The End of Chem1051 Classes!
– Now, Review Classes
2 more examples….
• Practice lots of examples, a few each day, do not
try and study all of 1051 in a couple of days, it will
not work!
• Read the questions! And recognise what the ‘asker’
wants!
• Get help if you need it; Resource Room,
Tutors/TAs, Tutorial, Me (e-mail: fkerton@mun.ca)
• Gold can be plated out of a solution containing
Au3+. What mass of gold (in grams) will be plated
by the flow of 5.5 A of current for 25 minutes?
• Silver can be plated out of a solution containing
Ag+. How much time (in minutes) would it take to
plate 12 g of silver using a current of 3.0 A?
• Career advice - I am happy to provide this,
arrange an appointment or just try and find me.
Slide 50
Quantitative Aspects of Electrolysis
Problem – Calculating quantities in electrolysis
1 mol e- = 96485 C
PROBLEM: A technician is plating a faucet (tap) with 0.86 g of Cr from an
electrolytic bath containing aqueous Cr2(SO4)3. If 12.5 min is
allowed for the plating, what current is needed?
PLAN:
mass of Cr needed
SOLUTION:
Cr3+(aq) + 3e-
3mol e-/mol Cr
mol of e- transferred
9.65x104C/mol
0.86g (mol Cr) (3 mol e-)
C/soC omol e- omol product og product
Cr(s)
= 0.050 mol e-
(52.00 gCr) (mol Cr)
0.050 mol
e-
(9.65x104 C/mol
e-)
=
4.8x103 C
echarge (C)
divide by time
4.8x103 C
(min)
12.5 min
(60s)
Charge (C) = current (C/s) x time (s)
Mass of product?
divide by M
mol of Cr needed
Slide 49
MASS
MASS(g)
(g)
of
ofsubstance
substance
oxidized
oxidizedor
or
reduced
reduced
M(g/mol)
AMOUNT
AMOUNT(MOL)
(MOL)
of
ofsubstance
substance
oxidized
oxidizedor
or
reduced
reduced
AMOUNT
AMOUNT(MOL)
(MOL)
of
electrons
of electrons
transferred
transferred
Faraday
constant
(C/mol e-)
= 6.4C/s = 6.4 A
balanced
half-reaction
current (A)
CHARGE
CHARGE(C)
(C)
time(s)
Slide 48
CURRENT
CURRENT(A)
(A)
Slide 47
Problem – Making qualitative predictions
continued
• Predict the half-reaction occurring at the anode and
the cathode for electrolysis of each of the following:
(b) Ag+(aq) + e2H2O(l) + 2e-
Ag(s)
E0 = -0.80V
E0 = -0.42V
H2(g) + 2OH-(aq)
Ag+ is the cation of an inactive metal and therefore will be reduced to Ag
at the cathode.
Ag(s)
Ag+(aq) + e-
– A mixture of molten AlBr3 and MgBr2
– An aqueous solution of LiI
The N in NO3- is already in its most oxidized form so water will have to be
oxidized to produce O2 at the anode.
O (g) + 4H+(aq) + 4e2H O(l)
2
(c)
Mg2+(aq)
+
2e-
2
E0
Mg(s)
= -2.37V
Mg is an active metal and its cation cannot be reduced in the presence of
water. So as in (a) water is reduced and H2(g) is produced at the cathode.
The S in SO42- is in its highest oxidation state; therefore water must be
oxidized and O2(g) will be produced at the anode.
Slide 46
Problem – Predicting half- and overall reactions in
electrolysis
Slide 45
Complications in Electrolytic Cells
PROBLEM: What products form during electrolysis of aqueous solution of the
following salts: (a) KBr; (b) AgNO3; (c) MgSO4?
PLAN:
Compare the potentials of the reacting ions with those of water,
remembering to consider the 0.4 to 0.6V overvoltage.
The reduction half-reaction with the less negative potential, and the oxidation halfreaction with the less positive potential will occur at their respective electrodes.
SOLUTION:
(a) K+(aq) + e2H2O(l) +
K(s)
2e-
H2(g) +
2OH-(aq)
E0 = -2.93V
E0 = -0.42V
The overvoltage would make the water reduction -0.82 to -1.02 but the
reduction of K+ is still a higher potential so H2(g) is produced at the cathode.
