) |Ag is 0.788V. Calculate the solubility
product of AgI.
5.
Consider the cell Ag|AgBr(s)|Br–||Cl–| AgCl(s)| Ag at 25º C . The solubility product constants
of AgBr & AgCl are respectively 5 × 10–13 & 1 × 10–10 . For what ratio of the concentrations of
Br– & Cl– ions would the emf of the cell be zero ?
6.
The pKsp of Agl is 16.07 . If the Eº value for Ag+| Ag is 0.7991 V . Find the Eº for the half cell
reaction AgI (s) + e–  Ag + I–.
7.
For the galvanic cell : Ag|AgCl(s)| KCl (0.2M) || K Br (0.001 M)| AgBr(s) | Ag,
Calculate the EMF generated and assign correct polarity to each electrode for a spontaneous
process after taking into account the cell reaction at 250C.
[Ksp(AgCl) = 2.8 × 10–19 ; Ksp(AgBr) = 3.3×10–13]
8.
Given, E° = –0.268
0.268 V for the Cl– | PbCl2 | Pb couple and – 0.126 V for the Pb2+ | Pb couple,
determine Ksp for PbCl2 at 25°C?
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9.
Calculate the equilibrium constant for the reaction:
3Sn(s) + 2Cr2O72– + 28H+  3Sn4+ + 4Cr3+ + 14H2O
E0 for Sn/Sn2+ =0.136 V E0 for Sn2+/Sn4+ = – 0.154 V
E0 for Cr2O72–/Cr3+ = 1.33 V
10
One of the methods of preparation of per disulphuric acid, H2S2O8, involve electrolytic
oxidation of H2SO4 at anode (2H2SO4  H2S2O8 + 2H+ + 2e–) with oxygen and hydrogen as
by–products.
products. In such an electrolysis, 9.722 L of H2 and 2.35 L of O2 were generated at STP.
What is the weight of H2S2O8 formed?
11
A current of 3 amp was passed for 2 hour through a solution of CuSO4,3 g of Cu2+ ions were
deposited as Cu at cathode. Calculate percentage current efficiency of the process.
12.
Dal lake has water 8.2 ×1012 litre approximately. A power reactor produces electricity at the
rate of 1.5×106 coulomb per second at an appropriate voltage.How
voltage.How many years would it take to
electrolyse the lake?
13.
The equivalent conductance of 0.10 N solution of MgCl2 is 97.1 mho cm2 equi–1 at 25°C. a cell
with electrode that are 1.5 cm2 in surface area and 0.5 cm apart is filled with 0.1 N MgCl2
solution. How much current will flow when potential difference between the electrodes is 5
volt.
14.
When a solution of specific conductance 1.342 ohm–1 metre–1 was placed in a conductivity cell
with parallel electrodes, the resistance was found to be 170.5 ohm. Area of electrodes is
1.86×10–4 m2. Calculate separation of electrodes.
15.
The specific conductance at 25°C of a saturated solution of SrSO4 is 1.482×10–4 ohm–1
cm–1while that of water used is 1.5×10–6 mho cm–1. Determine at 25°C the solubility in gm per
litre of SrSO4 in water. Molar ionic conductance of Sr2+ and SO42– ions at infinite dilution are
59.46 and 79.8 ohm–1 cm2 mole–1 respectively. [ Sr = 87.6 , S = 32 , O = 16 ]
16.
The EMF of the cell M | Mn+ (0.02M) || H+ (1M) | H2(g) (1 atm), Pt at 25°C is 0.81V. Calculate
the valency of the metal if the standard oxidation of the metal is 0.76V
17.
From the standard potentials shown in the following diagram, calculate the potentials E10 and
E02 .
E10
B 3
BrO
0.54V
BrO–
0.45V 1
1.07V
Br–
Br
2 2
E 02
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18.
Calculate the EMF of the cell,
Zn – Hg(c1M) | Zn2+ (aq)| Hg – Zn(c2M)
at 25°C, if the concentrations of the zinc amalgam are: c1 = 10g per 100g of mercury and c2 =
1g per100 g of mercury.
19.
Calculate the quantity of electricity that would be required to reduce 12.3 g of nitrobenzene to
aniline, if the current efficiency for the process is 50 percent. If the potential drop across the
cell is 3.0 volts, how much energy will be consumed?
20.
How long a current of 2A has to be passed through a solution of AgNO3 to coat a metal surface
of 80cm2 with 5m thick layer? Density of silver = 10.8g/cm3.
21.
10g solution of CuSO4 is electrolyzed using 0.01F of electricity. Calculate:
(a)The weight of resulting solution (b)Equivalents of acid or alkali in the solution.
22.
Cadmium amalgam is prepared by electrolysis of a solution of CdCl2 using a mercury cathode.
How long should a current of 5A be passed in order to prepare 12% Cd-Hg
amalgam on a
Cd
cathode of 2gm Hg (Cd=112.4)
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EXERCISE # (O-I)
1.
GALVANIC CELL, ITS REPRESENTATION & SALT BRIDGE
In a galvanic cell
(A) chemical reaction produces electrical energy
(B) electrical energy produces chemical reaction
(C) reduction occurs at anode
(D) oxidation occurs at cathode
2.
Which of the following statement is wrong about galvanic cell ?
(A) Cathode is positive charged
(B) Anode is negatively charged
(C) Reduction takes place at the anode
(D) Electrons move from anode to cathode
3.
Which of the following has been universally accepted as a reference electrode at all
temperatures and has been assigned a value of zero volt?
(A) Standard platinum electrode
(B) Standard copper electrode
(C) Standard graphite electrode
(D) Standard hydrogen electrode
4.
Which is not true for a standard hydrogen electrode ?
st be 25°C
(A) The hydrogen ion concentration is 1 M (B) Temperature must
(C) Pressure of hydrogen is 1 atmosphere (D) All are correct
5.
The equation representing the process by which standard reduction potential of zinc can be
defined is :
(A) Zn2+(s) + 2e–  Zn(s)
(C) Zn2+(g) + 2e–  Zn(s)
6.
(D) Zn2+(aq.) + 2e–  Zn(s)
KCl can be used in salt bridge as electrolyte in which of the following cells?
(A) Zn | ZnCl2 || AgNO3 | Ag
(C) Cu | CuSO4 || AuCl3 | Au
7.
(B) Zn(g)  Zn2+(g) + 2e–
(B) Pb | Pb(NO3)2 || Cu(NO3)2 | Cu
(D) Fe | FeSO4 || Pb(NO3)2 | Pb
For the cell Pt (Cl2) | Cl– (aq) || Ag+ (aq) | Ag, cell reaction is given by
(A) 2Cl– + 2Ag+ 2Ag
2Ag + Cl2
(B) Cl2 + 2Ag 2Ag+ + 2Cl–
(C) 2Cl– + 2Ag 2Ag+ + Cl2
2Ag + 2Cl–
(D) Cl2 + 2Ag+ 2Ag
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8.
Which of the following statement is wrong about galvanic cell ?
(A) cathode is positive charged
(B) anode is negatively charged
(C) reduction takes place at the anode
(D) reduction takes place at the cathode
9
A standard hydrogen electrode has zero electrode potential because
(A) hydrogen is easier to oxidise
(B) electrode potential is assumed to be zero
(C) hydrogen atom has only one electron
(D) hydrogen is the lightest element.
10.
Which of the following is/are function(s) of salt - bridge ?
(A) It completes the electrical circuit
(B) It maintains electrical neutrality by flow of ions between the two compartments through
salt - bridge
(C) It minimises the liquid - liquid junction potential
(D) All of these
11.
KCl can't be used in salt bridge if electrolyte of a galvanic cell contains:
(A) Ag+- ions
(B) Pb2+- ions
(C) Hg22+ -ions
(D) All
12.
13.
STANDARD CELL POTINTIAL AND Electrochemical series
A standard reduction electrode potentials of four metals are
A = – 0.250 V,
B = – 0.140 V
C = – 0.126 V,
D = – 0.402 V
The metal that displaces A from its aqueous solution is ::–
(A) B
(B) C
(C) D
(D) None of the above
The standard electrode potentials for the reactions
Ag+ (a) + e–  Ag(s)
Sn2+ (a) + 2e–  Sn (s)
0.14 volt, respectively. The standard emf of the cell.
at 25 °C are 0.80 volt and –0.14
2+
+
Sn(s)|Sn (aq) (1M)||Ag (aq)(1M)|Ag(s) is :
(A) 0.66 volt
(B) 0.80 volt
(C) 1.08 volt
(D) 0.94 volt
14.
E°(Ni2+/Ni) = –0.25
0.25 volt, E° (Au3+ / Au) = 1.50 volt. The standard emf of the voltaic cell.
Ni(s) | Ni2+(aq) (1.0 M) || Au3+(aq) (1.0 M) | Au(s) is :
(A) 1.25 volt
(B) –1.75 volt
(C) 1.75 volt
(D) 4.0 volt
15.
E° for F2 + 2e— = 2F– is 2.8 V, E° for ½F2 + e– = F– is ?
(A) 2.8 V
(B) 1.4 V
(C) –2.8 V
(D) –1.4 V
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16.
From the following E° values of half cells,
(i) A + e  A–; E° = –0.24
0.24 V
(ii) B– + e  B2–;
3–
E° = +1.25 V
2–
(iii) C– + 2e  C ; E° = –1.25 V
(iv) D + 2e  D ; E° = +0.68 V
What combination of two half cells would result in a cell with the largest potential ?
(A) (ii) and (iii)
(B) (ii) and (iv)
(C) (i) and (iii)
(D) (i) and (iv)
17.
18.
Zn cannot displace following ions from their aqueous solution :
(A) Ag+
(B) Cu2+
(C) Fe2+
(D) Na+
The standard reduction potentials at 25°C for the following half reactions are :
 Zn(s), E°RP = –0.762V
Zn2+ (aq) + 2e– 
 Cr(s), E°RP = –0.740V
Cr3+ (aq) + 3e– 
 H2(g), E°RP = 0.00 V
2H+(aq) + 2e– 
 Fe2+(aq), E°RP = 0.77V
Fe3+(aq) + 2e– 
Which is the strongest reducing agent ?
(A) Zn
(B) Cr
19.
(C) H2(g)
(D) Fe2+ (aq)
Using the standard electrode potential values given below, decide which of the statements, I, II,
III and IV are correct. Choose the right answer from (A), (B), (C) and (D).
 Fe(s) ; E° = –0.44 V
Fe2+(aq) + 2e– 
 Cu(s) ; E° = +0.34 V
Cu2+(aq) + 2e– 
 Ag(s) ; E° = +0.80 V
Ag+(aq) + e– 
20.
I. Copper can displace iron from FeSO4 solution.
II. Iron can displace copper from CuSO4 solution.
III. Silver can displace copper from CuSO4 solution.
IV. Iron can displace silver from AgNO3 solution.
(A) I and II
(B) II and III
(C) II and IV
(D) I and IV
The reduction potential values are given below:
Al3+ / Al = –1.67 volt,
Mg2+ / Mg = –2.34 volt
Cu2+ / Cu = + 0.34 volt,
I2 / 2I– = + 0.53 volt
Which one is the best reducing
ducing agent ?
(A) Al
(B) Mg
(C) Cu
(D) I2
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21.
The following facts are available :–
:
2X– + Y2  2Y– + X2
2W– + Y2  NO reaction
2Z– + X2  2X– + Z2
Which of the following statements is correct :–
:
(A) E 0W  / W  E 0Y /Y  E 0X /X  E 0Z / Z
2
(B) E 0W  / W  E 0Y /Y  E 0X //XX  E 0Z / Z
(C) E 0W / W  E 0Y /Y  E 0X /X  E 0Z / Z
2
 E 0X //XX  E 0Z / Z
(D) E 0W  / W  E 0Y /Y
/Y
2
2
2
22.
2
2
2
2
2
2
2
Ag
CuSO4
(blue)
2
I
Fe
CuSO4
II
CuSO4
Blue colour solution changes to colourless (or fades) in :
(A) I, II, III
(B) I, II
(C) II, III
III
(D) I, III
Consider the following sets :
Cu
AgNO3
(colourless)
Cu
I
ZnSO 4
(colourless)
Cu
II
MgCl2
(colourless)
colourless solution changes to blue coloured solution in :
(A) I
(B) II
(C) III
24.
2
2
Consider the following sets :
Zn
23.
2
The thermodynamic efficiency of cell is given by
bynFEcell
nFEcell
H
(A)
(B)
(C) –
G
G
H
III
(D) I, III
(D) Zero
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25.

26.
27.
28.
If  G° of the cell reaction,
AgCl(s) + ½H2(g)  Ag(s) + H+ +Cl
+Cl– is –21.52 KJ then 
G° of 2AgCl(s) +H2(g)  2Ag(s) +2H+ +2Cl– is :
(A) –21.52 KJ
(B) –10.76 KJ
(C) –43.04 KJ
(D) 43.04 KJ
The standard reduction potentials for two half-cell
half cell reactions are given below,
2+
—
Cd (aq) + 2e  Cd(s),
E° = – 0.40V
+
—
Ag (aq) + e  Ag(s),
E° = 0.80V
The standard free energy change for the reaction
2Ag+(aq) + Cd(s)  2Ag(s) + Cd2+ (aq) is given by :
(A) 115.8 KJ
(B) –115.8 KJ
(C) –231.6 KJ
(D) 231.6KJ
n+1
For a reaction, M
NERNST EQUATION & ITS APPLICATIONS
+ ne  M+1, the Nernst equation has a form of :
(A) E = E° +
RT
n
nF
(C) E = E° –

RT
log [Mn 1]
nF
[M ]
[M ]
[Mn 1 ]
(B) E = E° –
RT
n
nF
(D) E = E° +

RT
log [Mn 1]
nF
[M ]
[M ]
[Mn 1 ]
The reduction potential of hydrogen electrode ( PH 2 = 1 atms; [H+] = 0.1 M) at 25°C will be –
(A) 0.00 V
(B) –0.059 V
(C) 0.118 V
(D) 0.059 V
29.
Which of the following represents the reduction potential of silver wire dipped into 0.1 M
AgNO3 solution at 25° C ?
(A) E°red
(B) (E°red + 0.059)
(C) (E°oxi – 0.059)
(D) (E°red – 0.059)3
30.
Which of the following will increase the voltage of the cell with following cell reaction
Sn(s) + 2Ag+(aq)  Sn+2(aq) + 2Ag(s)
(A) Decrease in the concentration of Ag+ ions
(B) Increase in the concentration of Sn+2 ions
(C) Increase in the concentration of Ag+ ions
(D) (A) & (B) both
31.
The emf of the cell
Ti / Ti+ (0.01M) | | Cu2+(0.1M) / Cu is 0.83 V
The emf of this cell will be increased by :
(A) Increase the concentration of Cu++ ions (B) Decreaseing the concentration of Ti+
(C) Increasing the concentration of both
(D) (A) & (B) both
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32.
The emf of the cell in which the following reaction,
Zn(s) + Ni2+(aq) (a = 0.1)  Zn2+(aq) (a = 1.0) + Ni(s)
occurs,, is found to be 0.5105 V at 298 K. The standard e.m.f. of the cell is ::
(A) –0.5105 V
(B) 0.5400 V
(C) 0.4810 V
(D) 0.5696 V
33.
Given electrode potentials :
Fe3+(aq) + e–  Fe2+(aq) ; E° = 0.771 volts I2(g) + 2e–  2I–(aq) ; E° = 0.536 volts
E°cell for the cell reaction,
2Fe3+(aq) + 2I–(aq)  2Fe2+(aq) + I2(g) is –
(A) (2 × 0.771 – 0.536) = 1.006 volts
(C) 0.771 – 0.536 = 0.235 volts
34.
(B) (0.771 – 0.5 × 0.536) = 0.503 volts
0.235 volts
(D) 0.536 – 0.771 = –0.235
The equilibrium constant for the reaction
Sr(s) + Mg+2 (aq)  Sr+2 (aq) + Mg(s) is 4 × 1012 at 25°C
The E° for a cell made up of the Sr/Sr+2 and Mg+2/Mg half cells
(log 2 = 0.3)
(A) 0.3717 V
(B) 0.7434 V
(C) 0.1858 V
35.
(D) 0.135 V
The standard emf for the cell reaction,
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) is 1.10 volt at 25 °C. The emf for the cell reaction when 0.1
M Cu2+ and 0.1 M Zn2+ solution are used at 25°C is :
(A) 1.10 volt
(B) 0.110 volt
(C) –1.10 volt
(D) –0.110 volt
36.
What is the potential of the cell containing two hydrogen electrodes as represented below
Pt | H2(g) | H+(aq)(10–8 M) || H+(aq)(0.001 M) | H2(g)|Pt
(A) – 0.295 V
(B) – 0.0591 V
(C) 0.295 V
(D) 0.0591 V
37.
At 25°C the standard emf of cell
cell having reactions involving two electrons change is found to be
0.295V. The equilibrium constant of the reaction is –
(A) 29.5 × 10–2
(B) 10
(C) 1010
(D) 29.5 × 1010
38.
Consider the cell, Cu|Cu+2||Ag+|Ag. If the concentration of Cu2+ and Ag+ ions becomes ten
times the emf of the cell :–
:
(A) Becomes 10 times
(B) Remains same
(C) Increase by 0.0295 V
(D) Decrease by 0.0295 V
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39.
By how much times will potential of half cell Cu+2/Cu change if, the solution is diluted to 100
times at 298 K :–
(A) Increases by 59 mV
(B) Decrease by 59 mV
(C) Increases by 29.5 mV
(D) Decreases by 29.5 mV
40.
For a reaction - A(s) + 2B+(aq)  A2+(aq) + 2B(s) KC has been found to be 1012. The E°cell is:
(A) 0.354 V
(B) 0.708 V
(C) 0.0098 V
(D) 1.36 V
41.
Determine the value of E° cell for the following reaction Cu+2(aq) + Sn+2(aq)  Cu(s) + Sn+4(aq) Equilibrium constant is 106
(A) 0.177
(B) 0.0177
(C) 0.215
(D) 1.77
42.
The standard emf of a galvanic cell involving cell reaction with n = 4 is found to be 0.295 V at
25°C. The equilibrium constant of the reaction would be,
(A) 1.0 × 1020
(B) 2.0 × 1011
(C) 4.0 × 1012
(D) 1.0 × 102
CONCENTRATION CELL & METAL, METAL INSOLUBLE SALTSALT ION ELECTRODE
43.
Which respresent a concertration cell ?
(A) Pt | H2 | HCl || HCl | PtH2
(B) Pt | H2 | HCl || Cl2 | Pt
2+
2+
(C) Zn | Zn || Cu | Cu
(D) Fe | Fe+2 || Cu2+ | Cu
44.
45.
Zn | Zn2+ (C1)|| Zn2+ (C2)|Zn. for this cell G is negative if (A) C1 = C2
(B) C1 > C2
(C) C2 > C1
(D) None
The chemical reaction
2AgCl(s) + H2 (g)  2HCl (aq) + 2Ag (s)
taking place in a galvanic cell is represented by the notation
(A) Pt(s) | H2 (g), 1 bar | 1 M KCl (aq) | AgCl(s) | Ag (s)
(B) Pt(s) | H2 (g), 1 bar | 1 M HCl (aq) | 1 M Ag+ (aq) | Ag (s)
(C) Pt(s) | H2 (g), 1 bar | 1 M HCl (aq) | AgCl (s) | Ag (s)
(D) Pt(s) | H2 (g), 1 bar | 1 M HCl (aq) | Ag (s) | AgCl (s)
46.
The cell Pt (H2) (1 atm) | H+(pH = ?) || I– (a = 1) | AgI(s), Ag(s) | Pt has emf, E298K = 0. The
standard electrode potential for the reaction AgI + e¯  Ag + I is – 0.151 volt. Calculate the
pH value.
(A) 3.37
(B) 5.26
(C) 2.56
(D) 4.62
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47.
wat
Using the information in the preceding problem, calculate the solubility product of AgI in water
º
at 25°C [ E (Ag ,Ag) = + 0.799 volt]
(A) 1.97 × 10–17
48.
49.
(B) 8.43 × 10–17
(C) 1.79 × 10–17
(D) 9.17 × 10–17
The solubility product of silver iodide is 8.3 × 10–17 and the standard reduction potential of Ag,
Ag+ electrode is + 0.8 volts at 25° C. The standard reduction potential of Ag,Agl/l¯ electrode
from these data is
(A) – 0.30 V
(B) + 0.15 V
(C) + 0.10 V
(D) – 0.15 V
THERMODYNAMIC FUNCTIONS OF CELL
 d(G) 
 dEcell 
G = H – TS and G
G = H + T 
then 
 is :

