Spontaneity of Redox Reactions
How is
related to other thermodynamic quantities such as G
and K. In an electrochemical cell, chemical energy is converted to
electrical energy. Electrical energy in this case is the product of the emf
of the cell and the total electrical charge (in coulombs) that passes
through the cell:
The total energy charge is determined by the number of moles of
electrons(n) that pass through the circuit. By definition
Where F is Faraday constant, is the electrical charge contained in 1 mole
of electrons. Expts. Show 1 faraday is equivalent to 96,485.3 coulombs
or 96,500 coulombs ( roughly).
Chang, R.(2002). Chemistry 7th
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1F = 96,500 C/mol
Because 1J = 1 C X 1V
We can also express the units of faraday as 1 F = 96,500 J/V. mol
The measured emf is the max. voltage that a cell can achieve. This value is
used to calculate the max. amount of electrical energy obtained from the
chemical reaction. This energy is used to do the electrical work (Wele)
hence
Wmax =W ele= - n F Ecell
The –ve sign on the right hand side indicates that the electrical work done
by the system on the surroundings.
G is the free energy available to do work , hence G = Wmax
We can now write G = -n F Ecell both n and F are positive quantities and
G is negative quantity for a spontaneous process, so Ecell must be positive
quantity.
Under standard state conditions the above eqn. becomes
G0 = -n F E0cell ………………..(1)
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From our study on second law of thermodynamics we know the
relation between free energy and equilibrium constant (K)
G = - RT ln K …………(2)
Combining eqn. (1) and (2)
-n F E0cell= - RT ln K
Solving for E0cell =
…………(3)
when T = 298K, eqn (3) can be simplified by substituting for R and
F:
E0cell =
=
the eqn can be rewritten using base 10 logarithm
E0cell =
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Thus, if any of the three quantities G , K or E0cell is known, the
other quantities can be calculated.
Chang, R.(2002). Chemistry 7th
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Eg. Calculate the equilibrium constant for the following reaction at 25
C:
Sn(s) +2Cu2+ (aq)
 Sn2+(aq) +2 Cu+ (aq)
Solution : to calculate the K first calculate E0cell of the Sn2+/Sn and
Cu2+ /Cu couples. The half cell reaction are
Oxidation:
Sn(s)

Sn2+(aq) + 2eReduction: 2Cu2+ (aq) +2e- 
2 Cu+ (aq)
From table we find
E0 Sn2+/Sn = -0.14 V and E0 Cu2+/Cu+ = 0.15 V. Thus
E0 cell = E0 Cu2+/Cu+ - E0 Sn2+/Sn=0.15 V – (-0.14)= 0.29V
E0 cell
=
In the overall reaction we find n=2. Therefore
lnK =22.6 K=e22.6 K = 6.5 X 109
Practice: Calculate the equilibrium constant for the following reaction at
25 C : Fe2+(aq) +2Ag(s)  Fe(s) +2Ag+(aq)
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Calculate the standard free energy change for the following reaction at
25˚ C.
2Au(s) + 3 Ca2+ (1M)  2Au3+(1M) + 3Ca(s)
To calculate G0 for the reaction we can use G0 = -n F E0 , first we
need to calculate E0 .
Calculate G0 for the following reaction at 25˚ C.
2 Al 3+ (aq) + 3Mg(s)  2Al(s) + 3Mg 2+ (aq)
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The Effect of Concentration on Cell Emf
The discussion so far were on redox reactions in their standard state
conditions, but standard states are difficult to maintain and sometimes
impossible. Nevertheless, there is a mathematical relationship between
the emf of a cell and the concentration of reactants and products in a
redox reaction under non standard state conditions.
The Nernst Equation
Consider a redox reaction of the type
aA + bB  cC + dD
from equation G =  G˚ + RT ln Q ( from thermodynamics)
Because G =  nFE and G˚ =  nFE˚, the equation can be
expressed as
 nFE =  nFE˚ + RT ln Q dividing this eqn by –nF, we get
E=E˚ 
lnQ
(The Nernst Equation)
where Q is the reaction quotient
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E = E˚ 
ln Q
At 298 K, the Nernst equation becomes E = E˚ 
(nat log)
or E = E˚ 
log Q (base 10 log)
ln Q
At equilibrium, there is no net transfer of electrons, so E=0, Q=K,
where K is the equilibrium.
The Nernst equation enables us to calculate E as a function of reactant
and product concentrations in a redox reaction. For example, for the
Daniell cell Zn(s) + Cu 2+ (aq)  Zn2+ (aq) +Cu(s)
E = E˚ 
ln
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If the ratio [Zn2+/Cu2+] is less than 1, ln [Zn2+/Cu2+] is a negative number, so
the second term on the right- hand side of the above equation is positive.