2Br-(aq)
2H2O(l)
Br2(g) + 2eO2(g) + 4H+(aq) + 4e-
•
•
•
•
Overpotential.(1)
Competing reactions.(2)
Non-standard states.(3)
Nature of electrodes.(4)
E0 = 1.07V
E0 = 0.82V
The overvoltage would give the water half-cell more potential than
the Br-, so the Br- will be oxidized. Br2(g) forms at the anode.
Slide 44
Slide 43
The tin-copper reaction as the basis of a voltaic (galvanic) and an
electrolytic cell.
Comparison of Voltaic and Electrolytic Cells
Electrode
Cell Type
'G
Ecell
Name
Process
Sign
Voltaic
<0
>0
Anode
Oxidation
-
Voltaic
<0
>0
Cathode
Reduction
+
Electrolytic
>0
<0
Anode
Oxidation
+
Electrolytic
>0
<0
Cathode
Reduction
-
voltaic cell
Slide 42
20-7 Electrolysis: Causing
Non-spontaneous Reactions to Occur
electrolytic cell
Oxidation half-reaction
Sn(s) Sn2+(aq) + 2eReduction half-reaction
Cu2+(aq) + 2eCu(s)
Overall (cell) reaction
Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
0.0592 V [Ag+]sat’d AgI
log
[Ag+]0.10 M soln
n
Galvanic Cell: also known as voltaic
Ecell = Ecell° -
Eocell= 1.103 V
Electrolytic Cell:
Zn2+(aq) + Cu(s) ĺ Zn(s) + Cu2+(aq)
Slide 41
Example 20-10
Ecell = Ecell° -
Zn(s) + Cu2+(aq) ĺ Zn2+(aq) + Cu(s)
Oxidation half-reaction
Cu(s)
Cu2+(aq) + 2eReduction half-reaction
Sn2+(aq) + 2eSn(s)
Overall (cell) reaction
Sn(s) + Cu2+(aq) Sn2+(aq) + Cu(s)
0.417 =0 Eocell = -1.103 V
x
0.0592 V
log
0.100
n
0.0592 V
(log x – log 0.100)
1
log x = log 0.100-
0.417
= -1 – 7.04 = -8.04
0.0592
x = 10-8.04 = 9.1 x 10-9
Batteries and Corrosion (Sections 20-5 and 20-6) omitted
Slide 40
Ksp = x2 = 8.3 x 10-17
Slide 39
Example 20-10
Measurement of Ksp
Using a Voltaic Cell to Determine Ksp of a Slightly
Soluble Solute.
Ag|Ag+(sat’d AgI)||Ag+(0.10 M)|Ag(s)
With the data given for the reaction on the previous slide,
calculate Ksp for AgI.
Ag+(0.100 M) + e- ĺ Ag(s)
Ag(s) ĺ Ag+(sat’d) + e-
AgI(s) ĺ Ag+(aq) + I-(aq)
Let [Ag+] in a saturated Ag+ solution be x:
Ag+(0.100 M) ĺ Ag+(sat’d M)
Ag+(0.100 M) ĺ Ag+(sat’d M)
[Ag+]sat’d AgI
0.0592 V
0.0592 V
log
log Q = Ecell° Ecell = Ecell [Ag+]0.10 M soln
n
n
Slide 38
PROBLEM: A concentration cell consists of two Ag/Ag+ half-cells. In
half-cell A, electrode A dips into 0.0100M AgNO3; in half-cell
B, electrode B dips into 4.0x10-4M AgNO3. What is the cell
potential at 298 K? Which electrode has a positive charge?
PLAN: E0cell will be zero since the half-cell potentials are equal. Ecell
is calculated from the Nernst equation with half-cell A (higher
[Ag+]) having Ag+ being reduced and plating out, and in halfcell B Ag(s) will be oxidized to Ag+.