 dT  p
 dT 
(A)
S
nF
(B)
nE
S
(C) – nFFcell
(D) + nEFcell
50.
The efficiency of an hypothetical cell is about 84% which involves the following reaction :
A (s) + B2+ (aq) 
 A2+ (aq) + B (s) : H = – 285 kJ
Then, the standard electrode potential of the cell will be (Assume as S
 = 0)
(A) 1.20
(B) 2.40 V
(C) 1.10 V
(D) 1.24 V
51.
The temperature coefficient, of the emf i.e.
52.
 Ag(s) in which the cell
The standard emf of the cell, Cd(s) CdCl2(aq) (0.1 M) AgCl(s)
reaction is,
Cd(s) + 2AgCl(s) 
 2Ag(s) + Cd+2 (aq) + 2Cl–(aq) is 0.6915 V at 0°C and 0.6753 V at
25°C. The Hº
Hº of the reaction at 25°C is :
(A) – 176 kJ
(B) – 234.7 kJ
(C) + 123.5 kJ
(D) – 167.26 kJ
53.
dE
= – 0.00065 volt. deg–1 for the cell Cd | CdCl2
dT
(1M) || AgCl (s) | Ag at 25°C. Calculate the entropy changes S298K for the cell reaction, Cd +
2AgCl  Cd++ + 2Cl + 2Ag
(A) – 105.5 JK–1
(B) – 150.2 JK–1
(C) – 75.7 JK–1
(D) – 125.5 JK–1
ELECTROLYTIC CELL
When an electric current is passed through a cell containing an electrolyte, positive ions move
towards the cathode and negative ions towards the anode. What will happen if the cathode is
pulled out of the solution?
(A) The positive ions will start moving towards the anode and negative ions will stop moving.
(B) The negative ions will continue
continue to move towards the anode and the positive ions will stop
moving
(C) Both positive and negative ions will move towards the anode.
(D) None of these movements will take place.
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54.
55.
56.
Which of the substances Na, Hg, S, Pt and graphite can be used as electrodes in electrolytic
cells having aqueous solution ?
(A) Hg and Pt
(B) Hg, Pt and graphite
(C) Na, S
(D) Na, Hg, S
The products formed when an aqueous solution of NaBr is electrolyzed in a cell having inert
electrodes are :
(A) Na and Br2
(B) Na and O2
(C) H2, Br2 and NaOH (D) H2 and O2
Electrolysis of a CuSO4 produces :–
:
(A) An increase in pH
(C) Either decrease or increase
(B) A decrease in pH
(D) None
57.
A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at
the cathode and anode are respectively.
(A) H2, O2
(B) O2, H2
(C) O2, Na
(D) none
58.
When an aqueous solution of lithium chloride is electrolysed using graphite electrodes
(A) Cl2 is liberated at the anode.
(B) Li is deposited at the cathode
(C) as the current flows, pH of the solution remains constant
(D) as the current flows, pH of the solu
solution decreases.
59.
The amount of an ion discharged during electrolysis is not directly proportional to :
(A) resistance
(B) time
(C) current strength
(D) electrochemical equivalent of the element
60.
Number of electrons involved in the electrodeposition
electrodeposition of 63.5 g of Cu from a solution of
23
CuSO4 is: (NA = 6 × 10 )
(A) 6 × 1023
(B) 3 × 1023
(C) 12 × 1023
(D) 6 × 1022
61.
When one coulomb of electricity is passed through an electrolytic solution the mass deposited
on the electrode is equal to :
(A) equivalent weight
(B) molecular weight
(C) electrochemical equivalent
(D) one gram
62.
Electro chemical equivalent of a substance is 0.0006; its e wt. is :
(A) 57.9
(B) 28.95
(C) 115.8
(D) cannot be calculated
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63.
W g of copper deposited in a copper voltameter when an electric current of 2 ampere is passed
for 2 hours. If one ampere of electric current is passed for 4 hours in the same voltameter,
copper doposited will be :
(A) W
(B) W/2
(C) W/4
(D) 2W
64.
When the same electric current is passed through the solution of different electrolytes in series
the amounts of elements deposited on the electrodes are iinn the ratio of their:
(A) atomic number (B) atomic masses
(C) specific gravities (D) equivalent masses
65.
The amount of electricity that can deposit 108 g. of silver from silver nitrate solution is:
(A) 1 ampere
(B) 1 coulomb
(C) 1 Faraday
(D) 2 ampere
66.
The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity
from aqueous H2SO4 and fused MgSO4 are :
(A) 1 : 8
(B) 1 : 12
(C) 1 : 16
(D) None of these
67.
A current of 9.65 amp. flowing for 10 minute deposits
deposits 3.0 g of a metal. The equivalent wt. of
the metal is :
(A) 10
(B) 30
(C) 50
(D) 96.5
68.
The same amount of electricity was passed through two separate electrolytic cells containing
solutions of nickel nitrate [Ni(NO3)2]and chromium nitrate [Cr(NO3)3] respectively. If 0.3 g of
nickel was deposited in the first cell, the amount of chromium deposited is :
(at. wt. of Ni = 59, at. wt. of Cr = 52)
(A) 0.1 g
(B) 0.17 g
(C) 0.3 g
(D) 0.6 g
69.
The electric charge for electro deposition of 1 equivalent of a substance is :
(A) one ampere per second
(B) 96500 coulomb per second
(C) one ampere for one hour
(D) charge on one mole of electrons
70.
3.17 g., of a substance was deposited by the flow of 0.1 mole of electrons. The equivalent
weight of the substance is :
(A) 3.17
(B) 0.317
(C) 317
(D) 31.7
71.
A current of 9.65 amp. passing for 16 min. 40 sec. through a molten tin salt deposits 5.95 g. of
tin The oxidation state of the tin in the salt is : (at. wt of Sn = 119)
(A) +4
(B) +3
(C) +2
(D) +1
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72.
The time required for a current of 3 amp. to decompose electrolytically 18 g of H2O is:
(A) 18 hour
(B) 36 hour
(C) 9 hour
(D) 18 seconds
73.
1 mole of Al is deposited by X coulomb of electricity passing through aluminium nitrate
solution. The number of moles of silver deposited by X coulomb of electricity from silver
nitrate solution is :
(A) 3
(B) 4
(C) 2
(D) 1
74.
An ion is reduced to the element when it absorbs 6 × 1020 electrons. The number of gm
equivalents of the ion is :
(A) 0.10
(B) 0.01
(C) 0.001
(D) 0.0001
75.
When an electric current is passed through acid diluted water, 112 ml. of hydrogen gas at STP
collects at thee cathode in 965 second. The current passed, in ampere is :
(A) 1.0
(B) 0.5
(C) 0.1
(D) 2.0
76.
How many coulombs of electric charge are required for the oxidation of 1 mole of H2O to O2 ?
(A) 9.65 × 104 C
(B) 4.825 × 105 C
(C) 1.93 × 105 C
(D)) 1.93 × 104 C
77.
A factory produces 40 kg. of calcium in two hours by electrolysis. How much aluminium can
be produced by the same current in two hours :–
:
(At wt. of Ca = 40, Al = 27)
(A) 22 kg.
(B) 18 kg.
(C) 9 kg.
(D) 27 kg.
78.
The cost of electricity required to deposit 1 g of Mg is Rs. 5.00. How much would it cost to
deposit 9 g of Al (At wt. Al = 27, Mg = 24)
(A) Rs. 10
(B) Rs. 27
(C) Rs. 40
(D) Rs. 60
79.
Calculate the volume of hydrogen at STP obtained by passing a current of 0.536
ampere
0.
through acidified water for 30 minutes.
(A) 0.112 litre
(B) 0.224 litre
(C) 0.056 litre
(D) 0.448 litre
80.
An electric current is passed through silver voltameter connected to a water voltameter in
ameter weighed 0.108g more at the end of the electrolysis.
series. The cathode of the silver volt
voltameter
The volume of oxygen evolved at STP is :
(A) 56cm3
(B) 550 cm3
(C) 5.6 cm3
(D) 11.2 cm3
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81.
4.5g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a certain
tity of electric charge. The volume of hydrogen produced at STP from H+ ions in solution
quantity
by the same quantity of electric charge will be –
(A) 44.8L
(B) 11.2L
(C) 22.4L
(D) 5.6 L
82.
The time required to coat ameter surface of 80 cm2 with 5 × 10–3 cm thick layer of silver
(density 1.08 g cm–3) with the passage of 9.65A current through a silver nitrate solution is :
(A) 10 sec.
(B) 40 sec.
(C) 30 sec.
(D) 20 sec.
83.
One gm metal M+2 was discharged by the passage of 1.2 × 1022 electrons. Wh
What is the atomic
weight of metal?
(A) 25
(B) 50
(C) 100
(D) 75
84.
One mole of electron passes through each of the solution of AgNO3, CuSO4 and AlCl3 when
Ag, Cu and Al are deposited at cathode. The molar ratio of Ag, Cu and Al deposited are
(A) 1 : 1 : 1
(B) 6 : 3 : 2
(C) 6 : 3 : 1
(D) 1 : 3 : 6
85.
The density of A is 10 g cm–3. The quantity of electricity needed to plate an area 10 cm × 10 cm
to a thickness of 10–2 cm using ASO4 solution would be (Atomic mass of A = 193)
(A) 5000 C
(B)) 10000 C
(C) 40000 C
(D) 20000 C
86.
During electrolysis of an aqueous solution of sodium sulphate, 2.4 L of oxygen at STP was
liberated at anode. The volume of hydrogen at STP, liberated at cathode would be
(A) 1.2 L
(B) 2.4 L
(C) 2.6 L
(D) 4.8 L
87.
The charge required for the oxidation of one mole Mn3O4 into MnO 24 in presence of alkaline
medium is
(A) 5 × 96500 C
88.
89.
(B) 96500 C
(C) 10 × 96500 C
(D) 2 × 96500 C
COMMERCIAL CELLS & CORROSION
A fuel cell uses CH4(g) and forms CO32– at the anode. It is used to power a car with 80 Amp.
for 0.96 hr. How many litres of CH4(g) (STP) would be required ? (Vm = 22.4 L/mol) (F =
96500). Assume 100% efficiency.
(A) 5
(B) 6
(C) 7
(D) 8
When an acid cell is charged then :
(A) voltage of cell increases
(C) resistance of cell increases
(B) electrolyte of cell dilutes
(D) none of the above
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90.
When lead stroage battery is charged :
(A) lead dioxide dissolves
(B) sulphuric acid is regenerated
(C) the lead electrode
trode becomes coated with lead sulphate
(D) the amount of sulphuric acid decrease
Which is not correct method for prevention of iron from Rusting (A) Galvanisation
(B) Connecting to sacrificial electrode of Mg
(C) Making medium alkaline
(D) Making medium acidic
91.
CONDUCTIVITIES AND CELL CONSTANT
92 .
Electrolytic conduction differs from metallic conduction from the fact that in the former
(A) The resistance increases with increasing temperature
(B) The resistance decreases with increasing temperature
te
(C) The resistance remains constant with increasing temperature
(D) The resistance is independent of the length of the conductor
93.
Which of the following solution of KCl has the lowest value of specific conductance :
(A) 1 M
(B) 0.1 M
(C) 0.01 M
(D) 0.001 M
94.
Which of the following solutions of KCl has the lowest value of equivalent conductance ?
(A) 1 M
(B) 0.1 M
(C) .01 M
(D) .001 M
95.
The molar conductance at infinite dilution of AgNO3, AgCl and NaCl are 115, 120 and 110
respectively. The molar conductance of NaNO3 is :–
(A) 110
(B) 105
(C) 130
(D) 150
96.
The specific conductance of a 0.01 M solution of KCl is 0.0014 ohm–1 cm–1at 25° C. Its
equivalent conductance (cm2 ohm–1 equiv–1) is :–
(A) 140
(B) 14
(C) 1.4
(D) 0.14
97.
The resistance of 0.01 N solution of an electrolyte was found to be 200 ohm at 298 K using a
conductivity cell of cell constant 1.5 cm–1.
cm 1. The equivalent conductance of solution is :–
:
2
–1
2
–1
(A) 750 mhocm eq
(B) 75 mho cm eq
–1
2
–1
(C) 750 mho cm eq
(D) 75 mho–1cm2 eq–1
98.
The resistance of 0.1 N solution of a acetic acid is 250 ohm. When measured in a cell of cell
–
) of 0.1 N acetic acid is
constant 1.15 cm–1. The equivalent conductance (in ohm–1 cm2 equiv.–1
(A) 46
(B) 9.2
(C) 18.4
(D) 0.023
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99.
If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant
of 0.4 cm–1 then its molar conductance in ohm–1 cm2 mol–1 is :
(A) 10
(B) 102
(C) 103
(D) 104
100.
APPLICATION OF KOHLRAUSCH'S LAW
The equivalent conductivity of 0.1 N CH3COOH at 25 °C is 80 and at infinite dilution 400. The
degree of dissociation of CH3COOH is :
(A) 1
(B) 0.2
(C) 0.1
(D) 0.5
101.
The conductivity of a saturated solution of BaSO4 is 3.06 × 10–6 ohm–1 cm–1and its molar
–
conductance is 1.53 ohm–1
cm2 mol–1. The Ksp of BaSO4 will be :
(A) 4 × 10–12
(B) 2.5 × 10–9
(C) 2.5 × 10–13
(D) 4 × 10–6
102.
1 cm2 eq–1
Equivalent conductances of Ba+2 and Cl–
Cl ions are 127 & 76 ohm–1
eq respectively.
Equivalent conductance
nductance of BaCl2 at infinite dilution is (A) 139.5
(B) 101.5
(C) 203
(D) 279
103.
The resistance of 0.5 M solution of an electrolyte in a cell was found to be 50 . If the
electrodes in the cell are 2.2 cm apart and have an area of 4.4 cm2 then the molar conductivity
(in S m2 mol–1) of the solution is
(A) 0.2
(B) 0.02
(C) 0.002
(D) None of these
104.
Equivalent conductance of 0.1 M HA(weak acid) solution is 10 Scm2equivalent–1 and that at
infinite dilution is 200 Scm2equivalent–1 Hence pH of HA solution is
(A) 1.3
(B) 1.7
(C) 2.3
(D) 3.7
105.
If x is specific resistance of the electrolyte solution and y is the molarity of the solution, then ^m
is given by
y
xy
1000x
1000
(A)
(B) 1000
(C)
(D)
y
x
xy
1000
106.
The dissociation constant of n-butyric
n
acid is 1.6 × 10–5 and the molar conductivity at infinite
dilution is 380 × 10–4 Sm2mol–1. The specific conductance of the 0.01 M acid solution is
(A) 1.52 × 10–5 Sm–1
(B) 1.52 × 10–2 Sm–1
(C) 1.52 × 10–3 Sm–1
(D) None
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EXERCISE # (O-II)
Single correct :
1.
Consider the reaction,
Cl2(g) + 2Br–(aq)  2Cl–(aq) + Br2(g)
The emf of the cell when [Cl–]=[Br–]=0.01M and Cl2 gas at 1 atm pressure while Br2(g) at 0.01
atm will be (E° for the above reaction is = 0.29 volt) :
(A) 0.54 volt
(B) 0.35 volt
(C) 0.24 volt
(D) –0.29 volt
2.
How much will the reduction potential of a hydrogen electrode change when its solution
initially at pH = 0 is neutralised to pH = 7 ?
(A) increase by 0.059V
(B) decrease by 0.059V
(C) increase by 0.413V
(D) decrease by 0.413V
3.
If the pressure of H2 gas is increased from 1 atm to 100 atm keeping H+ concentration constant
at 1 M, the change in reduction potential of hydrogen half cell at 25°C will be
(A) 0.059 V
(B) 0.59 V
(C) 0.0295 V
(D) 0.118 V
4.
A silver wire dipped in 0.1 M HCl solution saturated with AgCl develops oxidation potential of
–0.209 V. If E 0Ag/Ag = – 0.799 V, the Ksp of AgCl in pure water will be
(A) 3 × 10–11
(B) 10–11
(C) 4 × 10
(D) 3 × 10–11
5.
Salts of A (atomic weight = 7), B (atomic weight =27) and C (atomic weight = 48) were
electrolysed under identical conditions using the same quantity of electricity. It was found that
when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The
valencies of A, B and C respectively are
(A) 3, 1 and 2
(B) 1, 3 and 2
(C) 3, 1 and 3
(D) 2, 3 and 2
6.
During
g electro refining of Cu by electrolysis of an aqueous solution of CuSO4 using copper
electrodes, if 2.5 g of Cu is deposited at cathode, then at anode
(A) decrease of more than 2.5 g of mass takes place
(B) 445 ml of O2 at STP is liberated
(C) 2.5 g off copper is deposited
(D) a decrease of 2.5 g of mass takes place
7.
The conductivity of a saturated solution of Ag3PO4 is 9 × 10–6 S m–1 and its equivalent
–
conductivity is 1.50 × 10–4
S m2 equivalent–1. The Ksp of Ag3PO4 is
(A) 4.32 × 10–18
(B) 1.8 ×10–9
(C) 8.64 × 10–13
(D) None of these
C.O.: NAIVEDHYAM, Plot No. SP-11,
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8.
Equal volumes of 0.015 M CH3COOH & 0.015 M NaOH are mixed together. What would be
molar conductivity of mixture if conductivity of CH3COONa is 6.3 ×10–4 S cm–1
(A) 8.4 S cm2 mol–1 (B) 84 S cm2 mol–1 (C) 4.2 S cm2 mol–1 (D) 42 S cm2 mol–1
9.
0
For the fuel cell reaction 2H2(g) + O2(g)  2H2O() ; f H 298
(H2O, ) = –285.5 kJ/mol
What is S0298 for the given fuel cell reaction ?
Given: O2(g) + 4H+(aq) + 4e–  2H2O()
(A) – 0.322 J/K
10.
(B) – 0.635 kJ/K
E° = 1.23 V
(D) – 0.322 kJ/K
(C) 3.51 kJ/K
Consider the following Galvanic cell.
Voltmeter
H2(g)
Cl2 (g)
+
–
K NO3
Pt(s)
HCl
Anode
HCl
Cathode
By what value the cell voltage change when concentration of ions in anodic and cathodic
compartments both increased by factor of 10 at 298 K
(A) +0.0590
(B) –0.0590
(C) –0.1180
(D) 0
Multiple correct:
During discharging of lead storage battery, which of the following is/are true ?
11.
(A) H2SO4 is produced
(B) H2O is consumed
(C) PbSO4 is formed at both electrodes
electr
(D) Density of electrolytic solution decreases
12.
Which of the following arrangement will produce oxygen at anode during electrolysis ?
(A) Dilute H2SO4 solution with Cu electrodes.
(B) Dilute H2SO4 solution with inert electrodes.
(C) Fused NaOH
H with inert electrodes.
(D) Dilute NaCl solution with inert electrodes.
13.
If 270.0 g of water is electrolysed during an experiment performed by miss Abhilasha with
75% current efficiency then
(A) 168 L of O2 (g) will be evolved at anode at 1 atm & 273 K
(B) Total 504 L gases will be produced at 1 atm & 273 K.
(C) 336 L of H2 (g) will be evolved at anode at 1 atm & 273 K
(D) 45 F electricity will be consumed
C.O.: NAIVEDHYAM, Plot No. SP-11,
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14.
Pick out the correct statements among the following from inspection of standard reduction
redu
potentials (Assume standard state conditions).
E 0Cl /Cl = +1.36 volt
Cl2 (a) + 2e 
2Cl– (a)
2
Br2 (a) + 2e

