Chapter 21
Electrochemistry: Chemical Change and
Electrical Work
21-1
Electrochemistry: Chemical Change and Electrical Work
21.1 Half-reactions and electrochemical cells
21.2 Voltaic cells: Using spontaneous reactions to generate
electrical energy
21.3 Cell potential: Output of a voltaic cell
21.4 Free energy and electrical work
21.5 Electrochemical processes in batteries
21.6 Corrosion: A case of environmental electrochemistry
21.7 Electrolytic cells: Using electrical energy to drive a
non-spontaneous reaction
21-2
Key points about redox reactions
 Oxidation (electron loss) always accompanies reduction
(electron gain).
 The oxidizing agent is reduced and the reducing agent
is oxidized.
 The number of electrons gained by the oxidizing agent
always equals the number lost by the reducing agent.
21-3
A summary of redox terminology
OXIDATION
One reactant loses electrons.
Reducing agent is oxidized.
Oxidation number increases.
REDUCTION
Other reactant gains electrons.
Oxidizing agent is reduced.
Oxidation number decreases.
Figure 21.1
21-4
Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
Zn loses electrons.
Zn is the reducing
agent and becomes
oxidized.
The oxidation number
of Zn increases from 0
to +2 (each Zn loses 2
electrons).
Hydrogen ion gains
electrons.
Hydrogen ion is the oxidizing agent
and becomes reduced.
The oxidation number of H decreases
from +1 to 0 (each H+ ion gains one
electron).
Half-reaction method for balancing redox reactions
This method divides the overall redox reaction into oxidation and
reduction half-reactions.

Each reaction is balanced for mass (atoms) and charge.

One or both are multiplied by an integer to make the number of
electrons gained and lost equal.

The half-reactions are recombined to give the balanced redox
equation.
Advantages

The separation of half-reactions reflects the actual physical
separations in electrochemical cells.
 The half-reactions are easier to balance, especially if they involve
acid or base.
 It is usually unnecessary to assign oxidation numbers to those
species not undergoing change.
21-5
Balancing redox reactions in acidic solution
Cr2O72-(aq) + I-(aq)
Cr3+(aq) + I2(aq)
1. Divide the reaction into half-reactions
Determine ONs for the species undergoing redox.
+6
-1
2
Cr2O7 (aq) + I-(aq)
Cr2O72I-
Cr3+
I2
+3
0
+
3
Cr (aq) + I2(aq)
Cr is going from +6 to +3
I is going from -1 to 0
2. Balance atoms and charges in each half-reaction
14H+(aq) + Cr2O72net: +12
6e- + 14H+(aq) + Cr2O7221-6
2 Cr3+ + 7H2O(l)
net: +6
Add 6 e - to
left.
2 Cr3+ + 7H2O(l)
6e- + 14H+(aq) + Cr2O72-
2 Cr3+ + 7H2O(l)
2I -
+ 2e-
I2
Cr(+6) is the oxidizing agent and I(-1) is the reducing agent.
3. Multiply each half-reaction by an integer, if necessary
2 I-
I2 + 2e-
x 3
4. Add the half-reactions together
6e- + 14H+ + Cr2O726 I14H+(aq) + Cr2O72-(aq) + 6 I-(aq)
2Cr3+
+ 7H2O(l)
3 I2 + 6e2Cr3+(aq) + 3I2(aq) + 7H2O(l)
Do a final check on atoms and charges.
21-7
Balancing redox reactions in basic solution

Balance the reaction in acid and then add OH- to neutralize
the H+ ions.
14H+(aq) + Cr2O72-(aq) + 6 I-(aq)
+ 14OH-(aq)
14H2O + Cr2O72- + 6 I
2Cr3+(aq) + 3I2(s) + 7H2O(l)
+ 14OH-(aq)
2Cr3+ + 3I2 + 7H2O + 14OH-
Reconcile the number of water molecules.
7H2O + Cr2O72- + 6 I-
2Cr3+ + 3I2 + 14OH-
Do a final check on atoms and charges.