Slide 37
Concentration Cells
Ecell = Ecell° Ecell = Ecell°Ecell = 0 -
SOLUTION: Ag+(aq, 0.010M) half-cell A
Ecell = E0cell -
0.0592V
1
log
Ag+(aq, 4.0x10-4M) half-cell B
0.0592 V
log Q
n
2 H+(1 M) ĺ 2 H+(x M)
0.0592 V
x2
log
12
n
0.0592 V
x2
log
1
2
Ecell = - 0.0592 V log x
[Ag+]dilute
[Ag+]concentrated
Ecell = (0.0592 V) pH
Ecell = 0 V - 0.0592 log 4.0x10-2 = 0.0828V
Half-cell A is the cathode and has the positive electrode.
Slide 36
Slide 35
Concentration Cells
Problem – Predicting spontaneous reactions for
non-standard conditions
Two half cells with identical electrodes
but different ion concentrations.
• Will the cell reaction proceed spontaneously as
written for the following cell?
Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)
Sn(s)|Sn2+(0.50 M)||Pb2+(0.0010 M)|Pb(s)
2 H+(1 M) + 2 e- ĺ H2(g, 1 atm)
H2(g, 1 atm) ĺ 2 H+(x M) + 2 e2 H+(1 M) ĺ 2 H+(x M)
Slide 34
PROBLEM: In a test of a new reference electrode, a chemist constructs a
voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the
following conditions:
[Zn2+] = 0.010M
[H+] = 2.5M
PH = 0.30atm
Example 20-8
Ecell = Ecell° -
2
Calculate Ecell at 298 K.
Slide 33
0.0592 V
log Q
n
0.0592 V
[Fe3+]
Ecell = Ecell° log
[Fe2+] [Ag+]
n
PLAN: Find E0cell and Q in order to use the Nernst equation.
SOLUTION: Determining E0cell :
2H+(aq) + 2e-
H2(g)
E0 = 0.00V
Zn2+(aq) + 2e-
Zn(s)
E0 = -0.76V
Zn2+(aq) + 2e-
E0 = +0.76V
Zn(s)
Ecell =
E0cell
-
0.0592V
n
log Q
Q=
P x [Zn2+]
Ecell = 0.029 V – 0.018 V = 0.011 V
H2
[H+]2
Q=
(0.30)(0.010)
Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)
(2.5)2
Q = 4.8x10-4
Fe2+(aq) + Ag+(aq) ĺ Fe3+(aq) + Ag (s)
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
Slide 32
Slide 31
Example 20-8
20-4 Ecell as a Function of Concentration
Applying the Nernst Equation for Determining Ecell.
ǻG = ǻG° + RT ln Q
What is the value of Ecell for the voltaic cell pictured
below and diagrammed as follows?
-nFEcell = -nFEcell° + RT ln Q
Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)
Ecell = Ecell° -
RT
nF
ln Q
Convert to log10 and calculate constants
The Nernst Equation: Ecell = Ecell° -
0.0592 V
log Q
n
Slide 30
Summary of Important Thermodynamic, Equilibrium
and Electrochemistry Relationships
Slide 29
Problem – Relating Keq to Eocell for a redox reaction
PROBLEM:
Lead can displace silver from solution:
Pb(s) + 2Ag+(aq)
Pb2+(aq) + 2Ag(s)
As a consequence, silver is a valuable by-product in the industrial extraction
of lead from its ore. Calculate K and 'G0 at 298 K for this reaction.
PLAN: Break the reaction into half-reactions, find the E0 for each half-reaction
and then the E0cell.
SOLUTION:
2X
Pb2+(aq) + 2eAg+(aq) + e-
Pb(s)
Ag(s)
Pb(s)
Pb2+(aq) + 2e+
Ag (aq) + e
Ag(s)
E0cell = 0.025693V ln Keq
n
n x E0cell
(2)(0.93V)
lnK =
=
0.025693V
0.025693V
Slide 28
E0 = -0.13V
E0 = 0.80V
E0 = 0.13V
E0 = 0.80V
E0cell = 0.93V
'G0 = -nFE0cell = -(2)(96.5kJ/mol*V)(0.93V)
K = 2.75x1031
'G0 = -1.8x102kJ
Slide 27
Relationship Between E°cell and Keq
Relative Reactivities (Activities) of Metals
1. Metals that can displace H from acid
ǻG° = -RT ln Keq = -nFE°cell
E°cell =
RT
nF
2. Metals that cannot displace H from acid
ln Keq
3. Metals that can displace H from water
4. Metals that can displace other metals from
solution
Slide 26
The Behavior of Metals Toward Acids
Problem – Making qualitative predictions
• Peroxodisulfate salts such as Na2S2O8 are good
M(s) ĺ M2+(aq) + 2 e- E° = -E°M2+/M
2 H+(aq) + 2 e- ĺ H2(g)
2
H+(aq)
+ M(s) ĺ H2(g) +
Slide 25
oxidizing agents used in bleaching. Dichromates
E°H+/H2 = 0 V
such as K2Cr2O7 have been used as laboratory
M2+(aq)
oxidizing agents. Which is the better oxidizing
E°cell = E°H+/H2 - E°M2+/M = 0 -E°M2+/M = -E°M2+/M
agent in acidic solution under standard conditions,
Cr2O72- or S2O82-?