2Br (a)
E
I2(s) + 2e

2I– (a)
E 0I /I = +0.54 volt
S2 O
2
8
(a) + 2e 
(A) Cl2 can oxidise SO
2
4
–
(a)
0
Br2 / Br
= +1.09 volt
2
0
ES O2 /SO2  = +2.00 volt
2 8
4
from solution
–
(B) Cl2 can oxidise Br and I– from aqueous solution
(C) S2O2–8 can oxidise Cl–, Br– and I– from aqueous solution
(D) S2O82– is added slowly, Br– can be reduce in presence of Cl–
15.
The EMF of the following cell is 0.22 volt.
Ag(s) | AgCl(s) | KCl (1M) | H+(1M) | H2(g) (1atm) ; Pt(s).
Which of the following will decrease the EMF of cell.
(A) increasing pressure of H2(g) from 1 atm to 2 atm
(B) increasing Cl– concentration in Anodic compartment
(C) increasing H+ concentration in cathodic compartment
(D) decreasing KCl concentration in Anodic compartment.
Assertion & Reasoning type questions :
16.
Statement –1
1 : The voltage of mercury cell remains constant for long period of time.
ive species.
Statement –2
2 : It is because net cell reaction does not involve act
active
2 is correct explanation for
(A) Statement–1
1 is true, statement
statement–2 is true and statement–2
statement–1.
2 is NOT the correct explanation for
(B) Statement–1
1 is true, statement
statement–2 is true and statement–2
statement–1.
(C) Statement–1 is true, statement–2
statement is false.
(D) Statement–1
1 is false, statement
statement–2 is true.
17.
0.8 volt respectively,
Statement–1
1 : The SRP of three metallic ions A+,B2+, C3+ are –0.3, –0.5,
–
so oxidising power of ions is C3+ > A+ > B2+.
Statement–2
2 : Higher the SRP, higher the oxidising power.
2 is correct explanation for
(A) Statement–1
1 is true, statement
statement–2 is true and statement–2
statement–1.
2 is NOT the correct explanation for
(B) Statement–1
1 is true, statement
statement–2 is true and statement–2
statement–1.
(C) Statement–1 is true,
e, statement
statement–2 is false.
(D) Statement–1
1 is false, statement
statement–2 is true.
C.O.: NAIVEDHYAM, Plot No. SP-11,
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18.
Statement–1
1 : We can add the electrode potential in order to get electrode potential of net
reaction.
Statement–2
2 : Electrode potential is an intensive property.
2 is correct explanation for
(A) Statement–1
1 is true, statement
statement–2 is true and statement–2
statement–1.
(B) Statement–1
1 is true, statement
statement–2 is true and statement–2
2 is NOT the correct explanation for
statement–1.
(C) Statement–1
1 is true, statement–2
statement is false.
(D) Statement–1
1 is false, statement
statement–2 is true.
Match the column:
19.
Column I
(A)
(B)
(C)
(D)
Dilute solution of HCl
Dilute solution of NaCl
Concentrated solution of NaCl
AgNO3 solution
Column II
(Electrolysis product using inert electrode)
(P)
O2 evolved at anode
(Q)
H2 evolved at cathode
(R)
Cl2 evolved at anode
Ag deposition at cathode
(S)
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EXERCISE # (JEE–MAINS)
1.
Given that EO2 /H2O = 1.23 V;
[JEE Mains 2019 (Apr.)]
E S O 2 /SO 2 = 2.05 V
2
E
8

Br2 / Br 
4
= + 1.09 V
EAu3 /Au = + 1.4 V
The strongest oxidizing agent is :
(1) O2
(2) S2 O82
2.
(3) Br2
(4) Au3+
Calculate the standard cell potential (in V) of the cell in which following reaction takes place :
Fe2+ (aq) + Ag+ (aq)  Fe3+ (aq) + Ag (s)
[JEE Mains 2019 (Apr.)]
Given that
E ºAg  /Ag = x V
E ºFe2 /Fe = y V
E ºFe3 /Fe = z V
(1) x – z
(2) x + 2y – 3z
(3) x – y
(4) x + y – z
3.
The standard Gibbs energy for the given cell reaction in kj mol–1 at 298 K is :
Zn(s) + Cu2+(aq)  Zn2+ (aq) + Cu(s), E°=2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
[JEE Mains 2019 (Apr.)]
(1) –384
(2) 384
(3) –192
(4) 192
4.
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday
electricity. How many mole of Ni will be deposited at the cathode? [JEE Mains 2019 (Apr.)]
(1) 0.20
(2) 0.15
(3) 0.10
(4) 0.05
5.
Consider the statements S1 and S2 :
S1:
Conductivity always increases with decrease in the concentration of electrolyte.
S2:
Molar conductivity always increases with decrease in the concentration of electrolyte.
The correct option among the following is:
[JEE Mains 2019 (Apr.)]
(1) Both S1 and S2 are wrong
(2) Both S1 and S2 are correct
(3) S1 is wrong and S2 is correct
(4) S1 is correct and S2 is wrong
6.
Which one of the following graphs between molar
mola conductivity (  m) versus C is correct ?
[JEE Mains 2019 (Apr.)]
(1)
(2)
(3)
(4)
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7.
Given:
[JEE Mains 2019 (Apr.)]
3+
–
2+
Co + e  CO ; Eº = + 1.81 V
Pb4+ + 2e–  Pb2+ ; Eº = + 1.67 V
Ce+4 + 3e  Ce3+ ; Eº = + 1.61 V
Bi3+ + 3e–  Bi ; Eº = + 0.20 V
Oxidizing power of the species will increase in the order:
(1) Ce4+ < Pb4+ < Bi3+ < Co3+
(2) Co3+ < Pb4+ < Ce4+ < Bi3+
(3) Bi3+ < Ce4+ < Pb4+ < Co3+
(4) Co3+ < Ce4+ < Bi3+ < Pb4+
8.
The decreasing order of electrical conductivity of the following aqueous solutions is:
0.1 M Formic acid (A),
[JEE Mains 2019 (Apr.)]
0.1 M Acetic acid (B),
0.1 M Benzoic acid (C).
(1) A > B > C
(2) C > A > B
(3) A > C > B
(4) C > B > A
9.
The anodic half-cell
cell of lead-acid
lead acid battery is recharged using electricity of 0.05 Faraday. The
amount of PbSO4 electrolyzed in g during the process i : (Molar mass of PbSO4 = 303 g mol–1)
[JEE Mains 2019 (Jan.)]
(1) 22.8
10.
(2) 15.2
(3) 7.6
(4) 11.4
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the
reaction
[JEE Mains 2019 (Jan.)]
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)at 300 K is approximately.
(R = 8 JK–1 mol–1, F = 96000 C mol–1)
(1) e–80
11.
(2) e–160
(3) e160
(4) e320
[JEE Mains 2019 (Jan.)]
Consider the following reduction processes :
Zn2+ + 2e–  Zn(s); Eº = –0.76 V
Ca2+ + 2e–  Ca(s); Eº = –2.87 V
Mg2+ + 2e–  Mg(s); Eº = –2.36 V
Ni2+ + 2e–  Ni(s); Eº = –0.25
–
V
The reducing power of the metals increases in the order :
(1) Ca < Mg < Zn < Ni
(2) Zn < Mg < Ni < Ca
(3) Ni < Zn < Mg < Ca
(4) Ca < Zn < Mg < Ni
C.O.: NAIVEDHYAM, Plot No. SP-11,
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12.
In the cell Pt(s)|H2(g, 1bar)|HCl(aq)|AgCl(s)|Ag(s)|Pt(s) the cell potential is 0.92 V when a 10–6
molal HCl solution is used. The standard electrode potential of (AgCl/Ag,Cl–) electrode is :
2.303RT


 0.06 V at 298
298K
K
Given,
F


(1) 0.94 V
(2) 0.40 V
13.
[JEE Mains 2019 (Jan.)]
(3) 0.20 V
(4) 0.76 V
For the cell Zn(s) | Zn2+(aq) || Mx+ (aq) | M(s), different half cells and their standard electrode
[JEE Mains 2019 (Jan.)]
potentials are given below :
If EOZn 2 / Zn = – 0.76 V, which cathode will give a maximum value of E Ocell per electron
transferred?
(1) Fe2+ / Fe
14.
(2) Ag3+ / Au
(3) Fe3+ / Fe2+
(4) Ag+ / Ag
[JEE Mains 2019 (Jan.)]
Given the equilibrium constant :
KC of the reaction :
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) is
10 × 1015, calculate the E 0cell of this reaction at 298 K
[2.303
RT
at 298K is 0.059V]
F
(1) 0.4736 V
15.
(2) 0.04736 mV
(3) 0.4736 mV
(4) 0.04736 V
 dE  
The standard electrode potential E  and its temperature coefficient 
 for a cell are 2V
 dT 
and –5 × 10–4 VK–1 at 300 K respectively. The cell reaction is Zn(s) + Cu2+ (aq)  Zn2+ (aq) +
Cu(s) The standard reaction enthalpy   r H   at 300 K in kJ mol–1 is,
[Use R = 8 JK–1 mol–1 and F = 96,000 C mool–1]
(1) 192.0
(2) –384.0
(3) 206.4
16.
[JEE Mains 2019 (Jan.)]
(4) –412.8
om for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2mol–1, respectively. If the
conductivity of 0.001 M HA is 5×10–5 S cm–1, degree of dissociation of HA is:
[JEE Mains 2019 (Jan.)]
(1) 0.50
(2) 0.125
(3) 0.25
(4) 0.75
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17.
How long ( approximate) should water be electrolysed by passing through 100 amperes current
so that the oxygen released can completely burn 27.66 g of diborane ?
[JEE Mains 2018]
(Atomic weight of B = 10.8 u)
(1) 1.6 hours
(2)) 6.4 hours
(3) 0.8 hours
(4)) 3.2 hours
18.
Given
E 0Cl2 /C1–  1.36 V, E 0Cl3 /Cr  –0.74 V
E 0Cl2 /C12– /Cr
 1.33V, E 0MnO – /Mn 2  1.51V
/ Cr3
7
4
Among the following, the strongest reducing agent is
(1) Cr
(2)) Mn2+
(3) Cr3+
[JEE Mains 2017]
(4)) Cl
–
19.
Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday
electricity through on aqueous solution of potassium succinate, the total volume of gases (at
MAINS (ONLINE)
2016]
both cathode and anode) at STP (1 atm and 273 K) is :
[JEE-MAINS
(ON
(1) 8.96 L
(2) 2.24 L
(3) 4.48 L
(4) 6.72 L
20.
What will occur if a block of copper metal is dropped into a beaker containing a solution of 1M
MAINS (ONLINE) 2016]
ZnSO4
[JEE-MAINS
(1) The copper metal will dissolve and zinc metal will
wi be deposited
(2) No reaction will occur
(3) The copper metal will dissolve with evolution of oxygen gas
(4) The copper metal will dissolve with evolution of hydrogen gas
21.
Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper
[JEE-MAINS 2015]
deposited at the cathode is : (at. mass of Cu = 63.5 amu)
(1)2g
(2) 127 g
(3) 0 g
(4) 63.5 g
22.
A variable, opposite external potential (Eext) is applied to the cell
Zn|Zn2+ (1 M) || Cu2+ (1 M) | Cu, of potential 1.1 V. When Eext < 1.1 V and Eext > 1.1 V,
MAINS (ONLINE) 2015]
respectively electrons flow from :
[JEE-MAINS
(1) anode to cathode in both cases
(2) anode to cathode and cathode to anode
thode to anode and anode to cathode
(3) cathode to anode in both cases
(4) cathode
23.
At 298 K, the standard reduction potentials are 1.51 V for MnO–4|Mn2+, 1.36 V for Cl2|Cl–,
1.07 V for Br2|Br–, and 0.54 V for I2|I– .At pH = 3, permanganate is expected to oxidize
 RT

MAINS (ONLINE) 2015]
[JEE-MAINS
 0.059V  :

 F

(1)Cl– and Br–
(2) Cl–, Br– and I–
(3) Br– and I–
(4) I– only
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24.
Resistance of 0.2 M solution of an electrolyte is 50 .. The specific conductance of the solution
is 1.4 S m–1. The resistance of 0.5 M solution of the same electrolyte
electrolyte is 280 . The molar
2
–1
[JEE-MAINS 2014]
conductivity of 0.5 M solution of the electrolyte in S m mol is :
3
2
–4
(1) 5 × 10
(2) 5 × 10
(3) 5 × 10
(4) 5 × 10–3
25.
The equivalent conductance of NaCl at concentration C and at infinite dilution are C and ,
respectively. The correct relationship between C and  is given as : [JEE-MAINS 2014]
(where the constant B is positive)
(1) C =  – (2)
26.
C (2) C =  + (2)
C (3) C =  + (2) C (4) C =  – (2) (3)
Given :
E0Cr3 /Cr = 0.74 V ; E 0MnO /Mn
= 1.51 V
/ Mn 2 
4
E Cr O2  /Cr3 = 1.33 V ; E 0Cl/Cl = 1.36 V
0
2 7
Based on the data given above, strongest oxidising agent will be :
(1) Cl–
(2) Cr3+
(3) Mn2+
(4) MnO4–
27.
The standard reduction potentials for Zn2+ / Zn, Ni2+ / Ni and Fe2+ / Fe are – 0.76, – 0.23 and –
when[AIEEE 2012]
0.44 V respectively. The reaction X + Y+2  X2+ + Y will be spontaneous when
(1) X = Zn, Y = Ni (2) X = Ni, Y = Fe
(3) X = Ni, Y = Zn (4) X = Fe, Y = Zn
28.
[AIEEE 2011]
The reduction potential of hydrogen half
half-cell will be negative if :–
+
+
(1) p(H2) = 2 atm [H ] = 1.0 M
(2) p(H2) = 2 atm and [H ] = 2.0 M
+
(3) p(H2) = 1 atm and [H ] = 2.0 M
(4) p(H2) = 1 atm and [H+] = 1.0 M
29.
Resistance of 0.2 M solution of an electrolyte is 50 .. The specific conductance of the solution
is 1.3 S m–1. If resistance of the 0.4M solution of the same electrolyte is 260 , its molar
[AIEEE 2011]
conductivity is :–
–1
2
–1
–4
2
–
(1) 6250 S m mol
(2) 6.25 × 10 S m mol
–4
2
–1
1
(3) 625 × 10 S m mol
(4) 62.5 S m2 mol–1
30.
The Gibbs energy for the decomposition of Al2O3 at 500 °C is as follows : [AIEEE 2010]
2
4
Al2O3  Al + O2, rG = +966 kJ mol–1
3
3
The potential difference needed for electrolytic reduction of Al2O3 at 500 °C is at least ::
(4) 2.5 V
(1) 5.0 V
(2) 4.5 V
(3) 3.0 V
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31
Given :
0
[AIEEE-2009]
0
E Fe3 /Fe = –0.036V, E Fe2 / Fe = –0.439V. The value of standard electrode potential for the
change.
Fe+3(aq) + e–  Fe+2(aq) will be :–
(1) 0.770 V
(2) –0.27 V
32.
(4) 0.385 V
(3) –0.072 V
Given : E 0Cr3 /Cr = –0.72
0.72 V, E 0Fe2  /Fe = –0.42 V. The potential for the cell, [AIEEE 2008]
Cr(s) | Cr3+(aq) (0.1 M) || Fe2+(aq) (0.01 M)|Fe(s) is
(1) 0.26 V
(2) 0.339 V
(3) – 0.339 V
33.
(4) – 0.26 V
The cell Zn | Zn+2(aq)(1M) || Cu+2(aq)(1M) | Cu (E°cell = 1.10 V) was allowed to be completely
  Zn 2  
discharged at 298 K. The relative concentration of Zn to Cu ,   2   is :
  Cu  
[AIEEE 2007]
4
(1) 9.65 × 10
(2) Antilog (24.08) (3) 37.3
(4) 1037.3
+2
34.
+2
Given the data at 25°C,
Ag(s) + I–(aq)  AgI(s) + e–, E° = 0.152V
Ag(s)  Ag+(aq) + e–, E° = – 0.800 V
What is the value of log Ksp for AgI ?
(Where Ksp= solubility product)
RT


 0.59V 
 2.303
F


(1) –8.12
(2) +8.612
[AIEEE 2006]
(3) –37.83
(4) –16.13
35.
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is
100.. The conductivity of this solution is 1.29 Sm–1. Resistance of the same cell when filled
with 0.02M of the same solution is 520
520.. The molar conductivity of 0.02M solution of the
[AIEEE 2006]
electrolyte will be.
–4
2
–1
–4
2
–
–1
(1) 124 × 10 Sm mol
(2) 1240 × 10 Sm mol
4
2
–1
(3) 1.24 × 10 Sm mol
(4) 12.4 × 10–4 Sm2 mol–11
36.
The molar conductivities,  0NaOAC and  0HCl at infinite dilution in water at 25°C are 91.0 and
426.2 S cm2/mol respectively. To calculate  0HOAc the additional value required is :
(1) KCl
(2) NaOH
(3) NaCl
[AIEEE 2006]
(4) H2O
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37.
For a spontaneous reaction the G, equilibrium constant (K) and E 0Cell will be respectively
(1) –ve, < 1, –ve
(2) –ve, >1, –ve
(3) –ve, >1, +ve
[AIEEE 2005]
(4) +ve, >1, –ve
38.
Aluminium oxide may be electrolysed at 1000°C to furnish aluminium metal ( 1 Faraday =
[AIEEE 2005]
96500 Coulombs).The
The cathode reaction is
3+
–
Al + 3e  Al
To prepare 5.12 kg of aluminium metal by this method would require,
(1) 5.49 × 104 C of electric charge
(2) 5.49 × 101 C of electric charge
(3) 5.49 × 107 C of electric charge
(4) 1.83 × 107 C of electric charge
39.
The highest electrical conducitivity of the following aqueous solution is of [AIEEE 2005]
(1) 0.1 M fluoroacetic acid
(2) 0.1 M difluoroacetic acid
(3) 0.1 M acetic acid
(4) 0.1 M chloroacetic acid
40.
Electrolyte
KCl
KNO3
HCl
NaOAC
NaCl
Calculate  HOAc using appropriate molar
infinite dilution in H2O at 25°C
(1) 390.7
(2) 217.5
41.
 (S cm2 mol–1)
149.9
145.0
426.2
91.0
126.5
conductances of the electrolytes listed above at
(3) 517.2
Consider the following E0 values,
0
E Fe3 /Fe2 = +0.77V
[AIEEE 2005]
(4) 552.7
[AIEEE 2004]
0
ESn 2 /Sn = –0.14 V
Under standard conditions the potential for the reaction,
Sn(s) + 2Fe3+ (aq) 2Fe
2Fe2+(aq) + Sn2+(aq) is:
(1) 0.91V
(2) 1.40V
(3) 1.68V
(4) 0.63V
42.
The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at 25°C. The
equilibrium constant of the reaction is (F = 96,
96,500C mol–1 ; R = 8.314 JK–1 mol–1)
[AIEEE 2004]
10
5
1
(1) 1.0 × 10
(2) 1.0 × 10
(3) 1.0 × 10
(4) 1.0 × 1030
43.
The limiting molar conductivities 0 for NaCl, KBr and KCl are 126,152 and 150Scm2 mol–1
[AIEEE 2004]
respectively. The m0 for NaBr is :
2
–1
2
–1
2
–1
(1) 278 S cm mol
(2) 176 S cm mol
(3) 128 S cm mol
(4) 302 S cm2 mol–1
C.O.: NAIVEDHYAM, Plot No. SP-11,
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44.
In a cell that utilises the reaction,
[AIEEE 2004]
+
2+
Zn(s) + 2H (a) Zn (aq) + H2(g) addition of H2SO4 to cathode compartment, will :
(1) increase the Ecel l and shift equilibrium to the right
(2) lower the Ecell and shift equilibrium to the right
(3) lower the Ecell and shift equilibrium to the left
(4) increase the Ecell and shift equilibrium to the left
45.
0.41, + 1.57, +0.77 and + 1.97V respectively.
The E 0M3 /M2 values for Cr, Mn, Fe and Co are –0.41,
For which one of these metals the change in oxidation state from +2 to +3 is easiest?
[AIEEE 2004]
(1) Fe
(2) Mn
(3) Cr
(4) Co
46.
electron change, the standard e.m.f. of the cell is found to be
For a cell reaction involving a tw
two-electron
0.295V at 25° C. The equilibrium constant of the reaction at 25°C will be : [AIEEE 2003]
(1) 10
(2) 1 × 1010
(3) 1 × 10–10
(4) 29.5 × 10–2
47.
Standard reduction electrode potentials
potentials of three metals A, B and C are respectively + 0.5V, –
[AIEEE 2003]
3.0V and –1.2
1.2 V. The reducing powers of these metals are :
(1) C > B > A
(2) A > C > B
(3) B > C > A
(4) A > B > C
48.
For the redox reaction :
Zn(s) + Cu2+ (0.1M)  Zn
Z 2+ (1M) + Cu(s) taking place in a cell,
RT