21-8
The redox reaction between dichromate and iodide anions
Cr2O72-
Figure 21.2
21-9
I-
Cr3+ + I2
Sample Problem 21.1
PROBLEM:
Balancing redox reactions by the half-reaction
method
Permanganate ion is a strong oxidizing agent and its deep purple
color makes it useful as an indicator in redox titrations. It reacts in
basic solution with the oxalate anion to form carbonate anion and
solid manganese dioxide. Balance the skeleton ionic reaction that
occurs between KMnO4 and Na2C2O4 in basic solution.
MnO4-(aq) + C2O42-(aq)
PLAN:
Proceed in acidic solution and then neutralize with base.
SOLUTION:
+7
MnO44H+ + MnO4+ 3e-
21-10
MnO2(s) + CO32-(aq)
MnO4-
R
MnO2
C2O4
2-
O
+4
MnO2
+3
C2O42-
MnO2 + 2H2O
C2O42- + 2H2O
CO32-
+4
2 CO 23
2CO32- + 4H+
+ 2e-
Sample Problem 21.1
4H+ + MnO4- +3e-
(continued)
MnO2+ 2H2O
C2O42- + 2H2O
x3
x2
8H+ + 2MnO4- + 6e-
2MnO2+ 4H2O 3C2O42- + 6H2O
8H+ + 2MnO4- + 6e3C2O42- + 6H2O
2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l)
+ 4OH2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq)
21-11
2CO32- + 4H+ + 2e-
6CO32- + 12H+ + 6e-
2MnO2 + 4H2O
6CO32- + 12H+ + 6e2MnO2(s) + 6CO32-(aq) + 4H+(aq)
+ 4OH2MnO2(s) + 6CO32-(aq) + 2H2O(l)
Types of Electrochemical Cells
Voltaic (or galvanic) cell: uses a spontaneous reaction (∆G < 0)
to generate electrical energy.
Electrolytic cell: uses electrical energy to drive a non-spontaneous
reaction (∆G > 0).
Contain two electrodes (anode and cathode) dipped into an
aqueous electrolyte solution.
The oxidation half-reaction occurs at the anode; the reduction
half-reaction occurs at the cathode.
21-12
General characteristics of voltaic and electrolytic cells
21-13
VOLTAIC CELL
ELECTROLYTIC CELL
Energy does
is released
from
system
work on
its
spontaneous
redox reaction
surroundings
Energy is absorbed
tosupply)
drive a
surroundings
(power
non-spontaneous
redox reaction
do work on the system
(cell)
oxidation half-reaction:
oxidation half-reaction:
X
X+ + ereduction half-reaction:
Y+ + eY
AA + ereduction half-reaction:
B+ + eB
overall (cell) reaction:
X + Y+
X+ + Y (DG < 0)
overall (cell) reaction:
A- + B+
A + B (DG > 0)
Figure 21.3
The spontaneous reaction between zinc and copper(II) ion
Zn(s) + Cu2+(aq)
21-14
Zn2+(aq) + Cu(s)
Figure 21.4
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A voltaic cell based on the zinc-copper reaction
oxidation half-reaction:
Zn(s)
Zn2+(aq) + 2e-
reduction half-reaction:
Cu2+(aq) + 2eCu(s)
Figure 21.5
21-15
overall (cell) reaction:
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Notation for a voltaic cell
components of
anode compartment
components of
cathode compartment
(oxidation half-cell)
(reduction half-cell)
phase of lower phase of higher
oxidation state oxidation state
phase of higher
oxidation state
phase of lower
oxidation state
phase boundary between half-cells
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)
examples:
Zn(s)
Zn2+(aq) + 2e-
Cu2+(aq) + 2e-
Cu(s)
graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite
inert electrode
21-16
A voltaic cell using inactive electrodes
oxidation half-reaction:
2I-(aq)
Figure 21.6
21-17
I2(s) + 2e-
reduction half-reaction:
MnO4-(aq) + 8H+(aq) + 5eMn2+(aq) + 4H2O(l)
overall (cell) reaction:
2MnO4-(aq) + 16H+(aq) + 10I-(aq)
2Mn2+(aq) + 5I2(s) + 8H2O(l)
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Sample Problem 21.2
PROBLEM:
PLAN:
Diagramming voltaic cells
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.