When E°M2+/M < 0, E°cell > 0. Therefore ǻG° < 0.
Metals with negative reduction potentials react with acids
All metals below hydrogen in table of E o should displace H2(g)
from acidic solutions (Pb through Li)
Slide 24
Slide 23
Problem – Applying the criterion for spontaneous
change in a redox reaction
Spontaneous Change
PROBLEM: (a) Combine the following three half-reactions into three balanced
equations (A, B, and C) for spontaneous reactions, and
calculate E0cell for each.
(b) Rank the relative strengths of the oxidizing and reducing agents:
(1) NO3-(aq) + 4H+(aq) + 3e(2) N2(g) + 5H+(aq) + 4e-
N2H5+(aq)
(3) MnO2(s) +4H+(aq) + 2ePLAN:
NO(g) + 2H2O(l)
E0 = 0.96V
E0 = -0.23V
Mn2+(aq) + 2H2O(l)
E0 = 1.23V
Put the equations together in varying combinations so as to produce
(+) E0cell for the combination. Since the reactions are written as
oxidation, reverse the sign of
– Reaction proceeds spontaneously as written.
• E°cell = 0
– Reaction is at equilibrium.
• E°cell < 0, (Ecell -ve)
reductions, remember that as you reverse one reaction for an
E0.
• ǻG < 0 for spontaneous change.
• Therefore:
• E°cell > 0, (Ecell +ve)
Balance the number of electrons
– Reaction proceeds in the reverse direction spontaneously.
gained and lost without changing the E0.
In ranking the strengths, compare the combinations in terms of E0cell.
Slide 22
Combining Half Reactions
Fe3+(aq) + 3e- ĺ Fe(s)
Fe2+(aq) + 2e- ĺ Fe(s)
Fe3+(aq) + 3e- ĺ Fe(s)
20-3 Ecell, ǻG, and Keq
• Cells do electrical work.
E°Fe3+/Fe = ?
– Moving electric charge.
E°Fe2+/Fe = -0.440 V ǻG° = +0.880 J
Fe3+(aq) + 3e- ĺ Fe2+(aq)
Slide 21
Zelec = nFE
• Faraday constant, F = 96,485 C mol-1
E°Fe3+/Fe2+ = 0.771 V ǻG° = -0.771 J
E°Fe3+/Fe = +0.331 V ǻG° = +0.109 V
ǻG = -nFE
ǻG° = -nFE°
ǻG° = +0.109 V = -nFE°
E°Fe3+/Fe = +0.109 V /(-3F) = -0.0363 V
Slide 20
Slide 19
Problem – Determining Eo from an Eocell measurement
PROBLEM: A voltaic cell houses the reaction between aqueous bromine
and zinc metal:
Br2(aq) + Zn(s)
Zn2+(aq) + 2Br-(aq)
E0cell = 1.83V
Calculate E0bromine given E0zinc = -0.76V
anode: Zn(s)
Zn2+(aq) + 2e-
E0Zn as Zn2+(aq) + 2e-
A new battery system currently under study for possible use in
electric vehicles is the zinc-chlorine battery. The overall reaction
producing electricity in this cell is Zn(s) + Cl2(g) o ZnCl2(aq).
What is Eocell of this voltaic cell?
PLAN: The reaction is spontaneous as written since the E0cell is (+).
Zinc is being oxidized and is the anode. Therefore the
E0bromine can be found using E0cell = E0cathode - E0anode.