E°Cell is 1.10 volt. ECell for the cell will be  2.303
 0.0591
F


(1) 1.07 volt
(2) 0.82 volt
(3) 2.14 volt
49.
[AIEEE 2003]
(4) 1.80 volt
When, during electrolysis of a solution of AgNO3 9650 coulombs of charge pass through the
electroplating bath, the mass of silver deposited on the cathode will be : [AIEEE 2003]
(1) 21.6g
(2) 108g
(3) 1.08g
(4) 10.8g
C.O.: NAIVEDHYAM, Plot No. SP-11,
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EXERCISE # (JEE-ADVANCED)
1.
For the electrochemical cell,
Mg(s)|Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M)|Cu(s)
the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x
M, the cell potential changes to 2.67 V at 300 K. The value of x is ____.
F
(given,
= 11500 KV–1) , where F is the Faraday constant and R is the gas constant,
R
ln(10) = 2.30)
[JEE Adv. 2018]
2.
Consider an electrochemical cell: A(s) | An+ (aq, 2 M) || B2n+ (aq, 1 M)| B(s). The value of ΔHθ
for the cell reaction is twice that of ΔGθ at 300 K. If the emf of the cell is zero, the ΔSθ (in J
K–1 mol–1) of the cell reaction per mole of B formed at 300 K is ____.
(Given: ln(2)
(2) = 0.7, R(universal gas constant) = 8.3 J K–1 mol–1. H, S and G are enthalpy,
[JEE Adv. 2018]
entropy and Gibbs energy, respectively.)R
3.
The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by
using a conductivity cell consisting of plantinized Pt electrodes. The diatcane between the
elecrodes is 120 cm with an area of cross section of 1cm2 . The conductance of this solution
was found to be 5 × 10–7 S. The pH of the solution is 4. The value of limiting molar
conductivity   0m  of this weak monobasic acid in aqueous solution is Z ×102 S cm–1mol–1.
[JEE Adv. 2017]
The value of Z is
4.
For the following cell,
Zn(s) ZnSO 4  aq  CuSO 4  aq  Cu (s)
when the concentration of Zn2+ is 10 times the concentration of Cu2+ , then expression for G
(in J mol–1) is [F is faraday constant ; R is gas constant; T is temperature; Eº (cell ) = 1.1V]
[JEE Adv. 2017]
(A) 2.303RT – 2.2 F
(B) – 2.2 F
(C) 1.1F
(D) 2.303RT + 1.1 F
5.
For the following electrochemical cell at 298K,
Pt(s) | H2(g, 1bar) | H+ (aq, 1M) || M4+(a), M2+(a) | Pt(s)
 M 2  aq. 
Ecell = 0.092 V when
= 10x
 M 4  aq  
RT
Given : E0M4 /M2 = 0.151 V ; 2.303
= 0.059 V
F
The value of x is (A) –2
(B) –1
(C) 1
[JEE-Adv. 2016]
(D) 2
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6.
193 kJ mol–1 is used for the
All the energy released from the reaction X  Y, rGº = –193
oxidizing
M+ and M+  M3+ + 2e¯, Eº = –0.25 V.
Under standard conditions, the number of moles of M+ oxidized when one mole of X is
converted to Y is [F = 96500 C mol–1]
[JEE-Adv. 2015]
7.
The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the
8.
molar conductivity of a solution of a weak acid HY (0.1 M). If 0X––  0Y–, the difference in
their pKa values , pKa(HX) – pKa(HY), is (consider degree of ionization of both acids to be
[JEE-Adv. 2015]
<<1).
[JEE-Adv. 2014]
In a galvanic cell , the salt bridge –
(A) Does not participate chemically in the cell reaction
(B) Stops the diffusion of ions from one electrode to another
(C) Is necessary for the occurence of the cell reaction
(D) Ensures mixing of the two electrolytic solutions
An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List
List-I. The
variation in conductivity
ty of these reactions is given in List-II.
List II. Match List-I
List with List-II and
select the correct answer using the code given below the lists :
[JEE-Adv. 2013]
List-I
List-II
(P)
(C2H5)3N + CH3COOH
(1)
Conductivity decreases and then increases
X
Y
(Q)
KI(0.1M) + AgNO3(0.01M) (2)
Conductivity decreases and then does not change
X
Y
much
(R)
CH3COOH + KOH
(3)
Conductivity increases and then does not change
X
Y
much
ity does not change much and then
(S)
NaOH + HI
(4)
Conductivity
X
Y
increases
Codes :
S
P
Q
R
S
P
Q
R
1
(A)
3
4
2
1
(B)
4
3
2
2
(C)
2
3
4
1
(D)
1
4
3
9.
10.
The standard reduction potential data at 25ºC is given below
Eº (Fe3+, Fe2+) = +0.77 V ;
Eº (Fe2+, Fe) = –0.44 V ;
Eº (Cu2+, Cu) = +0.34 V ;
Eº (Cu+, Cu) = +0.52 V ;
Eº (O2(g) + 4H+ + 4e–  2H2O] = +1.23 V ;
Eº [(O2(g) + 2H2O + 4e–  4OH–)] = +0.40 V ;
Eº (Cr3 +, Cr) = –0.74
0.74 V ;
[JEE-Adv. 2013]
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Eº (Cr2+, Cr) = –0.91 V ;
Match Eº of the redox pair in List
List-I with the values given in List--II and select the correct
answer using the code given below the lists :
List-I
List-II
3+
(P)
Eº(Fe , Fe)
(1)
– 0.18 V
(Q)
Eº(4H2O  4H+ + 4OH+)
(2)
– 0.4 V
(R)
Eº(Cu2+ + Cu  2Cu+)
(3)
– 0.04 V
3+
2+
(S)
Eº(Cr , Cr )
(4)
– 0.83 V
Codes :
S
P
Q
R
S
P
Q
R
1
(A)
4
1
2
3
(B)
2
3
4
(C)
1
2
3
4
(D)
3
4
1
2
Paragraph for Question 11 and 12
The electrochemical cell shown below is a concentration cell.
M | M2+ (saturated solution of a sparingly soluble salt, MX2) | | M2+ (0.001 mol dm–3) | M The
emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes.
[JEE 2012]
The emf of the cell at 298 K is 0.059V.
11.
–
The value of G (kJ mol–1
) for the given cell is (take If = 96500 C mol–1)
(A) –5.7
(B) 5.7
(C) 11.4
(D) –11.4.
12.
The solubility product (Ksp ; mol3 dm–9) of MX2 at 298 K based on the information available
for the given concentration cell is (take 2.303 × R × 298/F = 0.059 V)
(A) 1 × 10–15
(B) 4 × 10–15
(C) 1 × 10–12
(D) 1 × 10–12
13.
AgNO3 (a) was added to an aqueous KCl solution gradually and the conductivity of the
solution was measured. the plot of conductance (()) versus the volume of AgNO3 is
[JEE 2011]
Volume
(P)
(A) (P)
14.
(B) (Q)
(C) (R)
2+
 2Fe
(a)+
2H2O()
(D) (S)
[JEE 2011]
Consider the following cell reaction :
2Fe(s)+O2(g)+4H+(aq)
Volume
(S)
Volume
(R)
Volume
(Q)
E° = 1.67 V
At[Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25°C is –
(A) 1.47 V
(B) 1.77 V
(C) 1.87 V
(D) 1.57 V
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15.
16.
Paragraph for Questions 15 to 16
The concentration of potassium ions inside a biological cell is at least twenty times
time higher than
the outside. The resulting potential difference across the cell is important in several processes
such as transmission of nerve impulses and maintaining the ion balance. A simple model for
[JEE 2010]
such a concentration cell involving a metal M is :
+
+
M(s)| M (aq ; 0.05 molar) || M (aq ; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential |Ecell| = 70 mV.
For the above cell :–
G0 > 0 (D) Ecell > 0 ; G0 < 0
(A) Ecell < 0 ; G > 0 (B) Ecell > 0 ; G < 0 (C) Ecell < 0 ; G0
+
If the 0.05 molar solution of M is replaced by a 0.0025 molar M+ solution, then the magnitude
of the cell potential would be ::–
(A) 35 mV
(B) 70 mV
(C) 140 mV
(D) 700 mV
17.
For the reaction of NO3¯ ion in an aqueous solution,
solution, E° is +0.96 V. Values of E° for some
metal ions are given below
V2+ (aq) + 2e¯  V
E° = – 1.19 V
3+
Fe (aq) + 3e¯  Fe
E° = – 0.04 V
3+
Au (aq) + 3e¯  Au
E° = + 1.40 V
2+
[JEE 2009]
Hg (aq) + 2e¯  Hg
E° = + 0.86 V
The pair(s) of metal
etal that is(are) oxidised by NO3¯ in aqueous solution is(are)
(A) V and Hg
(B) Hg and Fe
(C) Fe and Au
(D) Fe and V
18.
Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere
current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C
[JEE 2008]
mol–1)
4
4
4
(A) 9.65 × 10 sec
(B) 19.3 × 10 sec
(C) 28.95 × 10 sec (D) 38.6 × 104 sec
Paragraph for Question Nos. 19
1 & 20 (2 questions)
Redox reactions play a pivoted role in chemistry and biology. The values of standard redox
potential (E°) of two half-cell
half cell reactions decide which way the reaction is expected to proceed.
A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited.
Given below are
re a set of half-cell
half cell reactions (acidic medium) along with their E° (V with
respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to
Questions 14-16.
I2 + 2e–  2I–
E° = 0.54
–
Cl2 + 2e  2Cl–
E° = 1.36
3+
–
2+
Mn + e  Mn
E° = 1.50
3+
–
2+
Fe + e  Fe
E° = 0.77
+
–
[JEE 2007]
O2 + 4H + 4e  2H2O
E° = 1.23
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19.
Among the following, identify the correct statement.
(A) Chloride ion is oxidised by O2
(B) Fe2+ is oxidised by iodine
(C) Iodine ion is oxidised by chlorine
(D) Mn2+ is oxidised by chlorine
20.
While Fe3+ is stable, Mn3+ is not stable in acid solution because
(A) O2 oxidises Mn2+ to Mn3+
(B) O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+
(C) Fe3+ oxidises H2O to O2
(D) Mn3+ oxidises H2O to O2
21.
22.
23.
(3 questions)
Paragraph for Question Nos. 221 to 23(3
Chemical reactions involve interaction of atoms and molecules. A large number of
atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical
arying with their atomic/molecular masses. To handle such large numbers
compound varying
conveniently, the mole concept was introduced. This concept has implications in diverse areas
such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following
follo
example illustrates a typical case, involving chemical/ electrochemical reaction, which requires
a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared
and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of
the electrodes (atomic mass : Na = 23,Hg = 200; 1 Faraday = 96500 coulombs)
[JEE 2007]
The total number of moles of chlorine gas evolved is
(A) 0.5
(B) 1.0
(C) 2.0
(D) 3.0
If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this
solution is
(A) 200
(B) 225
(C) 400
(D) 446
The total charge (coulombs) required for complete electrolysis is
(A) 24125
(B) 48250
(C) 96500
(D) 193000
Question No. 24 to 26(3 questions)
Tollen’s reagent is used for the detection of aldehyde when a solution of AgNO3 is added to
glucose with NH4OH then gluconic acid is formed
Ag+ + e–  Ag ; E0red = 0.8 V
C6H12O6 + H2O C6H12O7 (Gluconic acid) + 2H+ + 2e– ; E0red = – 0.05 V
Ag(NH3)2+ e–  Ag(s) + 2NH3 ; E0 = – 0.337 V
[Use 2.303 ×
RT
F
= 0.0592 and
= 38.92 at 298 K]
F
RT
[JEE 2006]
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24.
2Ag+ + C6H12O6 + H2O  2Ag(s) + C6H12O7 + 2H+
Find ln K of this reaction
(A) 66.13
(B) 58.38
(C) 28.30
(D) 46.29
25.
When ammonia is added to the solution, pH is raised to 11. Which half-cell
reaction is affected
half
by pH and by how much?
(A) Eoxd will increase by a factor of 0.65 from E0oxd
(B) Eoxd will decrease
ease by a factor of 0.65 from E0oxd
(C) Ered will increase by a factor of 0.65 from E0red
(D) Ered will decrease by a factor of 0.65 from E0red
26.
Ammonia is always is added in this reaction. Which of the following must be incorrect?
(A) NH3 combiness with Ag+ to form a complex.
(B) Ag(NH3)+2 is a weaker oxidising reagent than Ag+.
(C) In absence of NH3 silver salt of gluconic acid is formed.
(D) NH3 has affected the standard reduction potential of glucose/gluconic acid electrode.
27.
14
. If 10–7 mole of AgNO3
We have taken a saturated solution of AgBr.Ksp of AgBr is 12 × 10–14
are added to 1 litre of this solution find conductivity (specific conductance) of this solution in
[JEE 2006]
terms of 10–7 S m–1 mol–11.
1
–3
3
2
–1
–3
2
–1
0
0
0
Given :   = 6 × 10 S m mol ;   = 8 × 10 S m mol ;   = 7 × 10–3 S m2
 Ag 
 Br 
 NO3 
mol–1
28.
[JEE 2005]
The half cell reactions for rusting of iron are:
1
2H+ + O2 + 2e–  H2O ; E0 = + 1.23 V, Fe2+ + 2e–  Fe ; E0 = –0.44 V
2
0
G (in kJ) for the reaction is :
(A) –76
(B) –322
(C) –122
(D) –176
29.
(a) Calculate G 0f of the following reaction
Ag+ (aq) + Cl– (aq) AgCl(s)
Given : G 0f (AgCl) = –109
109 kJ/mole, G 0f (Cl–) = –129 kJ/mole, G 0f (Ag+) = 77 kJ/mole
Represent the above reaction in form of a cell
Calculate E0 of the cell. Find log10KSP of AgCl
(B) (b) 6.539 × 10–2 g of metallic Zn (amu = 65.39) was added to 100 ml of saturated solution
 Zn 2 
of AgCl. Calculate 
log10 , given that
2
 Ag  
Ag+ + e–  Ag
E0 = 0.80 V ;
Zn2+ + 2e–  Zn E0 = –0.76V
[JEE 2005]
Also find how many moles of Ag will be formed ?
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30.
Zn | Zn2+ (a = 0.1M) || Fe2+ (a = 0.01M)|Fe. The emf of the above cell is 0.2905 V. Equilibrium
constant for the cell reaction is
[JEE 2004]
0.32/0.0591
0.32/0.0295
0.26/0.0295
0.32/0.295
(A) 10
(B) 10
(C) 10
(D) e
31.
Find the equilibrium constant at 298 K for the reaction,
Cu2+(aq) + In2+(aq)  Cu+(aq) + In3+(aq)
Given that E 0Cu 2 |Cu  = 0.15V, E 0In3 |In  = –0.42V, E 0In 2  |In  = –0.40 V
[JEE 2004]
32.
In the electrolytic cell, flow of electrons is from:
(A) Cathode to anode in solution
(B) Cathode to anode through external supply
(C) Cathode to anode through internal supply
(D) Anode to cathode through internal supply.
33.
Two students use same stock solution of ZnSO4 and a solution of CuSO4. The e.m.f of one cell
is 0.03 V higher than the other. The conc. of CuSO4 in the cell with higher e.m.f value is 0.5 M.
 2.303 RT