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:
oxidation half-reaction:
Cr(s)
Cr3+(aq)
+ 3e-
reduction half-reaction:
Ag+(aq) + eAg(s)
overall (cell) reaction:
Cr(s) + Ag+(aq)
Cr3+(aq) + Ag(s)
21-18
voltmeter
e-
salt bridge
Cr
Ag
K+
NO3Cr3+
Ag
+
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
Why does a voltaic cell work?
The spontaneous reaction occurs as a result of the different
abilities of materials (such as metals) to give up their electrons
and the ability of the electrons to flow through the circuit.
Ecell > 0 for a spontaneous reaction
1 volt (V) = 1 joule (J)/ coulomb (C)
21-19
Table 21.1
Voltages of Some Voltaic Cells
voltaic cell
21-20
voltage (V)
common alkaline battery
1.5
lead-acid car battery (6 cells = 12V)
2.0
calculator battery (mercury)
1.3
electric eel (~5000 cells in 6-ft eel = 750V)
0.15
nerve of giant squid (across cell membrane)
0.070
Determining an unknown Eohalf-cell with the standard
reference (hydrogen) electrode
oxidation half-reaction:
Zn(s)
Zn2+(aq) + 2e-
overall (cell) reaction:
Zn(s) + 2H3O+(aq)
Zn2+(aq) + H2(g) + 2H2O(l)
21-21
reduction half-reaction:
2H3O+(aq) + 2eH2(g) + 2H2O(l)
Figure 21.7
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Calculating an unknown Eohalf-cell from Eocell
Sample Problem 21.3
PROBLEM:
A voltaic cell houses the reaction between aqueous bromine and
zinc metal:
Br2(aq) + Zn(s)
Zn2+(aq) + 2Br-(aq)
Eocell = 1.83 V
Calculate Eobromine given Eozinc = -0.76 V
PLAN:
The reaction is spontaneous as written since the Eocell is (+). Zinc is
being oxidized and is the anode. Therefore the Eobromine can be
found using Eocell = Eocathode - Eoanode.
SOLUTION:
anode: Zn(s)
Zn2+(aq) + 2e -
EoZn as Zn2+(aq) + 2e-
E = +0.76
Zn(s) is -0.76 V
Eocell = Eocathode - Eoanode = 1.83 = Eobromine - (-0.76)
Eobromine = 1.86 - 0.76 = 1.07 V
21-22
21-23

By convention, electrode potentials are written as reductions.

When pairing two half-cells, you must reverse one reduction half-cell to
produce an oxidation half-cell. Reverse the sign of the potential.

The reduction half-cell potential and the oxidation half-cell potential are
added to obtain the Eocell.

When writing a spontaneous redox reaction, the left side (reactants)
must contain the stronger oxidizing and reducing agents.
example:
Zn(s)
stronger
reducing agent
21-24
+
Cu2+(aq)
Zn2+(aq)
stronger
weaker
oxidizing agent oxidizing agent
+
Cu(s)
weaker
reducing agent
Sample Problem 21.4
Writing spontaneous redox reactions and ranking
oxidizing and reducing agents by strength
PROBLEM: (a) Combine the following three half-reactions into three
spontaneous, balanced equations (A, B and C), and calculate
Eocell for each.
(b) Rank the relative strengths of the oxidizing and reducing agents:
PLAN:
(1) NO3-(aq) + 4H+(aq) + 3e(2) N2(g) + 5H+(aq) + 4e-
NO(g) + 2H2O(l)
N2H5+(aq)
(3) MnO2(s) + 4H+(aq) + 2e-
Mn2+(aq) + 2H2O(l)
Eo = 0.96 V
Eo = -0.23 V
Eo = 1.23 V
Put the equations together in varying combinations so as to produce
(+) Eocell for the combination. Since the reactions are written as
reductions, remember that as you reverse one reaction for an
oxidation, reverse the sign of Eo. Balance the number of electrons
gained and lost without changing the Eo.
In ranking the strengths, compare the combinations in terms of Eocell.