SOLUTION:
Problem – Calculating Eocell for a reaction
E = +0.76
Zn(s) is -0.76V
E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76)
E0bromine = 1.83 - 0.76 = +1.07 V
Slide 18
Standard Reduction Potentials
Slide 17
Measuring Standard Reduction Potential
anode
Slide 16
cathode
cathode
anode
Slide 15
Standard Cell Potential
Reduction Couples
Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) E°cell = 0.340 V
Cu2+(1M) + 2 e- ĺ Cu(s)
E°Cu2+/Cu = ?
E°cell = E°right - E°left
Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) E°cell = 0.340 V
anode
cathode
E°cell = E°Cu2+/Cu - E°H+/H2
0.340 V = E°Cu2+/Cu - 0 V
Standard cell potential: the potential difference of a
cell formed from two standard electrodes.
E°Cu2+/Cu = +0.340 V
H2(g, 1 atm) + Cu2+(1 M) ĺ H+(1 M) + Cu(s)
E°cell = 0.340 V
E°cell = E°(right)cathode - E°(left)anode
Slide 14
Standard Electrode Potential, E°
Slide 13
Standard Hydrogen Electrode
2 H+(a = 1) + 2 e- l H2(g, 1 bar)
• E° defined by international agreement.
• The tendency for a reduction process to occur at
an electrode.
– All ionic species present at a=1 (approximately 1 M).
– All gases are at 1 bar (approximately 1 atm).
– Where no metallic substance is indicated, the potential is
established on an inert metallic electrode (ex. Pt).
Slide 12
E° = 0 V
Pt|H2(g, 1 bar)|H+(a = 1)
Slide 11
Problem - Representing a Redox Reaction by
Means of a Cell Diagram
20-2 Standard Electrode Potentials
• Cell voltages, the potential differences between
electrodes, are among the most precise scientific
measurements.
• The potential of an individual electrode is difficult to
establish.
• Arbitrary zero is chosen.
PROBLEM: Diagram, show balanced equations, and write the notation for a
voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3
solution, another half-cell with an Ag bar in an AgNO3 solution, and
a KNO3 salt bridge. Measurement indicates that the Cr electrode is
negative relative to the Ag electrode.
PLAN:
Identify the oxidation and reduction reactions and write each halfreaction. Associate the (-)(Cr) pole with the anode (oxidation) and
the (+) pole with the cathode (reduction).
SOLUTION:
e-
Oxidation half-reaction
Cr(s)
Cr3+(aq) + 3e-
The Standard Hydrogen Electrode (SHE)
Reduction half-reaction
Ag+(aq) + eAg(s)
Voltmeter
salt bridge
Cr
K+
Ag
NO3Cr3+
Ag+
Overall (cell) reaction
Cr3+(aq) + 3Ag(s)
Cr(s) + 3Ag+(aq)
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
Slide 10
Slide 9
Terminology
A voltaic (galvanic) cell based on the zinc-copper reaction.
• Galvanic cells.
– Produce electricity as a result of spontaneous
reactions, e.g. Cells on slides 4 & 6
• Electrolytic cells.
– Non-spontaneous chemical change driven by
electricity.
• Couple, M|Mn+
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e-
Reduction half-reaction
Cu(s)
Cu2+(aq) + 2e-
– A pair of species related by a change in number of e-.
Overall (cell) reaction
Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)
Slide 8
Slide 7
Terminology
Terminology
• Electromotive force, Ecell.
– The cell voltage or cell potential.
• Cell diagram.
– Shows the components of the cell in a symbolic way.
– Anode (where oxidation occurs) on the left.
– Cathode (where reduction occurs) on the right.
• Boundary between phases shown by |.
• Boundary between half cells
(usually a salt bridge) shown by ||.
Ecell = 1.103 V
Slide 6
An Electrochemical Cell
Slide 5
An Electrochemical Half-Cell
Anode
Cathode
Slide 4
Slide 3
Electrode Potentials and Their Measurement
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring • Madura
9th Edition
Cu(s) + 2Ag+(aq)
Cu(s) + Zn2+(aq)
Chapter 20: Electrochemistry
Cu2+(aq) + 2 Ag(s)
No reaction
Fran Kerton
Memorial University of Newfoundland,
St. John’s, Canada
A1B 3X7
Slide 2