Find out the conc. of CuSO4 in the other cell 
 0.06  .
F


34.
[JEE 2003]
Standard electrode potential data are useful for understanding the suitablilty of an oxidant in a
redox titration. Some half cell reactions and their standard potentials are given below:
MnO–4 (aq) + 8H+(aq) + 5e–  Mn2+ (aq) + 4H2O (l); E° = 1.51 V
[JEE 2002]
); E° = 1.38 V
Cr2 O72 (aq) + 14 H+ (aq) + 6e–  2Cr3+ (aq) +7H2O ();
Fe3+ (aq) + e–  Fe2+ (aq); E° = 0.77 V
Cl2 (g) + 2e–  2Cl– (aq); E° = 1.40 V
Identify the only incorrect statement regarding quantitative estimation of aqueous Fe(NO3)2
35.
(A) MnO 4 can be used in aqueous HCl
(B) Cr2O 72 can be used in aqueous HCl
(C) MnO 4 can be used in aqueous H2SO4
(D) Cr2 O 72 can be used in aqueous H2SO4
[JEE 2001]
The reaction,
–
ClO–3
–
3ClO (aq) 
(aq) + 2Cl (aq)
is an example of
(A) Oxidation reaction
(C) Disproportionation reaction
(B) Reduction reaction
(D) Decomposition reaction
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36.
The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is
(A) LiCl > NaCl > KCl
(B) KCl > NaCl > LiCl [JEE 2001]
(C) NaCl > KCl > LiCl
(D) LiCl > KCl > NaCl
37.
Saturated solution of KNO3 is used to make salt bridge because
(A) velocity of K+ is greater than that of NO–3
(B) velocity of NO–3 is greater than that of K+
(C) velocities of both K+ and NO–3 are nearly the same
(D) KNO3 is highly soluble in water
[JEE 2001]
38.
The standard potential of the following cell is 0.23 V at 15° C & 0.21 V at 35° C
Pt | H2(g) | HCl (aq) | AgCl(s) | Ag(s)
(i) Write the cell reaction.
(ii) Calculate H0 ,S0 for the cell reaction by assuming that these quantities remain unchanged
in the range 15°C to 35°C.
(iii)
Calculate the solubility of AgCl in water at 25°C. Given standard reduction potential of
[JEE 2001]
the Ag+/Ag couple is 0.80 V at 25°C.
39.
For the electrochemical cell, M | M+ || X– | X, E° (M+/M) = 0.44 V and E° (X/X–) = 0.33V.
[JEE 2000]
From this data , one can deduce that
+
–
(A) M + X M + X is the spontaneous reaction
(B) M+ + X–  M + X is the spontaneous reaction
(C) Ecell = 0.77 V
(D) Ecell = –0.77 V
40.
Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper
cathode. A constant current of 2 mA was passed for 16 mintue. It was found that after
electrolysis, the absorbance (concentration) of the solution was reduced to 50% of its original
value. Calculate the concentration of copper sulphate in the solution to begin with.
[JEE 2000]
41.
The following electrochemical cell has been set up
Pt(I) | Fe3+, Fe2+(a =1) || Ce4+ , Ce3+ (a = 1) | Pt(II)
E 0Fe3 / Fe2  = 0.77 V and E 0Ce4  /Ce3 = 1.61 V
If an ammetter is connected between the two platinum electrodes. predict the direction of flow
[JEE 2000]
of current. Will the current increase or decrease with time?
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42.
A gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y and 1 M Z at
[JEE 1999]
25°C. If the reduction potential of Z > Y > X, then
(A) Y will oxidise X and not Z
(B) Y will oxidise Z and X
(C) Y will oxidise both X and Z
(D) Y will reduce both X and Z.
43.
Calculate the equilibrium constant for the reaction, 2Fe3+ + 3I– 2Fe2+ + I3–. The standard
reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for Fe3+ / Fe2+ and
[JEE 1998]
I3– / I– couples.
44.
Find the solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the emf of
the cell Ag|Ag+ (satd.Ag2 CrO4 soln.) || Ag +(0.1 M) | Ag is 0.164 V at 298K. [JEE 1998]
C.O.: NAIVEDHYAM, Plot No. SP-11,
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ANSWER KEY
1.
EXERCISE # (S-I)
(b) Ag
(d) reduction
(f) Ag
(a) Cu
(c) oxidation
(e) Cu
(g) anode-Cu  Cu2+ + 2e– ; cathode-Ag+ + e– Ag
2.
(h) Cu + 2Ag+ Cu2+ + 2Ag
(j) Cu
solution
(i) Cu
(k) to complete circuit and maintain electrical neutrality in
(a) 2Ag + Cu2+  2Ag+ + Cu,
(b) MnO–4+ 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O
4.
(c) 2Cl– + 2Ag+  2Ag + Cl2,
(d) H2 + Cd2+  Cd + 2H+
(a) Zn | Zn2+ | | Cd2+ | Cd,
(b) Pt, H2 | H+ | | Ag+ | Ag ,
2
+
3
(c) Pt | Fe2+, Fe3+ | | Cr2O2–
7 , H , Cr + | Pt
1.61 V
5.
1.35 V
6.
– 0.80 V, NO
7.
0.53 V, disproportionation
0.756 V
0.756
10.
12.
14.
1.68 V
11.
Spontaneous,  48250 J
Mg
13.
Y>Z>X
(i) CuO : Cu is below hydrogen in series, so it can reduce from CuO to Cu.
(ii) Ag2O : Lower in series stability of oxide become lesser.
(iii) Lower S.R.P. metal can displace higher S.R.P. metals ions from solution.
15.
17.
19.
20.
21.
1.14 volt
16.
–1.30 ×103 kJ mol–1
E =1.159V
18.
0.059 volt
0
–
M)||Sn2+(1M)|Sn
E cell = +0.01V, Ecell = –0.0785V,
0.0785V, correct representation is Pb|Pb2+ (10–3
[Cu2+] = 2.97 × 10–12M for E =0
E = –0.81 V
22.
– 0.2214 V
(a) The spontaneous cell reaction : Zn + 2Ag+ (aq) Zn2+ (aq) + 2Ag (s)
3
23.
(b) 1.56 V
8.
– 0.0367 V
9.
(c) [Zn2+] = 4 × 10–4 M
(d) As we add KI to cathode chamber, some Ag+ will precipitate out as :
Ag+ + I– AgI
The above reaction reducing [Ag+] from cathode chamber. This will reduce Ecell according to
24.
Nernst’s equation.
log [Zn2+] / [Cu2+] = 37.22
25.
Kc = 7.6 × 1012
26.
Kc = 2.18 × 1026
27.
Kc = 2.868 × 10107, G0 = – 611.8 kJ
28.
32.
Kw  10–14
pH = 4
30.
E = 0.413 V
29.
E = 0.059
31.
0.0295 V
C.O.: NAIVEDHYAM, Plot No. SP-11,
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0744
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33.
(A) Hg2Cl2(s) + Cu(s) Cu2+(aq) + 2Cl– (aq) + 2Hg(l)
(B) 2Ag (s)  2IO3–  Zn 2  
 2AgIO3( s )  Zn (s)
(s )
(C) Mn (s)  2OH –  Cu 2  
 Mn(OH) 2( s )  Cu (s)
34.
–0.42 V
37.
35.
KSP = 10–5
ELECTROLYTE
1 NaCl (Molten) with Pt electrode
36.
–0.16V
ANODE Product CATHODE Product
Na
Cl2(g)
2 NaCl (aq) with Pt electrode
Cl2(g)
H2(g)
3 Na2SO4 (aq) with Pt electrode
4 NaNO3 (aq) with Pt electrode
O2(g)
H2(g)
O2(g)
5 AgNO3 (aq) with Pt electrode
6 CuSO4 (aq) with Inert electrode
7 CuSO4 (aq) with Copper electrode
O2(g)
H2(g)
Ag
O2(g)
Cu dissolve
Cu
Cu
39.
41.
43.
45.
47.
49.
53.
55.
56.
57.
58.
(a) 6.02 × 1022 electrons lost, (b) 1.89 × 1022 electrons gained, (c) (b) 1.80 × 1023 electrons
gained
40.
(a) 0.75 F, (b) 0.69 F, (c)1.1 F
(i) 54 gm, (ii) 16.35 gm
5
1.023 × 10 sec
42.
1.12 mol, 12.535 litre
44.
Rs. 0.75 x
0.112 litre
46.
2
(i) 2.17 gm ; (ii) 1336. 15 sec
n=4
A = 114, Q = 5926.8C
48.
60 %
50.
51.
2M
1.825g
52.
71.5 amp
2
–1
–1
400 S cm equivalent
54.
0.00040 S cm ; 2500 ohm cm
5
–6
(i) 6.25 × 10 ohm, (ii) 1.6 × 10 amp
0.0125 mho g equiv–1 m2, 1.25 × 10–3 mho cm–1
(i) 250 mho cm2 mol–1 , (ii) 125 mho cm2 equivalent–1
60.
10–10 mole2 /litre2
0.8
59.
9.4 ×10–4 gm/litre
61.
63.
 = 0.5 , k = 10 × 10–4
510 ×10–4 mho cm2 mol–1
38.
1.
3.
5.
7.
9.
11
13.
15.
– 0.46 V
E0 = –0.22 V
[Br–] : [Cl–] = 1 : 200
–0.037 V
K = 10268
42.2 %
0.1456 ampere
0.1934 gm/litre
62.
64.
(i) 400 S cm2 mol–1 (ii) 12 %
(i) 7 (ii) 1 × 10–14
EXERCISE # (S-II)
2.
(ii). 1.27 V, (iii) 245.1 kJ
4.
Ksp = 1.1 × 10–16
6.
– E0 = -0.14903V
8.
1.536 ×10–5 M3
10
43.456g
12.
1.9 × 106 year
14.
4.25×10–2 metre
16.
n=2
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
96
17.
19.
21.
0.52 V, 0.61 V
115800C, 347.4 Kj
Final weight = 9.6g, 0.01 Eq of acid
1.
8.
15.
22.
29.
36.
43.
50.
57.
64.
71.
78.
85.
92 .
99.
106.
1.
8.
15.
1.
8.
15.
22.
29.
36.
43.
A
C
A
D
D
C
A
D
A
D
C
D
B
B
C
A
B
B
A,D
2
3
4
2
2
3
3
2.
9
16.
23.
30.
37.
44.
51.
58.
65.
72.
79.
86.
93.
100.
2.
9.
16.
2.
9.
16.
23.
30.
37.
44.
C
B
A
A
C
C
C
D
A
C
A
A
D
D
B
D
D
A
2
3
2
3
4
3
1
18.
20.
22.
0.0295 V
t = 193 sec
t = 93.65 sec.
EXERCISE # (O-I)
4.
B
5.
11.
D
12.
18.
A
19.
25.
C
26.
32.
B
33.
39.
B
40.
46.
C
47.
53.
D
54.
60.
C
61.
67.
C
68.
74.
C
75.
81.
D
82.
88.
D
89.
95.
B
96.
102. C
103.
3.
10.
17.
24.
31.
38.
45.
52.
59.
66.
73.
80.
87.
94.
101.
D
D
D
C
D
C
C
D
A
B
A
C
C
A
D
3.
10.
17.
EXERCISE # (O-II)
A
4.
B
5.
C
11.
C,D 12.
A
18.
D
19.
3.
10.
17.
24.
31.
38.
45.
D
C
C
C
C
A
B
B
C
B
A
B
A
A
C
6.
13.
20.
27.
34.
41.
48.
55.
62.
69.
76.
83.
90.
97.
104.
C
D
B
B
A
A
D
C
A
D
C
C
B
A
C
7.
14.
21.
28.
35.
42.
49.
56.
63.
70.
77.
84.
91.
98.
105.
A
C
B
B
A
A
A
B
A
D
B
B
D
A
C
A
7.
A
B
6.
A,B 14.
B,C
B,C,D 13.
(A) P, Q (B) P, Q (C) Q, R, (D) P,S
EXERCISE # (JEE-MAINS)
1
4.
4
5.
3
3
11.
3
12.
3
4
18.
1
19.
1
3
25.
1
26.
4
1
32.
1
33.
4
3
39.
2
40.
1
3
46.
2
47.
3
6.
13.
20.
27.
34.
41.
48.
3
2
2
1
4
1
1
7.
14.
21.
28.
35.
42.
49.
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
3
1
4
1
1
1
4
97
1.
8.
15.
22.
28.
10
A,B
B
D
B
EXERCISE # (JEE-ADVANCED)
6
4.
A
5.
D
6.
2. –11.62
3.
9.
A
10.
D
11.
D
12.
B
13.
16.
C
17.
A,B,D 18.
B
19.
C
20.
23.
D
24.
A
25.
A
26.
D
27.
–6
0
(b) 52.8, 10 moles
29.
(a) E = 0.59 V, log10KSP = –10
10
32.
C
33.
0.05
34.
A
35.
4
7.
3
D
14.
D
21.
–1
55 S m
30.
D
B
B
C
B
31.
KC = 10
37.
40.
B
C
38.
H0 = – 49987 Jmol–1, S0 =– 96.5 J mol–1 K–1, s =1.47 ×10–5 M 39.
–5
KC = 6.26 × 107
7.95 × 10 M
41.
decrease with time
42.
A
43.
44.
Ksp = 2.287 × 10–12
36.
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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HINT AND SOLUTIONS
EXERCISE # S–I
1. 1.
(a)
anode
Cathode
Ag –– Ag+ + e– × 2
Cu2+ + 2e– –– Cu
2Ag + Cu2+ –– 2Ag+ + Cu
(b)
anode
Cathode
Fe2+ –– Fe3+ + e– × 5
5e– + 8H+ + MnO4– –– Mn2+ + 4H2O
8H+ + 5Fe2+ + MnO4– –– 5Fe3+ + Mn2+ + 4H2O
(c)
anode
Cathode
2Cl– –– Cl2 + 2e–
Ag+ + e– –– Ag × 2
2Cl– + 2Ag+ ––Cl2 + 2Ag
(d)
anode
Cathode
H2 –– 2H+ + 2e–
Cd2+ + 2e– –– Cd
H2 + Cd2+ ––2H+ + Cd
2.
E°cell = E° Co / Co2+ + E°Ce4+ / Ce3+
E° Co / Co2+ = – E°Co2+ / Co = +0.28 V
1.89 = 0.28 + E°Ce4+ / Ce3+
E° Ce4+ / Ce3+ = +1.61 V
3.
(a)
Zn (s) / Zn2+ (aq) | | Cd2+ (aq) / Cd (s)
(b)
Pt / H2 (g) / H+ (aq) | | Ag+ (aq) / Ag (s)
(c)
Pt / Fe2+ , Fe3+ | | Cr2O72– , Cr3+ , H+ / Pt
5.
–0.15 = 1.20 + ECl– /Cl
2
ECl– /Cl = –1.35 V
2
ECl
6.
2 /Cl

= 1.35 V
Ag 
e– + H+ 
Ag+ + e–
1
H2
2
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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Ag+ H+  Ag+ +
1
H2
2
E°cell = –0.8
0.06 1
log = –0.8 < 0 (reaction will not spontaneous)
1
1
2+
–
+
Cu + e Cu G1 = –1 × F × 0.15
....(1)
Ecell = –0.8 –
7.
Cu +2 + 2e–  Cu G2 = –2 × F × 0.34
–
Cu + e  Cu
+
....(2)
....(3)
(3) =(2) - (1)
G3= G2 – G1
= – 0.68 F + 0.15 F
G3 –nFE0 = –0.53 F
– 1 × F× E0 = – 0.53 F
E0 = 0.53 V > 0
Cu undergoes disproportion


8.
Fe2+ + 2e–  Fe
E0 = 0.44 V
G1 = –2 × F × (–0.44)
–
Fe + e  Fe
3+
2+
E0 = 0.77 V
G2 = –1 × F × 0.77
By adding above tworeaction
Fe3+ + 3e–  Fe
G3 = G1 + G2
–3 × F × E0 = 0.88F – 0.77F
 0.11 
E0 = – 
 V = –0.0367 V
 3 
9.
2e– + 2H+ + SO42– –– SO32– + H2O
1
3
3e– + 3H+ + SO32– –– S2O32– + H2O
2
2
1
5
4e– + 5H+ + SO42– –– S2O32– + H2O
2
2
(iii) = (ii) + (i)
G3o  Go2  G1o
(i)
(ii)
(iii)
4FE3o  2FEo2  2FE1o
E1o  E o2
0.936  0.576
=
= 0.756 V
E =
2
2
o
3
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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100
10.
3e– + Cr3+ –– Cr
OCl– + H2O + 3e––– Cl– + 2OH–
2Cr3+ + 3H2O + 3OCl– –– 2Cr3+ + 3Cl– + 6OH–
(iii) = –2 × (i) + 3 × (ii)
(i)
(ii)
(iii)
6E3o  2  3  E1o  3  2  Eo2
E3o = 0.74 + 0.94 = 1.68 V
11.
Mn2+ –– Mn3+ + e–
(i)
3+
+
–
2H2O + Mn –– MnO2 + 4H + e
(ii)
2Mn2+ + 2H2O –– Mn2+ + MnO2 + 4H+
(iii)
(iii) = (ii) – (i)
E3o  Eo2  E1o
0.5 = –1.0 + 1.5
E° > 0 , So reaction is spontaneous
G° = –nFE° = –1 × 96500 × 0.5 = –48250 J
12.
SRP value  , oxidizing power 
Reducing power 
mg is best reducing agent.
13.
SRP value  , Reducing power 
SRP X > Z > Y
Reducing power Y > Z > X
14.
(i)
CuO , 2u lies lelow hydrogen in electro chemical series.
(ii)
Ag2O –– Ag
E° Ag+ / Ag is high
(iii)
E° Zn2+ / Zn = –0.763 V
SRP value  , Reducing power 
15.
C + O2 –– CO2
(i)
3
2Al + O2 –– Al2O3 (s)
(ii)
2
Al2O3 (s) –– Al2O3 (melt)
(iii)

2 Al2O3 (melt) + 3C –– Al ( ) + 3CO2(g) (iv)
(iv) = 3 × (i) – 2 × (ii) + (iii)
Go4  3G1o  2Go2  G3o
= –3 × 395 + 2 × 1269 + 16= 1369 kJ
G° = –nFE°
1369 × 103 = –12 × 96500 × E°
E° = –1.14 V
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
101
16.
rG° = –nFE° = –12 × 96500 × 2.73
= – 3161340
rG° = 4 G0ƒ [Al(OH)4-] – 6 G0ƒ H2O – 4 G0ƒ OH–
–3161.340 = 4 × x – 6 × (–237.2) – 4 × (–157)
= 4x + 1423.2 + 628
x = –1303.13 kJ
17.
Daniel cell :- Zn (s) ZnSO4 m Cu 2 m2 Cu(s)
1
Zn + Cu+2  Zn+2 + Cu
0.059
log 0 
E = E0 –
n
0.0591 [Zn 2 ]
log
E = E0 –
2
[Cu 2 ]
18.
0.059
001
log
2
01
0.59
= 1.1 –
log102 = 1.1 +0 .059= 1.159 V
2
pH = 1
[H+] = 0.1
2H+ –– H2 + 2e–
0.0591
E=0–
log [H+]2
2
0.0591
log (0.1)2 = 0.0591 V
E=–
2
19.
Sn + Pb+2  Pb + Sn2+
E = 1.1 –
E°cell = 0.14V – 0.13V = 0.01V
Ecell = E0 –
0.06
[Sn 2 ]
log
2
[Pb2 ]
0.06
1
log 3
2
10
0.06
= 0.01 –
 3 = 0.01 – 0.09
2
= –0.08 V
= 0.01 –
Ecell < 0  incorrect
representation should be as

Pb|Pb+2||Sn+2|Sn
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
102
20.
Ecell = 0
Ecell = E0 –
0.0591
[Cu 2 ]
log
n
1
0.0591
0.0591
–0.34 =
log[Cu 2 ]
log[Cu 2 ]
2
2
0.68
log[Cu+2] = –
= –11.501
0.0591
log[Cu+2] = –11.501
[Cu+2] = 10–11.501
= 2.97 × 10–12 M
E0 =
21.
Ecell = E0 – (
0.0591
1
)
log
2
[Zn 2 ]
ZnSO4  Zn+2
0.1×0.8
0.1×0.2
2+
–
Zn + 2e  Zn
1
0.0591
Ecell = –0.76 –
log
0.1 0.2
2
= –0.76 – 0.03 × 1.7 = –0.81V
22.
Cu(OH)2 –– Cu2+ + 2OH–
[Cu2+] at
pH = 14
[Cu2+] × [OH–]2 = 10–19
[Cu2+] × (1)2 = 10–19
[Cu2+] = 10–19 M
Cu+ + 2e– –– Cu
0.0591
1
E = E° –
log
2
[Cu 2 ]
0.0591
= 0.34 –
log 1019
2
0.0591
= 0.34 –
× 19 = –0.2214 V
2
23.
(i)
anode
Cathode
Zn –– Zn2+ + 2e–
Ag+ + e– –– Ag × 2
Zn + Ag –– Zn2+ + Ag
(ii)
EoCell = E° Zn / Zn2+ + E° Ag2+ / Ag
= 0.76 + 0.8 = 1.56 V
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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103
0.0591 [Zn 2 ]
log
2
(0.1)2
[Zn2+] = 4 × 10–4 M
1.6 = 1.56 –
(iii)
(iv)
24.
IF 125 is added then
As+ + I– –– AgI
[Ag]  , ECell 
EoCell = E° Zn / Zn2+ + E° Cu2+ / Cu
= 0.76 + 0.34 = 1.1 V
0.0591
[Zn 2 ]
log
1.1 =
2
[Cu 2 ]
log
[Zn 2 ]
= 37.22
[Cu 2 ]
25.
At equilibrium E = 0
0.0591
0.0591
log K = (–0.68 + 1.44) –
logK
 E = E0 –
n
1
0.0591
0.76 =
logK
1
0.76 1
logK =
 12.859
0.0591
K = 1012 × 10-0.859 = 7.6 × 1012
26.
EoCell = –0.68 + 1.44 = 0.76 V
0.0591
log Keq
EoCell =
1
0.76 = 0.0591 log Keq
log Keq = 1012.86 = 7.6 × 1012
27. (i) E°cell
= 2.37 + 0.80
= 3.17 V
0.591
log K
2
0.591
3.17 =
log K
2
K = 10107.275
= 10107 × 100.275
= 1.88 × 10107
E0 =

log K =
6.34
= 107.275
0.0591
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(ii) G = 2 × 3.17 × 96500
= –611810 J
= – 611.8 KJ
0.0591
log K
2
0.0591
0.8277 = –
logK
log K = –28.01
2
G = G° + RT lnK
G° = –RT lnK
= + 8.314 × 298 × 28 log10
2 H2O 2H+ + 2OH– Kw2
2 H+ H2 + 2e
E° = 0
–
2 H2O H2 + 2OH + 2e–
G3 = G1 + G2
G3= G1
–RTnK= – RT In 10–28
K = 10–28
[H ]2
0.0591
log  a2
29.
Ecell = –
2
[H ]c
28.
E = E0 –
K= 10–28
H2  + 2e
2e– + 2H+  H2+
30.
[H+]a =
ka  c  103 c
[H+]c =
ka  c  103 c
Ecell = 
0.0591
10 3 c
log 5
2
10  c
Ecell = –
[H ]2
0.0591
log  a2
2
[H ]c
[OH]– =

0.0591
log102
2
=
0.0591
 2 = –0.0591 V
2
109 103 = 10–6
1014 1014

 108
[OH] 106
[H+] =
[H+] =
E =
kb  c
=
50
 101
500
0.0591
[108 ]
log 1 =  0.059 × log(10–7) =  0.0591 ×(–7) = 0.4137 V
1
10
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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105
0.0591
.0591
0.0591
[0.01]
= 
log101 =
1 = 0.0295 V
log
2
[0.1]
2
2
31.
E= 0 
32.
H2  2H+ + 2e–
2H+ + 2e–  H2 =
0.2367 = 0 
[H a ]2 Pa
[H c ]2 Pc
0.0591  x 2 1 
log 

