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21-1
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Chapter 21
Electrochemistry:
Chemical Change and Electrical Work
21-2
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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
Nonspontaneous Reaction
21-3
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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.
21-4
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Figure 21.1
A summary of redox terminology.
Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
OXIDATION
One reactant loses electrons.
Zn loses electrons.
Reducing agent is oxidized.
Zn is the reducing
agent and becomes
oxidized.
Oxidation number increases.
The oxidation number
of Zn increases from x
to +2.
REDUCTION
Other reactant gains
electrons.
Oxidizing agent is reduced.
Hydrogen ion gains
electrons.
Hydrogen ion is the oxidizing agent
and becomes reduced.
Oxidation number decreases. The oxidation number of H
decreases from +1 to 0.
21-5
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Half-Reaction Method for Balancing Redox Reactions
Summary: 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 some integer to make the number of
electrons gained and lost equal.
•The half-reactions are then recombined to give the balanced redox
equation.
Advantages:
•The separation of half-reactions reflects actual physical
separations in electrochemical cells.
•The half-reactions are easier to balance especially if they
involve acid or base.
•It is usually not necessary to assign oxidation numbers to
those species not undergoing change.
21-6
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Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq)
Cr3+(aq) + I2(aq)
1. Divide the reaction into half-reactions Determine the O.N.s for the species undergoing redox.
+6
-1
2Cr2O7 (aq) + I-(aq)
Cr3+
Cr2O72I-
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) + Cr2O72-
21-7
2 Cr3+
net: +6
2 Cr3+
+ 7H2O(l)
Add 6e- to left.
+ 7H2O(l)
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Balancing Redox Reactions in Acidic Solution
6e- + 14H+(aq) + Cr2O722 I-
2 Cr3+
I2
continued
+ 7H2O(l)
+ 2e-
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-
X3
4. Add the half-reactions together 6e- + 14H+ + Cr2O726 I14H+(aq) + Cr2O72-(aq) + 6 I-(aq)
2 Cr3+
3 I2 + 6e2Cr3+(aq) + 3I2(s) + 7H2O(l)
Do a final check on atoms and charges.
21-8
+ 7H2O(l)
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Balancing Redox Reactions in Basic Solution
Balance the reaction in acid and then add OH- so as to neutralize
the H+ ions.
14H+(aq) + Cr2O72-(aq) + 6 I-(aq)
2Cr3+(aq) + 3I2(s) + 7H2O(l)
+ 14OH-(aq)
14H2O + Cr2O72- + 6 I-
+ 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.
21-9
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Figure 21.2
The redox reaction between dichromate ion and iodide ion.
Cr2O72-
I-
Cr3+ + I2
21-10
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Sample Problem 21.1:
Balancing Redox Reactions by the Half-Reaction
Method
PROBLEM: 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 ion to form carbonate ion and solid
mangaese dioxide. Balance the skeleton ionic reaction that occurs
between NaMnO4 and Na2C2O4 in basic solution:
MnO4-(aq) + C2O42-(aq)
PLAN:
MnO2(s) + CO32-(aq)
Proceed in acidic solution and then neutralize with base.
SOLUTION:
MnO4+7
MnO44H+ + MnO4-
21-11
MnO2
+4
MnO2
C2O42+3
C2O42-
MnO2 + 2H2O
C2O42- + 2H2O
+3e-
+2e-
CO32+4
2 CO322CO32- + 4H+
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Sample Problem 21.1:
Balancing Redox Reactions by the Half-Reaction
Method
continued:
4H+ + MnO4- +3e-
MnO2+ 2H2O
C2O42- + 2H2O
2CO32- + 4H+ + 2e-
4H+ + MnO4- +3e-
MnO2+ 2H2O
C2O42- + 2H2O
2CO32- + 4H+ + 2eX3
X2
8H+ + 2MnO4- +6e-
2MnO2+ 4H2O
8H+ + 2MnO4- +6e3C2O42- + 6H2O
2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l)
+ 4OH2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq)
21-12
3C2O42- + 6H2O
6CO32- + 12H+ + 6e-
2MnO2+ 4H2O
6CO32- + 12H+ + 6e2MnO2(s) + 6CO32-(aq) + 4H+(aq)
+ 4OH2MnO2(s) + 6CO32-(aq) + 2H2O(l)
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Figure 21.3
General characteristics of voltaic and electrolytic cells.