21-25
Sample Problem 21.4 (continued)
SOLUTION:
(1) NO3-(aq) + 4H+(aq) + 3e-
(a) Rev (2) N2H5+(aq)
N2(g) + 5H+(aq) + 4e-
(1) NO3-(aq) + 4H+(aq) + 3e(2) N2H5+(aq)
(A)
(1) NO(g) + 2H2O(l)
(B) 2NO(g) + 3MnO2(s) + 4H+(aq)
21-26
4NO(g) + 3N2(g) + 8H2O(l)
Eo = -0.96 V
Mn2+(aq) + 2H2O(l)
NO3-(aq) + 4H+(aq) + 3e-
(3) MnO2(s) + 4H+(aq) + 2e-
Eocell = 1.19 V
x3
NO3-(aq) + 4H+(aq) + 3e-
(3) MnO2(s) + 4H+(aq) + 2e-
Eo = 0.96 V
Eo = +0.23 V
NO(g) + 2H2O(l) x4
N2(g) + 5H+(aq) + 4e-
4NO3-(aq) + 3N2H5+(aq) + H+(aq)
Rev (1) NO(g) + 2H2O(l)
NO(g) + 2H2O(l)
Eo = 1.23 V
Eocell = 0.27 V
x2
Mn2+(aq) + 2H2O(l)
x3
2NO3-(aq) + 3Mn2+(aq) + 2H2O(l)
Sample Problem 21.4
Rev (2) N2H5+(aq)
(continued)
N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) + 4H+(aq) + 2e(2) N2H5
+(aq)
N2(g) +
Eo = +0.23 V
Mn2+(aq) + 2H2O(l)
5H+(aq)
(3) MnO2(s) + 4H+(aq) + 2e(C) N2H5+(aq) + 2MnO2(s) + 3H+(aq)
+ 4eMn2+(aq) + 2H2O(l)
Eo = 1.23 V
Eocell = 1.46 V
x2
N2(g) + 2Mn2+(aq) + 4H2O(l)
(b) Ranking oxidizing and reducing agents within each equation:
(A): oxidizing agents: NO3- > N2
reducing agents: N2H5+ > NO
(B): oxidizing agents: MnO2 > NO3-
reducing agents: NO > Mn2+
(C): oxidizing agents: MnO2 > N2
reducing agents: N2H5+ > Mn2+
21-27
Sample Problem 21.4 (continued)
A comparison of the relative strengths of oxidizing and reducing
agents produces the following overall ranking.
oxidizing agents: MnO2 > NO3- > N2
reducing agents: N2H5+ > NO > Mn2+
21-28
Relative reactivities (activities) of metals
1. Metals that can displace H2 from acid
2. Metals that cannot displace H2 from acid
3. Metals that can displace H2 from water
4. Metals that can displace other metals from solution
21-29
The reaction of calcium in water
oxidation half-reaction:
Ca(s)
Ca2+(aq) + 2e-
Figure 21.8
21-30
reduction half-reaction:
2H2O(l) + 2eH2(g) + 2OH-(aq)
overall (cell) reaction:
Ca(s) + 2H2O(l)
Ca2+(aq) + H2(g) + 2OH-(aq)
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Free energy and electrical work
DG a -Ecell
-Ecell =
-wmax
DG = wmax = charge x (-Ecell)
DG = -n F Ecell
charge
In the standard state:
charge = n F
n = # mols eF = Faraday constant
F = 96,485 C/mol e-
1 V = 1 J/C
F = 9.65 x 104 J/V.mol e-
21-31
DGo = -n F Eocell
DGo = - RT ln K
Eocell = - (RT/n F) ln K
The interrelationship of DGo, Eo, and K
DG
DG
o
o
K
E
o
cel
l
DGo = -nFEocell
reaction at
standard-state
conditions
<0
>1
>0
spontaneous
0
1
0
at equilibrium
>0
<1
<0
non-spontaneous
DGo = -RT ln K
By substituting standard state
values into Eocell, we get:
E
o
cel
l
Figure 21.9
21-32
K
Eocell = (0.0592 V/n) log K (at 25 oC)
Eocell = -RT ln K
nF
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Calculating K and DGo from Eocell
Sample Problem 21.5
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 DGo at 25 oC for this reaction.