2
 11 
8 = –log x2
–4 = log x

x = 10–4
PH = 4
33.
(a)
anode
Cathode
Cu –– Cu+ + 2e–
2e– + Hg2Cl2 –– 2Hg + 2Cl–
Cu2+ + Hg2Cl2 –– Cu2+ + 2Hg + 2Cl–
(b)
anode
Cathode
IO3– + Ag –– AgIO3 + e– × 2
Zn2+–– Zn
2Ag(s) + 2IO3– + Zn2+ –– Zn + 2AgIO3
(c)
anode
Cathode
Mn + 2OH– –– Mn(OH)2 + 2e–
2e– + Cu2+ –– Cu
Mn + 2OH– + Cu2+ –– Mn(OH)2 Cu
34.
Ag / AgCl(s) / Cl– | | Br– / AgBr / Ag
0.2 M
0.001M
+
+
Ag / Ag | | Ag / Ag
Ag  
0.06
a
log
E=0–
1
Ag  
c
Ksp.AgCl 1010

[Ag+]a =
= 5 × 10–10 M
0.2
[Cl ]
[Ag+]c =
E=–
Ksp.AgBr 1013
 3 = 10–10
[Br  ]
10
0.06
5  1010
log
= –0.06 × 0.7 = –0.42 V
1
1010
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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106
35.
2e– + PbCl2 –– Pb + 2Cl–
E° Cl– / Pb(s) / Pb = E° Pb2+ / Pb –
0.06
1
log
2
Ksp
–0.27 = –0.12 + 0.03 log Ksp
log Ksp = –5
Ksp = 10–5
36.
e– + AgI (s) –– Ag + I–
0.06
1
log
1
Ksp
= 0.8 – 0.06 × 16
= 0.8 – 0.96 = –0.10 V
E° I– / AgI / Ag = E° Ag+ / Ag –
37.
38.
(1)
molten NaCl
anode
2Cl– –– S2O32– + 2H+ + 2e–
cathode
Na+ + e– –– Na
(2)
NaCl (aq)
anode
cathode
(3)
Na2SO4 (aq) with pt electrode
anode
2H2O –– O2 + 4H+ + 4e–
cathode
2H2O + e– –– H2 + 2OH–
(4)
AgNO3 (aq) with pt electrode
anode
H2O –– O2 + 4H+ + 4e–
cathode
Ag+ + e– –– Ag
(5)
CuSO4 (aq) with inest electrode
anode
H2O –– O2 + 4H+ + 4e–
cathode
Cu+ + 2e– –– Cu
(6)
CuSO4 (aq) with Cu electrode
anode
Cu –– Cu2+ + 2e–
cathode
Cu2+ + 2e– –– Cu
(a)
2Cl– –– Cl2 + 2e–
nf = 2
Equivalent of charge = Equivalent of Cl2
355
=
× 2 = 0.1
71
Mole of electron = 0.1
No. of electron = 0.1 Na
2Cl– –– 2Cl2 + 2e–
2e– + 2H2O –– H2 + 2OH–
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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107
(b)
Cu2+ + 2e– –– Cu
nf = 2
Equivalent of charge = Equivalent of Cu=
Mole of electron =
(c)
2
= 1.89 × 1022
63.5
Al3+ + 3e– –– Al
nf = 3
Equivalent of charge = Equivalent of Al=
Mole of electron = 0.3 × NA = 1.8 × 1023
39.
1
×2
63.5
2.7
× 3 = 0.3
27
(a)
equivalent of Al = 0.25 × 3 = 0.75
equivalent of charge = 0.75
1 equivalent charge = 1 F
0.75 equivalent charge = 0.75 F
(b)
equivalent of charge = equivalent of SO32–=
27.6
× 2 = 0.69
80
0.69 equivalent charge = 0.69 F
(c)
equivalent of charge = equivalent of Cu= 1.1 × 0.5 × 2 = 1.1
equivalent charge = 1.1 F
40.
0.5 mole electron = 0.5 F charge
equivalent of charge = equivalent of Zn= equivalent of Ag = 0.5
Moles of Zn × 2 = 0.5
1
Moles of Zn =
4
Amount of Zn deposited = 65 ×
1
= 16.25 g
4
Moles of Ag × 2 = 0.5
Amount of Ag deposited = 108 × 0.5 = 54 g
41.
Mn2+ –– MnO2
nf = 2
equivalent of MnO2 = equivalent of chrge
25.5  t
1000
× 0.85 =
×2
96500
87
t = 1.023 × 105 sec
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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108
42.
2Cl– –– Cl2 + 2e–
nf = 2
equivalent of Cl2 = equivalent of chrge
30  60  60
moles of Cl2 × 2 =
96500
moles of Cl2 = 0.5%
vol. of Cl2 produced at 1 atm , 273 K = 0.56 × 22.4 = 12.531
equivalent of NaOH = 1.12
moles of NaOH = 1.12
mass of NaOH = 1.12 × 40 = 44.8 g
43.
2e– + 2H2O –– H2 + 2OH–
nf = 2
2H2O –– O2 + 4H+ + 4e–
nf = 4
44.
equivalent of Al =
45.
CrO2(aq) + 6H+ (aq) + 6e– –– Ce(s) + 3H2O
nf = 6
24125
equivalent of Cr = equivalent of charge =
= 0.25
96500
Moles of Cr × 6 = 0.25
1
Moles of Cr =
24
1
Moles of Cr =
× 52 = 2.16 g
24
equivalent of charge = equivalent of Cr
12.5  t 1.5

6
96500 52
t = 1336.15 sec
(i)
(ii)
46.
x
x
×3=
27
9
Equivalent is produced by coast = x Rs
1 equivalent will produced by coast = 9 Rs
x
x
×2=
Equivalent of mg =
24
12
x
x
Coast to produce
equivalent mg =
× 9 = 0.75 x Rs
12
12
equivalent of Ag = equivalent of known metal
0.5094
0.2653
1 
 nf
108
108
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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109
47.
Pdn+ –– Pd
nf = n
equivalent of Pd = equivalent of charge
2.977
3  3600
n 
106.4
96500
n=4
48.
MF2 –– M
nf = 2
equivalent of M = equivalent of charge
1.95
10  330
2 
M
96500
M = 114.045
equivalent of charge = equivalent of Cu=
equivalent of charge =
1.95
× 2 = 0.061 F
63.5
1.95
× 2 × 96500 = 5926.8 C
63.5
31.75
×2=1
63.5
equivalent of NaOH = equivalent of Cu
=1
Amount of NaOH = 1 × 40 = 40 g
Actual amount of NaOH obtained = 0.6 × 1 = 0.6 mole
0.6
% efficiency =
× 100 = 60%
1
49.
equivalent of Cu =
50.
Ni2+ –– Ni
nf = 2
equivalent of Ni = equivalent of charge=
5  20  60
96500
6000 1

96500 2
6000 1
Amount of Ni =
 × 58.5 = 1.82 g
96500 2
Moles of Ni =
51.
If active electrodes are used of a metal. Then same metal is oxidized at anode and reduced at
cathode so these is no change in concentration of electrolytic solution.
52.
2HSO4– –– S2O32– + 2H+ + 2e–
nf = 2
equivalent of Ni = equivalent of charge
Q  60  60
× 0.75 = 1 × 2
96500
2  96500
Q=
= 71.48 C
3600  0.75
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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110
53.
N = 0.01N,
R = 200,
(/A) = 0.88 cm–1
 1  × 0.88
K= 

 200 
 1  × 0.88 1000 = 440 5cm2 eq–1
eq = 

0.01
 200 
54.
K 1000
0.1
–3
K = 0.4 × 10 = 0.0004 Scm–1
m = 4 =
1
1
=
= 2500 ohm cm.
K 0.0004
   2cm  1  1
 A   4cm2   2  cm
 
 
–7
K = 8 × 10 Scm–1

(i)
R =  
A
 1   
R =   
 K  A 
 1   1  100
R= 


105 = 6.25 × 105 ohm
7   
8

10
2
16

  
1
(ii)
V = iR 
i=
= 1.6 × 10–6 amp.
5
6.25 10
Resistivity =
55.
1
1
= 0.00125 5cm–1 = 1.25 × 10–3 mho cm–1

 800
56.
K=
57.
K
250
=
Seq–1 m2
N 1000
2
eq =m  0.0125 Seq–1 m2
BaCl2 K = 0.005 ohm– cm–1
eq =
Morality =
2.08
2.08

 4.16M
3
500cm
0.5
m =
0.005 1000
5

 1.2
4.16
4.16
m =
0.005 1000
5

 250 Scm2mol-1
0.02
0.02
eq =
250
=125 Seq–1 cm2
2
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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111
eq 120

 0.8
eq 150
58.
=
59.
KAgBr = 8.75 × 10–7 – 0.75 × 10–7 = 8× 10–7
m(AgBr)  m(AgNO3 )  m(KBr)  m(KNO3 ) = 130+140 – 110 = 160
For sparsely soluble like AgBr   = 1
A(AgBr)  m(AgBr)
8  107  1000
s
4
8  10
s=
= 5 × 10–6 M = 188 × 5× 10–6 g/L = 9.48 × 10–4 g/L
160
160 =
K = 1.12 × 10–6 ohm–1 cm–1
60.
 N(AgCl) = 54 + 58 = 112 Scm2/eq
1.12 106 103
112 =
S
6
1.12 10 103
s=
112
s = 10–5
AgCl  Ag+ + Cl–
Solubility product = s2 = 10–10 mol/ltr
61.
HF

C (1– 
H+ + A–
C
C
C = 0.002M

m 200

 0.5 
m 400
2
C2 0.002(0.5)

= 10 × 10–4
K=
1  0.5
1 
62.
m(CH3COOH)  m(CH3COONa)  mHCl  mNaCl = 425 + 100 – 125 = 400 Scm2 mol–1
For
CH3COOH

N=M
eq  m  48
=
48
 0.12
400
% = 12
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112
63.
Ba(OH)2  BaCl2  2NaOH  2NaCl
= 280 × 10–4 + 2(240 × 10–4) – (125 × 10–4) × 2 = 510 × 10–4
64.
H2 O = 3.5 × 10–2 + 2 × 10–2= 5.5 × 10–
K = 5.5 × 10–6 Sm–1
H2 O =
M=
5.5 106
= 5.5 × 10–2
1000  M
5.5 106
103  5.5 102
H  = 10–7
PH = 7
kw = 10–14
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113
EXERCISE # S–II
1.
–
H2a  2H + 2e

a
–
2Hc + 2e  H2c
0.0501
log Q
2
[H  ]a2 (PH 2 )c
Q=
[H  ]c2 (PH 2 )a
E = E° –
(PH2)a = (PH2)c = 1 atm
[H+]a =
Ka, CH3COOH  C
= 1.8 105  0.1
[OH–]c = 1.8 105  0.01
[H+]c =
1014
1.8 105  0.01
(1.8)2 1013
0.0591
E=0–
log
1028
2
= –0.45 V
2.
(i) at anode
Fe(s) + 2OH–  FeO(s) + H2O() + 2e–
at cathode
Ni2O3(s) + H2O() + 2e–  2 NiO(s) + 20H–
Fe(s) + Ni2O3(s) FeO (s) + 2 NiO(s)
(ii) Ecell = 0.87 + 0.4 = 1.27 v
(iii) G = –nFE°
= –2 × 96500 × 1.27 = –2.45 kJ
|G| = 2.45 kJ
3.
[Cu2+] [OH–]2 = 10–19
If [H+] = 10–14 [OH–] = 1
[Cu2+] = 10–19
Cu2+ + 2e–  Cu
0.0591
1
E = 0.34 –
log 19
2
10
0.059119
= 0.34 –
2
= –0.22 V
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114
4.
Ag | Ag I | KI || Ag NO3 | Ag
0.005 M
0.05 M
[Ag  ]a
0.0591
0.788 = 0 –
log
[Ag  ]c
1
[Ag  ]a
0.0591
0.788 = 
log
0.05
1
[Ag+]a = 2.5 × 10–15
Ksp (AgI) = 0.05 × 2.5 × 10–15
= 1.125 × 10–16
5.
Ag |Ag Br(s) | Br– || Cl– | AgCl(s) | Ag(s)
Ag | Ag+ || Ag+ | Ag
Ecell = 0
Ecell = –
[Ag  ]a
0.0591
log
[Ag  ]c
1
[Ag+]a =
Ksp.AgCl 11010

[Cl ]
[Cl ]
Ksp.AgBr 5 1013

[Ag ]c =
[Br  ]
[Br  ]
+
Ecell
[Br  ]
1010
0.0591
=–
log
×
5 1013 [Cl ]
1
1010 [Br  ]
1
5 1013 [Cl ]
[Cl ] 1000 200


[Br  ]
5
1
6.
E I– |Ag I| Ag = E° Ag+ | Ag +
0.06
log Ksp
1
E I– |Ag I| Ag = + 0.7991 – 0.06 × 16.07 = 0.150 V
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115
7.
Ag |Ag+|| Ag+ | Ag
Ksp. AgCl 2.8 1010

[Ag ]a =
[Cl ]
0.2
+
[Ag+]c
=
Ksp AgBr 3.3 1013

[Br  ]
0.001
Ecell = 0 –
2.8 103  0.001
0.0591
log
0.2  3.3
1
= –0.0591+ log 4.24
= –0.037 V
E° Pb |PbCl2| Cl– = E° Pb | Pb2+ –
0.268 = 0.126 –
0.0591
log Ksp
2
0.0591
log Ksp
2
Ksp = 1.536 × 10–5
9.
4 × E° Sn | Sn4+ = 2 × E° Sn | Sn2+ + 2 × E° Sn2+| Sn4+
E° Sn | Sn4+ =
0.136  0.154
= – 0.009 V
2
4+
2–
3+
Ecell = E° Sn| Sn + E° Cr2O7 | Cr
= – 0.009 + 1.33 = 1.321 V
10.
at cathode
2H+ + 2e–  H2(g)
at anode
2SO42-  S2O82- + 2e–
2H2O  O2 + 4H+ + 4e–
Eq of H2 = Eq of S2O82– + Eq of O2
9.722
2.35
w2
×2=
4
22.7
194
22.7
w = 42.874 g
11.
Cu2+ + 2e–  Cu
3
3  2  60  60
×2=
× 
63.2
96500

 = 0.4221
%  = 42.21 %
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
116
12.
Eq of charge/year =
1.5 106  3.15 107
= 4.9 × 108
96500
Eq of charge = Eq of H2O
8
nH2O × 2 = 4.9 × 10
8
nH2O = 2.45 × 10
2.45 108
= 4.41 × 106 L/year
55.5
V(L) =
t=
13.
8.2 1012
= 2 × 106 year
6
4.4110
eq =
K 1000
N
97.1 =
K 1000
0.1
K = 97.1 × 10–4 Scm–1
K=
 1
.
a R
97.1 × 10–4 =
R=
I=
14.
K=
1 0.5

R 1.5
104
= 34.24 
97.1 3
V
5
= 0.146 amp

R 34.24
1 
R a
1.342 =
1


170.5 1.86 10–4
 = 1.342 × 170.5 × 1.86 × 10–4
= 4.25 × 10–2 m
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
117
15.
Ktotal  KSrSO4  KH2O
1.482 × 10–4 = KSrSO4 + 1.5 × 10–6
–6
–6
KSrSO4 = 148.2 × 10 –1.5 × 10
= 146.7 × 10–6 s cm–1
2+
m = m Sr + m SO42–
= 139.28 s cm2 mol–1
m =
146.7 10–6 103
K 1000
s=
139.28
S
s = 1.053 × 10–3 M = 1.053 × 10–3 × 183.6 g/L = 0.1934 g/L
16.
M/Mn+ (0.02 M) || H+(1M) / H2(g)
1 atm
–
M  M + ne
n+
e– + H+ 
1
H2 × n
2
M + nH+  Mn+ +
n
H2
2
[Mn  ](PH2 )n/2
0.06
0.81 = 0.76 –
log
n
[H ]n
0.81 = 0.76 –
0.06
0.02 1
log
n
1
0.06
log 0.02 = –0.05
n
6
n = – log 0.02
5
n=
17.
6
log50  2
5
5 × E1º = 0.54 × 4 + 0.45 × 1
E1º = 0.522 V
6Eº2 = 0.54 × 4 + 0.45 × 1 + 1.07 × 1
Eº2 = 0.613 V
18.
E=0–
[Zn 2 ]a
0.0591
0.0591 10
log
= –
= – 0.0295 V
log
2
[Zn ]c
2
2
1
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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118
3
–3
C6 H5 NO2  C6H5 NH2
19.
nf  6
Equvalent of charge = Eq. of C6H5NO2
= 0.1 × 6 = 0.6
Q
 0.2
96500
Q = 115,800 C
0.6 =
E = Q × V = 115,800 × 3 J =347.4 kJ
Mass of Ag deposited = 80 × 5 × 10–4 × 10.8 J = 0.432
20.
Eq. of Ag =
0.432
 0.004
108
0.004 =
2  t(sec)
96500
t(sec) =
0.004  96500
= 193 sec
2
21. (i) Eq. of Cu deposit = Eq. of charge = 0.01
Mole of Cu deposit =
0.01
2
Mass of Cu deposit =
0.01
 63.5 = 0.3175
2
Mass of Cu remaining = 10 – 0.3175 = 9.6825 g
(ii)
Eq. of charge = Eq. of H2O electrolysed
2H2O  4H+ + O2 + 4e–
nf = 1
Equivalent of H+ = 0.01
22.
Amount of Cd deposited
x
100  12
x2
x=
24
88
24
2
5  t(sec)