VOLTAIC CELL
Energydoes
is released
from
System
work on
its
spontaneous
redox reaction
surroundings
Oxidation half-reaction
X
X+ + e-
21-13
ELECTROLYTIC CELL
Energy is absorbed tosupply)
drive a
Surroundings(power
nonspontaneous
redox reaction
do work on system(cell)
Oxidation half-reaction
AA + e-
Reduction half-reaction
Y++ e- Y
Reduction half-reaction
B++ eB
Overall (cell) reaction
X+ + Y; ∆G < 0
X + Y+
Overall (cell) reaction
A + B; ∆G > 0
A- + B+
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Figure 21.4
The spontaneous reaction between zinc and copper(II) ion.
Zn(s) + Cu2+(aq)
21-14
Zn2+(aq) + Cu(s)
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Figure 21.5
A voltaic cell based on the zinc-copper reaction.
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e-
Reduction half-reaction
Cu(s)
Cu2+(aq) + 2eOverall (cell) reaction
Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)
21-15
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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
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Figure 21.6
A voltaic cell using inactive electrodes.
Oxidation half-reaction
I2(s) + 2e2I-(aq)
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5eMn2+(aq) + 4H2O(l)
Overall (cell) reaction
2Mn2+(aq) + 5I2(s) + 8H2O(l)
2MnO4-(aq) + 16H+(aq) + 10I-(aq)
21-17
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Sample Problem 21.2:
Diagramming Voltaic Cells
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).
Voltmeter
e-
SOLUTION:
Oxidation half-reaction
Cr(s)
Cr3+(aq) + 3eReduction half-reaction
Ag+(aq) + eAg(s)
salt bridge
Cr
Ag
K+
NO3Cr3+
Ag+
Overall (cell) reaction
Cr(s) + Ag+(aq)
Cr3+(aq) + Ag(s)
21-18
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
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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
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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
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Figure 21.7
Determining an unknown E0half-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
H2(g) + 2H2O(l)
2H3O+(aq) + 2e-
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Sample Problem 21.3:
Calculating an Unknown E0half-cell from E0cell
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
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:
anode: Zn(s)
Zn2+(aq) + 2e
E0Zn as Zn2+(aq) + 2e-
-
E = +0.76
Zn(s) is -0.76V
E0cell = E0cathode - E0anode = 1.83 = E0bromine - (-0.76)
E0bromine = 1.86 - 0.76 = 1.07 V
21-22
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Table 21.2 Selected Standard Electrode Potentials (298K)
21-23
2F-(aq)
F2(g) + 2eCl2(g) + 2e2Cl-(aq)
MnO2(g) + 4H+(aq) + 2eMn2+(aq) + 2H2O(l)
NO3-(aq) + 4H+(aq) + 3eNO(g) + 2H2O(l)
Ag+(aq) + eAg(s)
Fe3+(g) + eFe2+(aq)
O2(g) + 2H2O(l) + 4e4OH-(aq)
Cu2+(aq) + 2eCu(s)
H2(g)
2H+(aq) + 2eN2(g) + 5H+(aq) + 4eN2H5+(aq)
Fe2+(aq) + 2eFe(s)
2H2O(l) + 2eH2(g) + 2OH-(aq)
Na+(aq) + eNa(s)
Li+(aq) + eLi(s)
E0(V)
+2.87
+1.36
+1.23
+0.96
+0.80
+0.77
+0.40
+0.34
0.00
-0.23
-0.44
-0.83
-2.71
-3.05
strength of reducing agent
strength of oxidizing agent
Half-Reaction
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Writing Spontaneous Redox Reactions
•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 E0cell.
•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)
stronger
oxidizing agent
Zn2+(aq)
weaker
oxidizing agent
+
Cu(s)
weaker
reducing agent
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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 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(3) MnO2(s) +4H+(aq) + 2ePLAN:
NO(g) + 2H2O(l)
N2H5+(aq)
Mn2+(aq) + 2H2O(l)
E0 = 0.96V
E0 = -0.23V
E0 = 1.23V
Put the equations together in varying combinations so as to produce
(+) E0cell for the combination. Since the reactions are written as
reductions, remember that as you reverse one reaction for an
oxidation, reverse the sign of E0. Balance the number of electrons
gained and lost without changing the E0.
In ranking the strengths, compare the combinations in terms of E0cell.