Under this condition E is greater than the standard emf E˚. If the ratio is
greater than 1, E is smaller than E˚.
Predict whether the following reaction would proceed spontaneously as
written at 298 K:
Co(s) + Fe 2+(aq)  Co2+ (aq) + Fe(s) given that [Co2+ ]= 0.15 M and
[Fe2+ ]= 0.68 M.
Reasoning and solution : We can use the value of E for the redox process to
determine the spontaneity of the reaction. To do so we need to know E˚ and
ln Q in the Nernst equation. The half reaction reactions are
Oxidation: Co(s)  Co2+ (aq) + 2eReduction: Fe 2+(aq) + 2e-  Fe(s)
The values of E˚Co2+/Co= - 0.28V and E˚ Fe2+/Fe= -0.44V. There fore the
standard emf is
E˚ = E˚ Fe2+/Fe - E˚Co2+/Co= -0.44V-(-0.28V) = -0.16V
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Lets find the value of E
E = E˚ -
= - 0.16 V = - 0.16+ 0.019V = - 0.14V
Because E is negative, the reaction is not spontaneous in the direction written.
Practice ques: Will the following reaction occur spontaneously at 25˚C, given
that the [Fe 2+]=0.06M and [Cd2+]= 0.010M?
Cd(s) + Fe 2+(aq) → Cd2+(aq) + Fe(s)
Chang, R.(2002). Chemistry 7th
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Now suppose you want to know what ratio of [Co2+]/[Fe2+] the reaction will
would be spontaneous we can use the equation
E = E˚ We set E=0, which corresponds to the equilibrium situation
0 = - 0.16 V = -12.5
= e-12.5= K =4 X 10 -6
Thus for the reaction to be spontaneous, the ratio [Co2+]/[Fe2+] must be
smaller 4 X 10 -6.
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Consider the electrochemical cell shown below. In a certain experiment, the
emf(E) of the cell is found to be 0.54V. Suppose that the [Zn2+]= 1.0M and
pH2=1 atm. Calculate [H+].
This can be solved using
the Nernst equation. Note
that the conc. of gas
expressed in atm.
Write the redox reaction to find n, we know the standard reduction
potential of E˚ Zn2+ /Zn= 0.76V, substitute the values. ( ans= 2X10-4M)
It may be useful to realize here that an electrochemical cell which involves the
measurement of H+ ions can be used to measure pH. This principle is used
in pH meter.
Home work: What is the emf of a cell consisting of a Cd2+/ Cd half cell and a
Pt/H+/H2 half cell if [Cd2+ ] = 0.20M, [H+]= 0.16M and pH2=0.80atm?
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Concentration cells
Electrode potentials depend on ion concentration, it is possible to
construct a cell from two half cell of the SAME material but differing in
ion conc. Such a cell is called concentration cell.
Two zinc electrodes in two different aq. Solution of zinc sulphate. Eg 1.0M
and 0.1M
There fore reduction occur in 1.0M compartment and oxidation occur in
0.1M compartment. The cell diagram
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And the half-reactions are
Oxidation:
Zn(s) → Zn 2+ (0.10M) +2eReduction: Zn 2+ (1.0M) +2e-→ Zn(s)
Overall: : Zn 2+ (1.0M) → Zn 2+ (0.10M)
The emf of the cell is
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The
of this cell is zero ( same electrode, same ions) so
the emf E of the cell is
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The potential is small which eventually becomes zero when
the conc. Becomes same.
A biological cell can be compared to conc. Cell. Membrane
potential develop due to unequal conc of same ions on
different sides of membrane. Nerve impulses, heart beat,
communication in living cells.
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Batteries
An electrochemical cell or a series of combined electrochemical cells that
can be used as a source of direct electric current at a constant voltage. The
operation of battery is similar in principle to the elctrochemical cells
(galvanic, daniell, voltaic), a battery has the advantage of being completely
self contained . ( doesn’t require salt bridge or other components).
Types of batteries
The dry cell battery or Leclanche’ cell
The mercury battery
The lead storage battery
The solid –state lithium battery
Fuel cells
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Corrosion: Deterioration of metals by an electrochemical process.
Corrosion causes enormous damage-buildings, bridges, ships, automobiles etc.
Examples-Rusting of iron, tarnish on silver, green patina on copper or brass
etc.
What happens in rusting?
In the formation of rust on iron, oxygen and water must be present. The reaction
involved are quite complex and not completely understood, the main steps are
believed to be as follows.
The metal surface acts as the anode where oxidation occurs
Fe (s) → Fe2+ (aq) + 2eE=- 0.45 V
The electrons given up by iron reduce atmospheric oxygen to water at cathode
which is another region of the same metal surface:
O2 (g) + 4H+(aq) + 4e- → 2H2O(l)
E= 1.23V
(note that the E of the reduction half reaction will be less than 1.23V because the
H+ is not 1 M but even at pH 7 the reduction half reaction is 0.81 V, which
means the cell potential is positive and the reaction is spontaneous.)