88 112.4
96500
t = 93.65 sec
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
119
1.
EXERCISE # O–I
Chemical reaction produces electrical energy.
2.
Reduction takes place at cathode (not anode) in galvanic cell.
3.
Standard Hydrogen electrode is universally accepted as reference electrode at all temperatures
and assigned a value of a0 Volt.
4.
Temperature on be varied also in standard Hydrogen Electrode.
5.
Zn2+(aq) + 2e– –– Zn(s)
6.
KCl can be used as electrolyte in (A) because if salt bridge is not present, then the two
electrolytes ZnCl2 and AgNo3 will mix and give ppt of AgCl. Also electroneutrabty will not be
maintained in that case.
7.
Oxidation reaction
Reduction reaction
net Cell reaction
 2Cl– –– Cl2 + 2e–
 2Ag+ + 2e– –– 2Ag
2Ag+ + 2Cl– –– 2Ag + Cl2
8.
In galvanic cell, redn takes place at cathode
9.
It is taken as reference, so electrode potential assumed to be zero.
10.
Salt bridge is responsible for maintaining electrical circuit, neutrality and minimizing liquidliquid junction potential.
11.
All the given cations makes ppt with Cl– ion
12.
E° = –0.250
An+ + ne– –– A
To make this reaction spontaneous
Ecell > 0
Eredn  Eoxyn  0
–0.250 + E oxyn > 0
 E oxyn > 0.25 V
There fore metal D can displace A from it’s aqueous soln
13.
E° = EoOxyn  EoRe dn
o
o
= ESn/Cn
2  E
Ag /Ag
= 0.80 + 0.14 = 0.94 Volt
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
120
14.
E° = EoOxyn  EoRe dn
= EoNi/Ni2  EoAu3 /Au
= 0.25 + 1.5 = 1.75 Volt
15.
E° = is an intensive property, it does not depends on moles.
So E° will be same when moles is halfed
16.
E° = EoOxdn  EoRedn
For option A largest EoCell = 1.25 + 1.25 = 2.5 V
For option B largest EoCell = 1.25 + 0.68 = 0.57 V
For option C largest EoCell = 1.25 + 0.24 = 1.01 V
For option D largest EoCell = 0.68 + 0.24 = 0.92 V
 Clearly, option A has largest EoCell
17.
Zinc cannot displace those metals from it’s aqueous solution which have more oxidation
polential or less reduction potential So Na+ is the correct answer.
18.
The metal with less reduction potential is strangest reducing agent.
19.
Metals having lower reduction potential reduce metals having high reduction potential.
20.
Metal with lower reduction potential reduce metal with high reduction potential.
21.
2X– + Y2 –– 2Y– + X2
 X– can reduce Y
–
–
2Z + X2 –– 2X + Z2
 Z– can reduce X
2W– + Y2 ––X
 W– can not reduce Y but y can reduce W
Arrange Reduction potential Z– < X– < Y– < W–
So oxidation potential is reverse Z– < X– < Y– < W–
22.
Zinc and iron can displace copper from it’s CuSO4 solution.
23.
Cu + AgNO3 –– Cu(NO3)2 + Ag
Blue
24.
efficiency =
25.
G is extanove property i.e. it dipoles on moles.
26.
Ecell = EOxyn  ERe dn
used energy G nFEcell


total energy H
H
= 0.4 + 0.8 = 1.2
G = –nFE = –2 × 96500 × 1.2 = –231.6 KJ
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
121
27.
By Nerst equation
RT [M  ]
In
E = E° –
nF [Mn 1 ]
28.
2H+ + 2e– –– H2
E° = 0
0.1 M
1 atm
PH
0.0591
log  22
E = E° –
n
[H ]
0.0591
1
E=0–
= –0.059 V
log
2
(0.1)2
29.
Ag+ + e– –– Ag
0.1
0.0591
1
E = EoRe dn 
log
1
0.1
o
E = ERe dn – 0.059
30.
E = EoRe dn 
31.
2Ti + Cu2+ ––2Ti+ + Cu
0.1 M
0.01 M
0.0591
[Ti]2
log
E = E° –
2
[Cu 2 ]
T E  [Ti+] or [Cu2+]
32.
33.
0.0591
[Sn 2 ]
log
2
[Ag  ]2
If [Sn2+]  or [Ag+]  , then E 
0.0591 [Zn 2 ]
log
2
[Ni 2 ]
0.0591
0.5105 = E° –
log10
2
E° = 0.54 Volt
E = E° –
EoCell  EoOxydn  EoRe dn
EoCell  EoI/I2  EoFe3 /Fe2 = –0.536 + 0.771 = 0.235 Volt
34.
at eqm E = 0
0.0591
E° =
log Keqm
n
0.0591
E° =
log (4 × 1012)
n
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
122
E° =
35.
36.
0.0591
× 12.6 = 0.3717 Volt
n
0.0591
0.1
log
2
0.1
0.0591
E = 1.1 +
log1
2
E = 1.1 V
E = E° +
1
1
H2 + H+ –– H+ + H2
2
2
10–8 M
10–3 M
0.0591 108
log 3
E=0–
1
10
E = 0.295 V
0.0591
log Keqm
n
0.0591
0.295 =
log Keqm
2
Keqm = 1010
37.
E° =
38.
Cu + 2Ag+ –– Cu2+ + 2Ag
0.0591
[Cu 2 ]
log
E1 = E° –
2
[Ag 2 ]2
0.0592
[Cu 2 ]  10
log
2
[Ag 2 ]2  100
E2 – E1 = 0.0295 V
E2 = E° –
39.
Cu2+ + 2e– –– Cu
0.0591
1
E1 = E° –
log
2
[Cu 2 ]
0.0591
100
E2 = E° –
log
2
[Cu 2 ]
E2 – E1 = 59 × 10–3 V
40.
E° =
0.0591
log Kc
n
0.0591
E° =
log1012
2
E° = 0.354 V
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
123
41.
42.
0.0591
log Keq
n
0.0591
log106
E° =
2
E° = 0.177 V
E° =
0.0591
log Keq
n
0.0591
log Keq
0.295 =
4
Keq = 1020
E° =
43.
Concentration cell has two equivalent half cell
44.
G < 0
45.
Pt(s) / H2(g) , 1 bar / 1 MHCl (aq) / AgCl (s) / Ag(s)
E>0
C
0.0591
log 1
E = E° –
n
C2
If C2 > C1  E > 0
46.
For
47.
ECell = E H2/H+ + EI–/AgI/Ag
0 = EH2/H+ – 0.151
E H2/H+ = 0.151 V
H2 –– 2H+ + e–
0.0591
log[H+]
E=0–
1
+0.151 = 0.0591 pH
pH = 2.55
AgI + e– –– Ag + I–
0.0591
KsP
log 
1
[I ]
0.0591
Ksp
–0.151 = 0.799 +
log
1
1
0.0591 logKsp = –0.95
logKsp = –16.07
Ksp = 10–16.07 = 8.43 × 10–17
E°I–/AgI/Ag = E°Ag+/Ag +
48.
0.0591
1
0.0591
1
log
= 0.8 –
log
1
Ksp
1
8.3  10–17
1017
= 0.8 – 0.591 log
= –0.15 V
8.3
E°I–/AgI/Ag = EAg+/Ag +
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
124
dE
dT
dE S

dT nF
49.
S = nF
50.
efficiency =
84 =
| G |
× 100
H
| G |
× 100
285
G = –84 × 285 ×
1
100
G = –239.4 kJ
G = –nFE
–239.4 × 103 = –2 × 96500 × E
E = 1.24 V
51.
52.
dE
dT
= –2 × 96500 × 0.00065= –125.45 JK–1
S = nF
dEo
= –0.000648 Volt K–1
dT
G° = –nFE
G° = –2 × 96500 × 0.6753
G° = –130332.9 J
G° = H° – TS°
–130332.9 = H° + 298 × 0.000648
H° = –130.45 kJ
53.
If cathode is removed from a electro chemical cell. The cell will be not be completed and
polarity of anode end cup.
54.
55.
Hg, pt and graphite are inert electrodes
anode
2Br– –– Br2 + 2e–
cathode
2e– + 2H2O –– H2 + 2OH–
By product
NaOH
56.
anode
2H2O –– O2 + 4H+ + 2e–
cathode
Cu2+ + 2e– –– Cu
By product
NaOH

[H+]  , So pH 
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
125
57.
anode
cathode
2H2O –– O2 + 4H+ + 4e–
2e– + 2H2O –– H2 + 2OH–
58.
anode
cathode
2Cl– –– Cl2 + 2e–
2e– + 2H2O –– H2 + 2OH–
59.
w I and I 
1
1
sow 
R
R
60.
equivalent of charge =
61.
63.5
×2=2
63.5
2 equivalent charge = 2F = 2 mole electrons = 12 × 1023
E
96500
E = equivalent weight

96500 C liberates = E g
E
g

1 C will liberate =
96500
Z=
E
96500
E = 0.0006 × 96500 = 57.9
62.
Z=
63.
wQ
w1 I1 t1

w 2 I2 t 2
64.
65.
66.
w1 2  2

w 2 1 4

w1 w 2

E1 w 2

w2 = w
w1 E1

w 2 E2
108
equivalent of charge =
× 1 =1
108
=1F
–
+
2H + 2e –– H2
nf = 2
2+
–
Mg + 2e –– Mg
nf = 2
w1 E1
2
1



w 2 E2 24 12
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
126
67.
3
= 0.06
E
68.
9.65 10  60
= 0.06
96500
3
E=
= 50
0.06
equivalent of substance =

Ni2+ + 2e– –– Ni
nf = 2
3+
–
Cr + 3e –– Cr
nf = 3
w1 E1
0.3 59 3


 
w 2 E2
w2
2 52
w2 =
0.3 104
= 0.17 g
177
69.
equivalent of charge = equivalent of substance deposited
=1
1 equivalent of charge = 1 F = 96500 C
= charge of 1 mol e–
70.
0.1 =
71.
Snn+ –– Sn
3.17
E

E = 3.17
nf = 2
equivalent of Sn =
9.65  1000
= 0.1
96500
5.95
11.9
× n = 0.1 n =
=2
119
5.95
72.
18  2
3 t

18
96500
t = 18 hr
73.
equivalent of Al = equivalent of Ag= 1 × 3
moles of Ag × 1 = 3
moles of Ag = 3
74.
equivalent of substance = equivalent of charge=
6  1020
= 0.001
6  1023
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
127
75.
2H+ + 2e– –– H2
nf = 2
I  965
112

 2  I = 1 amp
96500 22400
76.
H2O –– 2H+ +
77.
equivalent of Ca = equivalent of Al
40  103
w
×2=
×2
40
27
w = 18 kg
78.
Coast of 1 equivalent of Mg = 5 × 12 = 60 Rs
9
Equivalent of Al =
×3=1
27
Coast of 1 equivalent of Al = 60 Rs
79.
H2O + 2e– –– H2 + 2OH–
nf = 2
0.536  30  60
Equivalent of H2 =
= 0.01
96500
Moles of H2 = 0.005
vol. of H2 = 22.4 × 0.005 = 0.112 L
80.
Equivalent of Ag = equivalent of O2=
1
O2 + 2e–
2
1
2F = 1 mol H2O =
mol O2
2
2F = 2 × 96500 C = 1.93 × 105 C
0.108
× 1 = 0.001
108
0.001
4
0.001
Vol. of O2 =
× 22700 = 5.6 mL
4
Moles of O2 =
81.
equivalent of H2 = equivalent of Al=
moles of H2 =
vol. of H2 =
4.5
× 3 = 0.05
27
0.5 1

2
4
1
× 22.7 = 5.676 L
4
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
128
82.
equivalent of charge = equivalent of Ag
9.65  t 80  5  103  1.08

96500
108
t = 40 sec
83.
1
1.2  1022
2 
 M = 100
M
6  1023
84.
Ag : Cu : Al = 1 :
85.
Q
10  10  103  10  2

96500
193
Q = 10,000 C
86.
Equivalent of H2 =
87.
Mn2O4 –– MnO42–
10
nf = 10
nf =
8
1 1
: =6:3:2
2 3
2.4
×4
2.27
Moles of H2 = 0.211
vol. of H2 = 0.211 × 22.7 = 4.8 L
10
10
×1=
8
8
10
10
Equivalent of charge =
× 96500 C
8
8
Equivalent of charge =
89.
Voltage of cell increases
90.
2PbSO4(s) + 2H2O(  ) –– Pb(s) + PbO2(s) + 2H2SO4(aq)
91.
At high pH, The passive layer is formed and prevent corrosiss
92.
93.
Resistance decrease with increase in temp as speed of ions increases
K  , Molarity 
94.
95.
K 1000
N
N  , eq 
eq =
m NaNO3 = m NaCl + m AgNO3 – m AgCl = 110 + 115 – 120= 105
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
129
K 1000 0.0014 1000
=
= 140 S cm2 eq–1
N
0.01
96.
eq =
97.
K=
1
× 1.5 = 7.5 × 10–3 S cm–1
200
eq =
K 1000 7.5 103 1000
=
= 750 S cm2 eq–1
0.01
N
K=
1
× 1.15 = 0.0046 S cm–1
250
eq =
K 1000
0.0046 1000
=
= 46 S cm2 eq–1
N
0.1
K=
1
× 0.4 = 0.01 S cm–1
40
98.
99.
K 1000
0.011000
=
= 1000 S cm2 mole–1
N
0.01
eq =
m
80 1



m 400 5
100.

101.
m 
K 1000
s
3.06  106  103
s=
= 2 × 10–3
1.53
ksp = s2 = 4 × 10–6
102.
m BaCl2 = eq BaCl2
= eq Ba2+ + eq Cl–= 127 + 76 = 203 S cm2 eq–1
103.
K=
1 2.2
= 0.01 S cm–1

50 4.4
m =
K 1000 0.01 1000
= 0.2 S cm2 mole–1

M
50
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
ONLINE PARTNER UNACADEMY
130
104.
m = 10 S cm2 mole–1
m = 200 S cm2 mole–1
=
10
= 0.05
200
[H+] = c = 5 × 10–2
pH = 2.2
105.
106.
K=
1
S cm2 mole–1
X
m =
1000
XY
Ka =
c2
1 

a=
kc
= 0.4
C
m = m ×  = 380 × 10–4 × 0.04= 1.52 × 10–3 S cm2 mole–1
1.52 × 10–3 =
K  102
0.01
K = 1.52 × 10–5 S m–1
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EXERCISE # O–II
Cl2(g) + 2Br–(eq) ––Cl–(eq) + Br2 (g)
[Cl ]2 PBr2
(0.01)2  0.01
Q=
= 0.01

PCl2 .[Br  ]2
1 (0.01)2
1.
0.06
log(0.01)
2
= 0.29 – 0.03 log10–2
= 0.35 V
E = E° –
Ans.
(B)
2.
2H+(eq) + 2e– –– H2(g)
Let PH2 = 1 bar
 When pH = 0  [H+] = 1 M
0.06
E = E° –
log 1 = E°
2
& when pH = ?  [H+] = 10–7 M
1
0.06
E = E° –
log 7 2 = E° – 0.42
2
(10 )
 Reduction pot. clecrease by 0.42 V
Ans. (D)
3.
2H+(eq) + 2e– –– H2(g)
When PH2 = 1 atm
0.06
1
log 2 = E°
2
1
When PH2 = 100 atm
E = E° –
0.06
100
log 2 = E° – 0.06
2
1
 change in reduction pot. is 0.06 V
Ans. (A)
E = E° –
Ag –– Ag+ + e–
0.06
log[Ag+]
E = E° –
1
0.06
Ksp
–0.209 = –0.799 –
log 
1
[Cl ]
0.06
Ksp
–0.209 = –0.799 –
log
1
0.1
–11
 Ksp = 10
Ans. (B)
4.
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5.
eq of A = eq of B = eq of C
2.1
2.7
7.2


7 / x 27 / y 48 / y
x:y:z::1:3:2
Ans. (B)
6.
During electro refining
At cathode : Cu2+ + 2e– –– Cu
At anode : Cu –– Cu2+ + 2e–
Due to liberation of O2 gas decrease in wt. of anode is more than 2.5 gm.
Ans. (A)
7.
m 
K
1000  Solubility
9  106
1.5 × 10 × 3 =
1000  S
–5
S = 2 × 10
Ksp = 33 × S4
= 27 × (2 × 10–5)4= 4.32 × 10–18
–4
Ans. (A)
8.
CH3COOH (eq) + NaOH (eq) –– CH3COONa (eq) + H2O (  )
0.015 × V
0.015 × V
0
0
0
0.015 × V
0.015  V 0.015M

[CH3COONa] =

2V
2
4
K  1000 6.3 10 1000

m =
= 84 S cm2 mole–1
M
0.015 / 2
2H2(g) + O2 (g) –– 2H2O (  )
G° = H° – TS°
2  96500 1.23
= –285.5 – 298 × S°
1000
S° = –0.322 kJ/K
Ans. (D)
9.
10.
Let initially concentration of electrolytes of cathode and anode us 1 M, then
Ecell = E°cell
If concentration of electrolytes of cathode and anode is 10 M, then
H2 (g) + Cl2 (g) –– 2H+ (aq) + 2Cl– (aq)
1 atm 1 atm
10 M
10 M
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0.06
(10)2  (10)2
log
2
11
E = E° – 0.12
 E decrease by 0.12 V
Ans. (C)
E = E° –

11.
2H2SO4 + Pb(s) + PbO2 (s) 
 2PbSO4(aq) + 2H2O (  )
Ans. (CD)
12.
13.
(A) : During electrolysis Cu at anode will oxidise as well as Cu2+ of solution will reduce at
cathode.
(B) : During electrolysis
At cathode : 2H+(g) + 2e– –– H2(g)
1
At anode : H2O (  ) –– O2 (g) + 2H+(g) + 2e–
2
(C) : During electrolysis
At cathode : Na+ + e– –– Na
1
At anode : 2OH– –– O2 (g) + H2O (  ) + 2e–
2
(D) : During electrolysis
At cathode : 2H+ (aq) + 2e– –– H2 (g)
At anode : 2Cl– (aq) –– Cl2 (g) + 2e–
1
O2(g)
2
At cathod
At anode
270
Mole of H2O electrolyzed =
= 15 mole
18
Equivalent of H2O electrolyzed = 15 × 2 = 30
H2O (  ) –– H2(g)
+
Equivalent of H2 gas released = Equivalent of O2 gas released = 30
30
mole of H2 gas =
= 15 mole
2
30
mole of O2 gas =
= 7.5 mole
4
Volume of O2 gas = 7.5 × 22.4 L = 168 L
Total volume of gas = (15 + 7.5) × 22.4 = 504 L
100
Faraday of electricity consumed = 30 ×
= 40 F
75
Ans. (ABC)
14.
Specie giving higher reduction potential can oxidize the specie having lower reduction
potential.
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15.
At cathode : 2H+(aq) + 2e– –– H2 (g)
At anode : Ag(s) + Cl– (aq) –– AgCl (s) + e–
Overll cell rxn : 2Ag(s) + 2H+ (aq) + 2Cl– (aq) –– H2(g) + 2AgCl (s)
Q
PH2
[H ]2 [Cl ]2
E = E° –
0.06
log Q
2
With increase in value of Q, EMF of cell decrease.
Ans. (AD)
16.
Cells whose cell reaction does not involve any specie whose active mass changes with progress
of reaction gave constant value of EMF.
17.
Lower the value of reduction potential higher will be reducing power.
18.
it is an intensive property
19.
(A):
During electrolysis of dil HCl
At cathode: 2e– + H2O (  ) ––H2 (g) + 2OH–
At anode:
2H2O (  ) ––O2 (g) + 4H+ + 4e–
(B) During electrolysis of dil NaCl
At cathode: 2e– + H2O (  ) ––H2 (g) + 2OH–
At anode:
2H2O (  ) ––O2 (g) + 4H+ + 4e–
(C) : During electrolysis of conc. NaCl
At anode: 2Cl–––Cl2 (g) + 2e–
At cathode: 2H2O (  ) ––O2 (g) + 4H+ + 4e–
(D) : During electrolysis of AgNO3
At cathode : 2Ag+ (aq) + e– –– Ag
At anode : 2H2O (  ) ––O2 (g) + 4H+ + 4e–
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EXERCISE # JEE MAIN
1.
2
8
2
Since S2O has highest value of reduction potential among given species therefore S2O8 is
strongest oxidizing agent.
2.
2

3
Fe(aq)
+ Ag(aq)
 Fe(aq)
+ Ag(s)
Eº = E0Ag /Ag – E0Fe3 /Fe2


Eqn.1: Fe+2 + 2e–  Fe, Eº = y
Gº = – 2Fy
+3
–
Eqn.2: Fe + 3e  Fe, Eº = z
Gº = – 3Fz
Eqn.2 – Eqn. 1, We get
Fe+3 + e–  Fe+2 , (Eº)1
Gº = – 3Fz + 2Fy
1
– 1 × F × (Eº) = – 3Fz + 2Fy


(Eº)1 = E0Fe3 /Fe2 = 3z – 2y
Now
Eº = E0Ag /Ag  E0Fe3 /Fe2
Now
= x – (3z – 2y)
= x – 3z + 2y
3.
G° = –nFE°
= –2 × 96000 × 2
= –384000 J = –384 kJ
4.
Reaction at cathode
Ni2+ + 2e– — Ni
0.1 mole
Deposited moles of Ni =
0.1
= 0.05
2
5.
Conductivity increases with increasing concentration of electrolyte and molar conductivity
decreases with increasing concentration of electrolyte.
6.
m = mº – A C
value of A will be same for NaCl and KCl
So Slope for both curve will be same.
But mº for KCl > mº for NaCl
So m KCl > m NaCl
(for a given Concentration)
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7.

oxidising power  S.R.P value.

correct order is : Bi+3 < Ce+4 < Pb+4 < CO+3
Eºred oxidising power 
Acidic Strength  Ka
order of Acidic Strength or ka is A > C > B
Acidic Strength ()  ka()  electrical conductivity ()
8.
As,

As,
9.
Anode
PbSO4 + 2e– Pb(s) + SO24
10.
Equivalent of PbSO4 electrolysed = 0.05
0.05
Moles of PbSO4 =
2
0.05
Mass of PbSO4 =
× 303 = 7.6 g
2
Ecell  2V
As,
G° = –nF Ecell
and also, G° = –RTlnKC
Equating (1) and (2)
... (1)
....(2)

 nF E aq =  RTlnKC
2 × 96000 = 8 × 300 lnKC

KC = e160

Correct option is 3
11.
SRP  ,
12.
Cell reaction :
Oxidising power 
Reducing power 

H2(g)  2AgCl(s) 
2H(eq)  2Ag(s)  2Cl(eq)
QC =
[H ]2[Cl ]2
(106 ) 2 (10 6 ) 2
=
= 10–24
1
PH2
0.06
log Q
n
0.06
0.92 = E° –
log 10–24
2
E = E° –
E° = 0.2V
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13.
  + E 
= 0.76 +  E 

Ecell = Eox

red cathode
anode

red cathode
A electrode having maximum SRP will produce maximum Ecell
14.
KC  10 × 1015
ECell  ?
G° = –nF ECell
...(1)
Also, G° –RTlnKC
...(2)
Equate (1) and (2)
 nF ECell =  RTlnKC

15.
2 × ECell =
S° = nF
2.303RT
log 1016
F
dE
dT
G° = –nFE°
G = H° – T S°
S° = –2 × 96000 × 5 × 10–4 = –96.5 J
G° = –2 × 96000 × 2 = – 384 kJ
G° = H° – T S°
– 384 = H° –
300  (96.0)
1000
H° = –384 – 28.8 = – 412.8 kJ mol–1
16.
m  m
HA
HCl
 m
NaA
– m
NaCl
= 425.9 + 100.5 – 126.4 = 400
K 1000 5 10–5 1000
m 

 50
M
0.001

17.
m 50 1

  0.125
m 400 8
B2H6+3O2→B2O3+3H2O
Moles of O2 required =3× moles of B2H6 =3
I t
= moles of O2 ×4 =12  t=3.2hrs
96500
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18.
SRP  ,
Oxidising power 
Reducing power 
19.
Catnode :
2e– + 2H2O  H2 + 2OH–
nf = 2
CH2 – COO–

Anode :
CH2 = CH2 + 2CO2 + 2e–
CH2 – COO–
nf = 2
nf = 1
Total equivalent of gases (CH2 = CH2 , CO2 , H2) = 0.2 + 0.2 + 0.2 = 0.6
0.2
0.2
+ 0.2 +
= 0.4
2
2
Total Vol. of gases = 0.4 × 22.4 = 8.96 L
Total moles of gases =
20.
Cu + ZnSO4 –– X
No reaction Cu lies below Zn in electro chemical series.
21.
Cu2+ + 2e–  Cu
2F
1 mole = 63.5 g.
22.
if Eext > E° Then electron flow from cathode to anode
23.
5e– + 8H+ + MnO4–  Mn2+ + 4H2O
0.06
[H  ]8 f
log  8
E = 1.51 –
5
[H ] i
0.06
(103 )8
log 7 8
E = 1.51 –
5
(10 )
0.06
E = 1.51 –
× 32 = 1.126 V
5
So it will oxidize Br– & I–
24.
 = 1.4 S/m.
R = 50 
M = 0.2
1 
=
×
R A

= 1.4 × 50 m–1.