21-25
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Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (2 of 4)
SOLUTION:
(1) NO3-(aq) + 4H+(aq) + 3e-
Rev (2) N2H5+(aq)
(a)
(1)
NO3-(aq)
+
4H+(aq)
(2) N2H5+(aq)
(A)
N2(g) + 5H+(aq) + 4e+
3e-
X4
(B) 2NO(g) + 3MnO2(s) + 4H+(aq)
E0cell = 1.19V
X3
4NO(g) + 3N2(g) + 8H2O(l)
E0 = -0.96V
Mn2+(aq) + 2H2O(l)
NO3-(aq) + 4H+(aq) + 3e-
(3) MnO2(s) +4H+(aq) + 2e-
21-26
E0 = +0.23V
NO3-(aq) + 4H+(aq) + 3e-
(3) MnO2(s) +4H+(aq) + 2e(1) NO(g) + 2H2O(l)
NO(g) + 2H2O(l)
N2(g) + 5H+(aq) + 4e-
4NO3-(aq) + 3N2H5+(aq) + H+(aq)
Rev (1) NO(g) + 2H2O(l)
NO(g) + 2H2O(l) E0 = 0.96V
Mn2+(aq) + 2H2O(l)
E0 = 1.23V
X2
E0cell = 0.27V
X3
2NO3-(aq) + 3Mn3+(aq) + 2H2O(l)
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Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (3 of 4)
Rev (2) N2H5+(aq)
N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e(2) N2H5+(aq)
E0 = +0.23V
Mn2+(aq) + 2H2O(l)
E0 = 1.23V
E0cell = 1.46V
N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e(C) N2H5+(aq) + 2MnO2(s) + 3H+(aq)
Mn2+(aq) + 2H2O(l)
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
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Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (4 of 4)
A comparison of the relative strengths of oxidizing and reducing
agents produces the overall ranking of
Oxidizing agents: MnO2 > NO3- > N2
Reducing agents: N2H5+ > NO > Mn2+
21-28
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Relative Reactivities (Activities) of Metals
1. Metals that can displace H from acid
2. Metals that cannot displace H from acid
3. Metals that can displace H from water
4. Metals that can displace other metals from
solution
21-29
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Figure 21.8
The reaction of calcium in water.
Oxidation half-reaction
Ca(s)
Ca2+(aq) + 2e-
Reduction half-reaction
H2(g) + 2OH-(aq)
2H2O(l) + 2e-
Overall (cell) reaction
Ca2+(aq) + H2(g) + 2OH-(aq)
Ca(s) + 2H2O(l)
21-30
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Free Energy and Electrical Work
∆G α -Ecell
-Ecell =
∆G = wmax = charge x (-Ecell)
-wmax
∆G = -n F Ecell
charge
In the standard state charge = n F
∆G0 = -n F E0cell
n = #mols eF = Faraday constant
F = 96,485 C/mol
e-
1V = 1J/C
F = 9.65x104J/V*mol e-
21-31
∆G0 = - RT ln K
E0cell = - (RT/n F) ln K
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Figure 21.9 The interrelationship of ∆G0, E0, and K.
∆G
∆G
0
K
0
E
0
cel
l
∆G0 = -nFEocell
Reaction at
standard-state
conditions
spontaneous
<0
>1
>0
0
1
0
at equilibrium
>0
<1
<0
nonspontaneous
∆G0 = -RT lnK
By substituting standard state
values into E0cell, we get
E
E0cell = (0.0592V/n) log K (at 298 K)
0
cel
l
21-32
K
E0cell = -RT lnK
nF
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Sample Problem 21.5:
PROBLEM:
Calculating K and ∆G0 from E0cell
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. Substitute into the equations found on slide
SOLUTION:
2X
E0cell =
log K =
Pb2+(aq) + 2eAg+(aq) + e-
Pb(s)
Pb2+(aq) + 2eAg+(aq) + eAg(s)
0.592V
n
n x E0cell
0.592V
21-33
E0 = -0.13V
E0 = 0.80V
Pb(s)
Ag(s)
log K
=
(2)(0.93V)
0.592V
E0 = 0.13V
E0 = 0.80V
E0cell = 0.93V
∆G0 = -nFE0cell = -(2)(96.5kJ/mol*V)(0.93V)
K = 2.6x1031
∆G0 = -1.8x102kJ
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The Effect of Concentration on Cell Potential
∆G = ∆G0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
RT
Ecell =
E0cell
-
ln Q
nF
•When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell
•When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell
•When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell
0.0592
Ecell =
21-34
E0cell
n
log Q
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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+] = 0.010M
[H+] = 2.5M
PH = 0.30atm
2
Calculate Ecell at 298 K.
PLAN: Find E0cell and Q in order to use the Nernst equation.
SOLUTION: Determining E0cell :
P x [Zn2+]
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 -
log Q
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
21-35
H2
[H+]2
Q=
(0.30)(0.010)
(2.5)2
0.0592V
n
Q=
Q = 4.8x10-4
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Figure 21.10
The relation between Ecell and log Q for the zinc-copper cell.
21-36
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Figure 21.11
A concentration cell based on the Cu/Cu2+ half-reaction.