The overall reaction is
2Fe (s) + O2(g)+ 4H+(aq)→ 2Fe2+(aq) +2H2O(l)
E= Ecat - Eand
= 1.23V-(-0.44V)= 1.67 V
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Notice the reaction occurs in acidic medium : the H+ ions are supplied in part
by the reaction of atmospheric CO2 with H2O to form H2CO3. the Fe2+ ions
is further oxidized by oxygen.
 Fe2+ (aq) + O2(g) + (4 + 2x) H2O(l)→ 2Fe2O3. xH2O + 8H+ (aq)
 The hydrated form of iron(III) oxide is called rust. As the amount of water
associated with iron oxide varies, we represent the formula as 2Fe2O3. xH2O
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The electrical circuit is completed by the migration of electrons and ions
causing rusting rapidly. The fast rusting of automobiles in cold placed is
due to salts (NaCl or CaCl2) spread on the roads to melt ice.
Aluminum has greater tendency to oxidize than Iron (compare reduction
potential). We might expect to see lot of corroded aluminum like soda
cans, our cooking pots etc. How ever this doesn’t happen because Al forms
insoluble Al2O3 layer when it is exposed in air. This layer is not porous (
like iron) and protects the underlying metal.
Copper forms a layer of green substance called patina (CuCO3) which
protects the underlying metal
Silver forms silver sulfide when it comes in contact with food stuff.
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Methods adopted to protect metals from corrosion.
Most methods are targeted to prevent rust formation.
Painting metal surface with paint. But sooner the paint is scratched the iron is
exposed and rusting starts.
Metals such as chromium, tin or zinc afford a more durable surface coating.
The steel used in making automobiles, for example, is coated by dipping it
in a bath of molten zinc, a process known as galvanizing.
As the potentials indicate
Fe 2+(aq) + 2e-  Fe(s)
E˚= -0.45V
Zn 2+ (1.0M) +2e-→ Zn(s)
E˚= -0.76 V
Zinc is oxidized more easily than iron, and therefore when the metal corrodes,
zinc is oxidizes instead of iron. An incipient oxidation of iron would
reverse the process immediately since Zn can reduce Fe2+.
As long as zinc and iron are in contact , Zn protects iron from oxidation even
if zinc layer is scratched.
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Cathodic protection is
The technique of protecting a metal from corrosion by connecting it to
second metal that is more easily oxidized.
It is unnecessary to cover the entire surface of the metal with second
metal, as in galvanizing. All it requires is an electrical contact with the
second metal. An under steel pipeline can be protected by connecting it
through an insulated wire to a stake of magnesium, which acts as a
sacrificial anode and corrodes instead of the iron. In effect the arrangement
is a galvanic cell in which the easily oxidized magnesium acts as the anode
and the pipeline behaves as the cathode and the moist soil acts as the
electrolyte.
Anode: Mg(s)
Mg2+ (aq) + 2eE˚= 2.37V*
Cathode: O2(g) + 4 H+(aq) +4e2H2O(l)
E˚ =1.23V*
*written as oxidation potential
PROBLEM
Mg is often attached to the hull of ships to protect to steel from rusting .
Write a balanced equation for the corrosion reactions that occur (a) in the
presence of Mg and (b) in the absence of Mg.
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Electrolysis
Process in which electrical energy is used to cause a nonspontaneous
chemical reaction to occur. An electrolytic cell is an apparatus used for
carrying out electrolysis. The same principles underlie electrolysis and the
processes that take place in a electrochemical cell.
Electrolysis of molten sodium chloride.
Sodium chloride is an ionic compound which can be electrolyzed to form
sodium metal and chlorine. The cell used for large scale electrolysis of
NaCl is called a Down’s cell.
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The reactions at the electrodes are
 Anode(oxidation) :
2Cl-(l)→ Cl2(g)+ 2 e Cathode(reduction: 2Na+(l) +2e-→ 2Na(l)
 Overall;
2Na+(l) + 2Cl-(l)→ 2Na(l)+ Cl2(g)
 This process is the major source of pure sodium metal and
chlorine gas.
 Theoretical aspects show that the E˚ value for the overall process is about
-4V, which means the reaction is non spontaneous process. There fore a
minium voltage of 4V must be supplied by the battery to carry out the
reaction. Inpractice however a higher voltage is required because of
inefficiencies in the electrolytic process because of over voltage.