A
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1
1 
×
=
 1.4  50  0.25 Sm–1
R A 280
0.25



 5  104 Sm2 mol–1
1000  M 1000  0.5
=
25.
c =   – B C
26.
Higher the SRP, better is oxidising agent
Hence MnO – is strongest oxidising agent
4
27.
X + Y2+ X2+ + Y
For reaction to be spontaneous Eº must be positive.
28.
2H+ (aq) + 2e–  H2 (g)
PH
0.0591
0.0591
2
log 2 2 ;
Ered = 0 –
log 2 ;
n
2
(1)
(H )
Ered is found to be negative for (3) option
Ered = Eºred –
29.
Ered = –
0.0591
log2
2
 = 1.3 S/m.
R = 50 
M = 0.2
1 
=
×
R A

= 1.3 × 50 m–1.

A
1
1 
=
×
=
 1.3  50  0.25 Sm–1
R A 260

0.25


 6.25  104 Sm2 mol–1
1000  M 1000  0.4
30.
2
4
Al2 O3 
Al + O2
3
3
rG
G
= +966 kJ mol–1 = 966 × 103 J mol–1
= – nFEcell
966 × 103 = – 4 × 96500 × Ecell
Ecell = 2.5 V
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31.
Fe3+ + 3e–  Fe
G1 = –3 × F × E0
Fe2+2e–Fe
G2 = –2 × F × E0
Fe3+ + e– Fe2+
G3 = G1 – G2
Fe3 /Fe
Fe2 /Fe
G3 = 3 × 0.036F – 2 × 0.439 × F = – 1 × E0 (Fe3+/Fe+2) × F
E0 (Fe3+/Fe+2) = 2 × 0.439 – 3 × 0.036= 0.878 – 0.108= 0.770 V
32.
Ecell = E0cell – log
= 0.3 –
33.
[Cr 3 ]2
[Fe2 ]3
0.059
(0.1)2
log
= 0.3 –0.04 = 0.26 V
6
(0.01)3
0 = + 1.1 –
log = 37.3
0.0591 [Zn 2 ]
log
2
[Cu 2 ]
[Zn 2 ]
[Zn 2 ]
= 1037.3

2
2
[Cu ]
[Cu ]
0.059
log Ksp
1
 log Ksp = – 16.11
34.
0.152 = – 0.8 –
35.
C = 0.1 M,
36.
CH3COONa + HCI  CH3COOH + NaCI
o
o
0CH3COONa +  HCI = 0CH3COOH +  NaCI
R = 100 
1

K = 1.29 Sm–1 =
×.
100
A
C = 0.02 M, R = 520 .
1
K=
× 129
520
1
 129
520
M =
= 124 × 10–4 Sm2mol-1
1000  0.02
or
From the reaction,
o
o
0CH3COOH = 0CH3COONa +  HCI –  NaCI
Thus to calculate the value of one should know the value of oNaCI along with and oHCI.
38.
5.12  1000
Q
3
27
96500
3
Q=54897×10 C = 5.4897×107C
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39.
Difluoroacetic acid will be strongest acid due to electron withdrawing effect of two fluoring
atoms so as it will show maximum electrical conductivity.
40.
° HOAC = ° NaOAC + ° HCl – ° NaCl
= 91 + 426.2 – 126.5= 390.7
41.
E° = E° Sn / Sn2+ + E°Fe3+ / Fe2+= 0.14 + 0.77 = 0.91 V
42.
E° =
43.
°m NaBr = °m KBr + °m NaCl – °m KCl = 152 + 126 – 150 = 12 Scm2 mol–1
44.
Zn(s) + 2H+(eq) –– Zn2+(eq) + H2(g)
On adding H2SO4, equilibrium shift to right
0.0591 [Zn 2 ]
log  2
E = E° –
n
[H ]
+
[H ] , E 
45.
Cr2 | Cr3  0.41V
0.06
10 g Keq
n
0.0591
0.591 =
10 g Keq
1
Keq = 1010
Mn2 | Mn3  1.57V
Fe2 | Fe3  0.77V
Co2 | Co3  1.97V
As Cr will have maximum oxidation potential value, therefore its oxidation will be easiest.
46.
SRP  ,
Oxidising power 
Reducing power 
47.
SRP  ,
Oxidising power 
Reducing power 

48.
Ecell
 Zn2 
0.0591
 E cell 
log
n
Cu 2 

Ecell  1.10 
48.
0.0591
1
log
1.07 V
2
0.1
9650
= 0.1mole
96500
number of equivalent of of Ag = 0.1mole
number of equivalent of of electrons =
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EXERCISE # JEE-ADVANCED
1.
Mg (s) + Cu2+ (aq) –– Cu (s) + Mg2+ (aq)
Initially :
E = 2.7 = E° –
When :
[Mg2+] = x
RT
 1
n    E° = 2.7
2  F  1
E = 2.67 = 2.7 –
300
x
 n  
2 11500
1
n x = 2.3  x = 10
2.
2A (s) + B2n+ (aq) –– 2A n+ (aq) + B(s)
Given H° = 2G°

G° = H° –S°

G° = 2G° – TS°
G° = TS°

G = G° + RT n Q = 0
 22 
G° = –8.3 × 300 × n   = 300 × S°
 1 
S° = – 8.3 × n 4 = –8.3 × 2 × 0.7 = –11.62 J/K
3.
=G×

120
= 5 × 10–7 ×
= 6 × 10–5 Scm–1
A
1
[H+] = c = 10–4 M   =
104
0.0015
K  1000 6  105  1000

 m 
= 40 Scm2mol–1
0.0015
0.0015


m

 m  m


m
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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143
4.
5.

m 

Z=6
40
= 6 × 102
4
10
0.0015
Zn (s) + Cu2+(aq) ––Zn2+ (aq) + Cu (s)
[Zn 2 ]
Qc =
= 10
[Cu 2 ]
G° = –2 × F × 1.1
G = G° + RT n Q = –2.2 F + RT n 10= 2.303 RT – 2.2 F
H2(g) + M4+(aq) –– 2H+(aq) + M2+(aq)
[H ]2 [M2 ] 1  [M2 ]
Q=
=
= 10x
4
4
1  [M ]
PH 2 [M ]
E = 0.092 = 0.151 –
6.
0.059
log10x  x = 2
2
M+  M3+ + 2e–
G0 = –nFE0 For 1 mole of M+
G0 = –2 × 96500 × (–0.25) J
= + 48250 J/mole = 48.25 KJ/mole
Energy released by conversion of 1 mole of
X  Y
G = –193 KJ
Hence mole of M+ convert =
193
=4
48.25
   
7.
X
Y

 
H



HX
 HY
Also
        
X
H
m
= ,
m 
Y
(1)
So

m (HX) = m
1
and

m (HY) = m
2
(Where 1 and 2are degrees of dissociation of HX and HY respectively.)
Now, Given that
m (HY) = 10 m (HX).





m
2 = 10 × m
1
2 = 10 1
(2)
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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144
Ka =

8.
C 2
,
1– 
but
 << 1,
therefore Ka = C2 .
2
K a (HX)
0.01  1 
1
0.01α12
=
=
×
=
.


2
K a (HY)
0.1
1000
0.1α 2
 10 

log (Ka (HX)) – log (Ka (HY)) = –3.

pKa (HX) – pKa (HY) = 3.
Salt bridge is introduced to keep the solutions of two electrodes separate, such that the ions in
electrode do not mix freely with each other. But it cannot stop the process of diffusion.
It does not participate in the chemical reaction. However, it is not necessary for occurence of
cell reaction, as we know that designs like lead accumulator, there was no salt bridge, but still
reactions takes place.
X
9.
Y
(P) (C2 H5 )3 N + CH3COOH 
CH3COO– (aq) + (C2H5)3NH+ (aq)
As CH3COOH is a weak acid, its conductivity is already less. On
addition of weak base, acid-base reaction takes place and new
ions are created. So conductivity increases.
(Q) KI (0.1 M) + AgNO3 (0.01 M) AgI  (ppt) + KNO3 (aq).
As the only reaction taking place is precipitation of AgI and in
place of Ag+, K+ is coming in the solution, conductivity remain
nearly constant and then increases.
(R) CH3COOH + KOH  CH3COOK (aq) + H2O
OH– (aq) is getting replaced by CH3COO–, which has poorer
conductivity. So conductivity dereases and then after the end
point, due to common ion effect, no further creation of ions take
place. So, conductivity remain nearly same.
(S) NaOH + HI  NaI (aq) + H2O
As H+ is getting replaced by Na+ conductivity dereases and
after end point, due to OH–, it increases.
So answer of 39 is : (P) – (3) ; (Q) – (4) ; (R) – (2) ; (S) – (1). Answer is (D).
10.
(P)
EºFe3+, Fe


1 × 0.77 + 2 × (– 0.44) = 3 × x
0.11
x=–
V ~ – 0.04 V.
3
(Q)
4H2O  4H+ + 4OH–
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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145
2H2O  O2 + 4H+ + 4e–
– 1.23 V
–
–
+ O2 + 2H2O + 4e  4OH
+ 0.4 V
_____________________________________
4H2O  4H+ + 4OH–
(R)

(S)

– 0.83 V
Eº(Cu2+ + Cu  2Cu+)
x × 1 + 0.52 × 1 = 0.34 × 2
x = 0.16 V.
Cu2+ + e–  Cu+
0.16 V
+
–
+ Cu  Cu + e
– 0.52 V
______________________________________________________________
Cu2+ + Cu  2Cu+
– 0.36 V
However, in the given option, – 0.18 V is printed.
Eº(Cr3+, Cr2+)
x × 1 + 2 × (– 0.91) = 3 × (– 0.74)
x – 1.82 = – 2.22
x = – 0.4 V
Hence, most appropriate is (D).
(P) – (3) ; (Q) – (4) ; (R) – (1) ; (S) – 2.
11.
G = – nFEcell = – 2 × 96500 × 0.059 × 10–3 kJ/mole
= – 11.4 kJ/mole.
12.
M|M2+ (aq) || M2+ (aq) | M
0.001 M
Anode :
M  M2+ (aq) + 2e–
Cathode:
M2+ (aq) + 2e–  M
____________________________
M2+ (aq)c  M2+ (aq)a
Ecell = 0 –
 M2 (aq)a 
0.059
log 

3
2
 10

0.059 = 
 M2 (aq)a 
0.059
log 

3
2
 10

C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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146
 M2 (aq)a 
– 2 = log 

3
 10

–2
–3
2+
10 × 10 = M (aq)a = solubility = s
Ksp = 4s3 = 4 × (10–5)3 = 4 × 10–15
13.
14.
E = Eº –
[Fe2 ]2
0.059
log  4
[H ] PO2
4
(103 )2
0.06
0.03
log
= 1.67 –
log107
3 4
(10
)

0.1
4
2
0.03
= 1.67 –
× 7 = 1.67 – 0.105 = 1.565 = 1.57 V.
2
= 1.67 –
16.
M (s) | M+ (aq, 0.05 M) || M+ (aq, 1 M) | M(s)
Anode :
M (s)  M+ (aq) + e–
Cathode :
M+ (aq) + e–  M (s)
_____________________
M+ (aq) |c  M+ (aq) |a
Ecell = E°cell –
0.0591
1
=0–
log
M (aq) |a
M (aq) |c
0.0591
log
1
= + ve = 70 mV and hence G = – nFEcell = – ve.
= 70 mV +
0.0591
log 20 = 140 mV.
1
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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147
17.
The species having less reduction potential with respect to NO3– (Eº = 0.96 V) will be oxidised
by NO3–. These species are V, Fe, Hg.
18.
Faraday law equivalents of H2 produced =
  t (sec)
96500
10  103  t
= 96500  2 = t
96500
t =19.3  104 sec
0.01  2 =
Cl2 + 2I–  I2 + 2Cl– .
Eº = 1.36 + (– 0.54) = 0.82 V (+ ve). Spontaneous.
20.
Mn3+ + e–  Mn2+,
1.50 volt.
+
–
2H2O  O2 + 4H + 4e ,
1.23 volt.
_____________________________________
4Mn3+ + 2H2O  4Mn2+ + O2 + 4H+,
Ecell = 1.5 – 1.23 = 0.27 volt. (+ ve).
3+
So Mn will oxidise H2O.
19.
21.
22.
23.
24.
Mol of NaCl = 4 × 0.5 = 2 mol.
1
1
No. of mole of Cl2 evolved = × mol of NaCl = × 2 = 1 mol.
2
2
Taking the 1 : 1 molar combination of Na–Hg amalgam.
weight = 2 × 23 + 2 × 200 = 446 g.
2Na+ + 2e–  2Na.
No. of Faraday required = 2.

total charge = 2 × 96500 = 193000 coulomb.
+
2Ag + C6H12O6 + H2  2Ag(s) + C6H12O7 + 2H+,
0.0592
0 = 0.75 –
log K.
2
Eº = 0.8 – 0.05 = 0.75 volt.
ln K = 2.303 × log K = 2.303 × 25.34 = 58.38.
25.
[H+] = 10–11 M.
Eoxide = – 0.05 –
or,
26.
0.0591
log(10–11)2
2
= – 0.05 + 0.65
H = 0.65 volt.
Standards electrode potential does not depend upon on concentration.
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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148
27.
28.
AgBr (s)  Ag+ + Br–
(s + 10–7) × s = Ksp = 12 × 10–14 .
s = 3 × 10–7 M.
[Ag+] = 4 × 10–7 M ; [Br–] = 3 × 10–7 M ; [NO3–] = 10–7 M.
Ktotal = Ag+ + Br– + NO3–
Ktotal = 4 × 10–4 × 6 × 10–3 + 3 × 10–4 × 8 × 10–3 + 1 × 10–4 × 7 × 10–3.
Ktotal = 24 + 24 + 7.
Ktotal = 55 Scm–1.
Fe  Fe2+ + 2e–,
0.44 V.
1
2H+ + O2 + 2e– H2O,
1.23 V.
2
___________________________________________________________________
1
O2 Fe2+ + H2O, Eº = 0.44 + 1.23 = 1.67 volt.
2
Gº = – 2 × 1.67 × 96500 = – 322.3 kJ.
(a)
Ag+(aq) + Cl–(aq) –– AgCl(s)
Gorxn = –109 – (–129) – 79
= –57 kJ/mole
G° = –nFE°
–57 × 100 = –1 × 96500 × E°
E° = 0.591 V
1
G° = –RT n
= –57 × 100
Ksp
log Ksp = –10
Fe + 2H+ +
29.
(b)
30.
2Ag+(aq) + Zn(s) –– Zn2+(aq) + 2Ag(s)
E° = 0.8 + 0.76 = 1.56 V
0.06
E° =
log Kq = 1.56
2
Kq = 1052
Zn + Fe2+ –– Fe + Zn2+
0.1
Q=
= 10
0.01
0.0591
E = E° –
logQ
2
0.0591
0.2905 = E° –
log(10)
2
E° = 0.32
0.0591
E° =
logK = 0.32
2
K = 100.32/0.0295
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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31.
In+  In3+ + 2e–,
E° = 0.42 V
In2+ + e  In+,
E° = – 0.4 V
______________________________
In2+  In3+ + e–
______________________________
Eº = 0.44 V
Eºcell = 0.15 + 0.44 = 0.59 V
0.059
0 = 0.59 –
log K.
1
K = 1010 .
33.
E + 0.03 = Eº –
E = Eº –
34.
[Zn 2 ]
0.06
log
.
0.5
2
[Zn 2 ]
0.06
log
. M = 0.05 M.
C
2
MnO4– ion can oxidise both Fe2+ to Fe3+ as well as Cl– to Cl2 . So Fe(NO3)2 cannot be estimated
quantitatively with MnO4– ion in HCl.
EºCell for the cell Pt, Cl2(g) (1 atm) | Cl– (aq) | | MnO4– (aq) | Mn2+ (aq). is euqal to (1.51 – 1.4)
= 0.11 V.
35.
Disproportionate reaction
36.
conductance 
38.
(i)
H2(g) + 2AgCl(s) –– 2Ag(s) + 2H+ (aq) + 2Cl–(aq)
(ii)
G° = H° – TS° = –nF°
1
Hydrated radius of ion
At 15°C : H° – 288 × S° = –2 × 96500 × 0.23
At 35°C : H° – 308 × S° = –2 × 96500 × 0.21
On solving
H° = –49587 J/mole
S° = –96.5 J/mole-K
39.
E°cell = 0.33 + 0.44 = 0.77 Volt
since E°cell is positive therefore reaction is spontaneous.
C.O.: NAIVEDHYAM, Plot No. SP-11, Old INOX, Indira Vihar, Kota (Raj.) 324005 Ph. 0744-2799900
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40.
Let initially [Cu2+] = C M
 moles of Cu2+ in the solution =
C  250
= 0.25 C moles
1000
After electrolysis
Moles of Cu2+ =
Eq. of Cu2+ =
0.25  C
2
0.25  C
× 2 = faraday of electricity.
2
0.25 × C = 2 × 10–3 × 16 × 60/69500
C
41.
2  103  16  60
= 7.95 × 10–5 M
0.25  96500
Cell reaction :
Fe2+ + Ce4+ –– Ce3+ + Fe3+
Q=1
E = E° = 1.61 – 0.77 = 0.84 V

42.
Direction of flow of current is cathode to anode which will decrease with time.
The specie having higher reduction potential oxidizes the specie having lower reduction
potential.
43.
E° = 0.77 – 0.54 = 0.23 =
0.06
logK
2
K = 6.26 × 107
44.
Ag+(cathode) –– Ag+(anode)
0.1 M
Q=
2S
2S
0.1
E = 0.164 = –
0.06
 2S 
log  
2
 0.1 
S = 9.23 × 10–3
Ksp = 4S3 = 2.287 × 10–12
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