Oxidation half-reaction
Cu(s)
Cu2+(aq, 0.1M) + 2e-
Reduction half-reaction
Cu(s)
Cu 2+(aq, 1.0M) + 2e-
Overall (cell) reaction
Cu2+(aq, 0.1M)
Cu2+(aq,1.0M)
21-37
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Sample Problem 21.7:
Calculating the Potential of a Concentration Cell
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 half-cell B Ag(s)
will be oxidized to Ag+.
SOLUTION: Ag+(aq, 0.010M) half-cell A
0.0592V
Ecell = E0cell -
1
log
Ag+(aq, 4.0x10-4M) half-cell B
[Ag+]dilute
[Ag+]concentrated
Ecell = 0 V -0.0592 log 4.0x10-2 = 0.0828V
Half-cell A is the cathode and has the positive electrode.
21-38
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Figure 21.12
The laboratory measurement of pH.
Pt
Glass
electrode
Reference
(calomel)
electrode
Hg
Paste of
Hg2Cl2 in
Hg
AgCl on
Ag on Pt
1M HCl
Thin glass
membrane
21-39
KCl
solution
Porous ceramic
plugs
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Table 21.3 Some Ions Measured with Ion-Specific Electrodes
21-40
Species Detected
Typical Sample
NH3/NH4+
Industrial wastewater, seawater
CO2/HCO3-
Blood, groundwater
F-
Drinking water, urine, soil, industrial stack gases
Br-
Grain, plant tissue
I-
Milk, pharmaceuticals
NO3-
Soil, fertilizer, drinking water
K+
Blood serum, soil, wine
H+
Laboratory solutions, soil, natural waters
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Figure 21.13
21-41
Alkaline Battery
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Figure 21.14
21-42
Mercury and Silver (Button) Batteries
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Figure 21.15
21-43
Lithium battery.
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Figure 21.16
21-44
Lead-acid battery.
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Figure 21.17
21-45
Nickel-metal hydride (Ni-MH) battery
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Figure 21.18
21-46
Lithium-ion battery.
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Figure 21.19
21-47
Fuel cell.
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Figure 21.20
21-48
The corrosion of iron.
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Figure 21.21
21-49
Enhanced corrosion at sea.
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Figure 21.22
The effect of metal-metal contact on the corrosion of iron.
faster corrosion
21-50
cathodic protection
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Figure 21.23
The use of sacrificial anodes to prevent iron corrosion.
21-51
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Figure 21.24
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)
21-52
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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Figure 21.25
The processes occurring during the discharge and
recharge of a lead-acid battery.
VOLTAIC(discharge)
ELECTROLYTIC(recharge)
21-53
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Table 21.4 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
-
21-54
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Sample Problem 21.8:
Predicting the Electrolysis Products of a Molten
Salt Mixture
PROBLEM: 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
21-55
anode
Mg(s) + Br2(g)
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Figure 21.26
The electrolysis of water.
Overall (cell) reaction
2H2O(l)
H2(g) + O2(g)
Oxidation half-reaction
2H2O(l) 4H+(aq) + O2(g) + 4e-
21-56
Reduction half-reaction
2H2O(l) + 4e2H2(g) + 2OH-(aq)
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Sample Problem 21.9:
Predicting the Electrolysis Products of Aqueous
Ionic Solutions
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) + 2e-
K(s)
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-
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.
21-57
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Sample Problem 21.9:
Predicting the Electrolysis Products of Aqueous
Ionic Solutions
continued
(b) Ag+(aq) + e2H2O(l) + 2e-
E0 = -0.80V
Ag(s)
H2(g) + 2OH-(aq)
E0 = -0.42V
Ag+ is the cation of an inactive metal and therefore will be reduced to Ag
at the cathode.
Ag(s)
Ag+(aq) + eThe 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-
Mg(s)
2
E0 = -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.
21-58
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Figure 21.27
A summary diagram for the stoichiometry of electrolysis.
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
ofelectrons
electrons
transferred
transferred
balanced
half-reaction
CHARGE
CHARGE(C)
(C)
time(s)
21-59
Faraday
constant
(C/mol e-)
CURRENT
CURRENT(A)
(A)
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Sample Problem 21.10:
Applying the Relationship Among Current, Time,
and Amount of Substance
PROBLEM: 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?
PLAN:
mass of Cr needed
SOLUTION:
Cr3+(aq) + 3e-
Cr(s)
divide by M
mol of Cr needed
3mol e-/mol Cr
mol of e- transferred
0.86g (mol Cr) (3 mol e-)
= 0.050 mol e-
(52.00 gCr) (mol Cr)
0.050 mol e- (9.65x104 C/mol e-) = 4.8x103 C
9.65x104C/mol echarge (C)
divide by time
current (A)
21-60
4.8x103 C
(min)
12.5 min
(60s)
= 6.4C/s = 6.4 A