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Electrolysis of water. Study on your own
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ELECTROLYSIS of of Aqueous Sodium Chloride
This is the most complicated electrolysis. In the electrolysis of
aqueous NaCl, the solution contains several species that could be
oxidized or reduced. The possible oxidations reaction at anode are
2Cl-(aq)→Cl2(g) + 2e2H2O(l)→O2(g) + 4H+ (aq) +4eComparing their standard electrode potential
Cl2(g) + 2e-→2Cl-(aq)
E= 1.36V
O2(g) + 4H+ (aq) +4e-→2H2O(l)
E= 1.23V
The E are not very different but the values suggest that
H2O should be preferentially oxidized at the anode. But in the
actual experiment the gas liberated is Cl2 and not O2. In studying
electrolytic process we sometimes find that the voltage required
for a reaction is higher than the electrode potential indicates.
Overvoltage is the difference between the electrode
potential and the actual voltage required to cause
electrolysis. The over voltage for O2 is quite high. Therefore
under normal operating conditions Cl2 is formed instead of O2.
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The reductions that might occur are
2H+(aq) + 2e- →H2(g)
E= 0.00V
2H2O(l)+2e- → H2(g) + 2OH- (aq)
E= -0.83V
Na+ (aq) + e-→ Na(s)
E= -2.71V
Reaction 3 is ruled out because it has a very negative
potential. Reaction 1 is preferred over reaction 2 but at pH
7 both are equally probable. Reaction 2 is used to describe
the cathode reaction because the concentration of H+ is too
low ( 1 X10-7 M) to make reaction 1 reasonable
Thus the half cell reactions are
Anode (oxidation): 2Cl-(aq)→Cl2(g) + 2eCathode (reduction) 2H2O(l)+2e- → H2(g) + 2OH- (aq)
Over all:
2H2O(l)+ 2Cl-(aq)→ H2(g) + Cl2(g) +2OH(aq)
The over all reaction shows that as the Cl- ion concentration
decreases OH- ion increases. In addition to H2 and Cl2 gas
produced, a useful by product NaOH is obtained by the
aqueous solution at the end of electrolysis
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Note that from the analysis of electrolysis : cat ions are likely to be
reduced at the cathode and the anions likely to be oxidized at anode and in
aqueous solutions water itself may be oxidized or reduced. The out come
depends on the nature of other species.
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Quantitative aspects of Electrolysis
Developed primarily by Faraday- he observed that the mass of the product
formed ( or reactant consumed) at an electrode is proportional to both the
amount of electricity transferred at the electrode and the molar mass of the
substance in question.
For E.g. In the electrolysis of molten NaCl, The cathode reaction tells us
that one Na+ ion accepts an electron from the electrode. To reduce one
mole of Na+ ions, we must supply Avogadro’s number (6.02 X 1023) of
electrons to the cathode. On the other hand, the stoichiometry of the anode
reaction shows that the oxidation of two Cl- ion yields one chlorine
molecule. Therefore, the formation of 1 mole of Cl2 results in the transfer
of 2 mols of electrons from Cl- ions to the anode. Similarly, it takes 2
moles of electrons to reduce 1 mole of Mg2+ ions and three moles of
Chang, R.(2002). Chemistry 7th
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In an electrolysis experiment, we measure the current ( in amperes, A) that
passes through an electrolytic cell in a given period of time. The
relationship between charge ( in coulombs, C) and current is
1C= 1A X 1s
That is, a coulomb is a quantity of electrical charge passing any point in 1
second when the current is 1 ampere. Steps involved in calculating the
quantities of subs produced in electrolysis.
For example: Consider electrolysis for molten CaCl2 in an electrolytic cell.
Suppose a current of 0.452 A is passed through the cell for 1.50 hr. How
much product will be formed at cathode and anode.
Solution write the half reactions.
Anode ( oxidation):
2Cl-(l) → Cl2(g) + 2 eCathode ( reduction): Ca2+(l) + 2 e-→ Ca (l)
Over all :
Ca2+(l) + 2Cl-(l) → Ca (l)+ Cl2(g) Chang, R.(2002). Chemistry 7th
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The quantities of calcium metal and chlorine gas formed depends depends
on current X time, or charge;
C= ?
C= 0.425A X 1.5 hr X 3600s/hr X1 C /1A.s= 2.44 X 103 C
1 mol e- =96,500 C and 2 mol of e- are required to reduce 1mol of Ca2+ ions,
the mass of Ca metal formed at the cathode is calculated as follows
No. of mols of electron = 2.44 X 103 C / 96,500 C mol-1
2 mols of e- reduce 1 mole of Ca2+
Therefore, 2.44 X 103 C / 96,500 C mol-1 will reduce
=1/2 X 2.44 X 103 C / 96,500 C mol-1 =0.01264 mols
Grams of Ca formed = 0.01264 mols X40.08g/mol
=0.5067g
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ed.Mc graw Hill.
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Chang, R.(2002). Chemistry 7th
ed.Mc graw Hill.
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