PLAN: Break the reaction into half-reactions, find the Eo for each half-reaction
and then the Eocell.
SOLUTION:
2x
Eocell =
log K =
Pb2+(aq) + 2eAg+(aq) + e-
Eo = -0.13 V
Eo = 0.80 V
Pb(s)
Pb2+(aq) + 2eAg+(aq) + eAg(s)
0.592V
log K
n
n x Eocell
0.592 V
21-33
Pb(s)
Ag(s)
=
(2)(0.93 V)
0.592 V
Eo = 0.13 V
Eo = 0.80 V
Eocell = 0.93 V
.
DGo = -nFEocell = -(2)(96.5 kJ/mol V)(0.93 V)
K = 2.6 x 1031
DGo = -1.8 x 102 kJ
Sample Problem 21.6
Using the Nernst equation to calculate Ecell
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+] = 0.010 M
[H+] = 2.5 M
PH = 0.30 atm
2
Calculate Ecell at 25
oC.
PLAN: Find Eocell and Q in order to use the Nernst equation.
SOLUTION: Determining Eocell :
2H+(aq) + 2eZn2+(aq) + 2eZn(s)
Ecell = Eocell -
Eo = 0.00 V
H2(g)
Zn(s)
Eo = -0.76 V
Zn2+(aq) + 2e-
Eo = +0.76 V
log Q
Ecell = 0.76 - (0.0592/2) log (4.8 x 10-4) = 0.86 V
21-34
Q=
H2
[H+]2
Q=
(0.30)(0.010)
(2.5)2
0.0592 V
n
P x [Zn2+]
Q = 4.8 x 10-4
The effect of concentration on
cell potential
DG = DGo + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell =
Eo
RT
cell
-
ln Q
nF
When Q < 1 and thus [reactant] > [product], ln Q < 0, so Ecell > Eocell
When Q = 1 and thus [reactant] = [product], ln Q = 0, so Ecell = Eocell
When Q >1 and thus [reactant] < [product], ln Q > 0, so Ecell < Eocell
Ecell =
21-35
Eo
0.0592
cell
n
log Q
The relation between Ecell and log Q for the zinc-copper cell
Figure 21.10
21-36
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A concentration cell based on the Cu/Cu2+ half-reaction
oxidation half-reaction:
Cu(s)
Figure 21.11
21-37
Cu2+(aq, 0.1 M) + 2e-
reduction half-reaction:
Cu 2+(aq, 1.0 M) + 2eCu(s)
overall (cell) reaction:
Cu2+(aq,1.0 M)
Cu2+(aq, 0.1 M)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sample Problem 21.7
PROBLEM:
PLAN:
Calculating the potential of a concentration cell
A concentration cell consists of two Ag/Ag+ half-cells. In half-cell A,
electrode A dips into 0.0100 M AgNO3; in half-cell B, electrode B
dips into 4.0 x 10-4 M AgNO3. What is the cell potential at 298 K?
Which electrode has a positive charge?
Eocell 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 half-cell B Ag(s)
will be oxidized to Ag+.
SOLUTION: Ag+(aq, 0.010 M) half-cell A
Ecell = Eocell -
0.0592 V
1
log
Ag+(aq, 4.0 x 10-4 M) half-cell B
[Ag+]dilute
[Ag+]concentrated
Ecell = 0 V - 0.0592 log 4.0 x 10-2
= 0.0828 V
Half-cell A is the cathode and has the positive electrode.
21-38
The laboratory measurement of pH
Pt
glass
electrode
reference
(calomel)
electrode
Hg
AgCl on
Ag on Pt
1 M HCl
Figure 21.12
21-39
thin glass
membrane
paste of
Hg2Cl2
in Hg
KCl
solution
porous
ceramic plugs
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21-40
The corrosion of iron
Figure 21.13
21-41
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Enhanced corrosion at sea
Figure 21.14
21-42
The effect of metal-metal contact on the corrosion of iron
faster corrosion
cathodic protection
Figure 21.15
21-43
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The use of sacrificial anodes to prevent iron corrosion
Figure 21.16
21-44
Figure 21.17
The tin-copper reaction as the basis of a voltaic
and an electrolytic cell
voltaic 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)
21-45
electrolytic cell
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)
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The processes occurring during the discharge and
recharge of a lead-acid battery
VOLTAIC (discharge)
Figure 21.18
21-46
ELECTROLYTIC (recharge)
.
21-47
Sample Problem 21.8
PROBLEM:
Predicting the electrolysis products of a molten
salt mixture
A chemical engineer melts a naturally occurring mixture of NaBr
and MgCl2 and decomposes it in an electrolytic cell. Predict the
substance formed at each electrode, and write balanced halfreactions and the overall cell reaction.
PLAN: Consider the metal and nonmetal components of each compound and
then determine which will recover electrons(be reduced; strength as an oxidizing
agent) better. This is the converse to which of the elements will lose electrons
more easily (lower ionization energy).
SOLUTION: Possible oxidizing agents: Na+, Mg2+
Possible reducing agents: Br-, ClNa, the element, is to the left of Mg in the periodic table, therefore the IE of Mg
is higher than that of Na. So Mg2+ will more easily gain electrons and is the
stronger oxidizing agent.
Br, as an element, has a lower IE than does Cl, and therefore will give up
electrons as Br- more easily than will Cl-.
Mg2+(l) + 2Br-(l)
cathode Br2(g)
anode
21-48
Mg(s) +
The electrolysis of water
oxidation half-reaction:
2H2O(l)
4H+(aq) + O2(g) + 4e-
reduction half-reaction:
2H2O(l) + 4e2H2(g) + 2OH-(aq)
overall (cell) reaction:
2H2O(l)
H2(g) + O2(g)
Figure 21.19
21-49
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Sample Problem 21.9
PROBLEM:
Predicting the electrolysis products of aqueous
ionic solutions
What products form during electrolysis of aqueous solutions 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 0.6 V 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) + e-
K(s)
2H2O(l) + 2e-
H2(g) + 2OH-(aq)
Eo = -2.93 V
Eo = -0.42 V
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-
Eo = 1.07 V
Eo = 0.82 V
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.
21-50
Sample Problem 21.9
(continued)
(b) Ag+(aq) + e2H2O(l) + 2e-
Ag(s)
H2(g) + 2OH-(aq)
Eo = -0.80 V
Eo = -0.42 V
Ag+ is the cation of an inactive metal and therefore will be reduced to Ag
at the cathode.
Ag+(aq) + eAg(s)
The N in NO3- is already in its most oxidized form so water will have to be
oxidized to produce O2 at the anode.
2H2O(l)
O2(g) + 4H+(aq) + 4e(c) Mg2+(aq) + 2e-
Mg(s)
Eo = -2.37 V
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.
21-51
A summary diagram for the stoichiometry of electrolysis
MASS (g)
of substance
oxidized or
reduced
M (g/mol)
AMOUNT (MOL)
of substance
oxidized or
reduced
AMOUNT (MOL)
of electrons
transferred
balanced
half-reaction
CHARGE (C)
time(s)
Figure 21.20
21-52
Faraday
constant
(C/mol e-)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
CURRENT (A)
Sample Problem 21.10
PROBLEM:
PLAN:
Applying the relationship among current, time,
and amount of substance
A technician is plating a faucet 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?
mass of Cr needed
SOLUTION:
Cr3+(aq) + 3e-
Cr(s)
divide by M
mol of Cr needed
3 mol e-/mol Cr
mol of e- transferred
9.65 x 104 C/mol echarge (C)
0.86 g (mol Cr) (3 mol e-)
(52.00 g Cr) (mol Cr)
0.050 mol e- (9.65 x 104 C/mol e-) = 4.8 x 103 C
4.8 x 103 C x (min)
12.5 min
divide by time
current (A)
21-53
= 0.050 mol e-
x
(60 s)
= 6.4 C/s = 6.4 A
dry cell
21-54
alkaline battery
21-55
mercury and silver (button) batteries
21-56
lead-acid battery
21-57
nickel-metal hydride (Ni-MH) battery
21-58
lithium-ion battery
21-59
21-60