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A summary of redox terminology.
Figure 21.1
Key Points About Redox Reactions
Zn(s) + 2H+(aq)
Zn2+(aq) + H2(g)
OXIDATION
•Oxidation (electron loss) always accompanies reduction
(electron gain).
One reactant loses electrons.
Zn loses electrons.
Reducing agent is oxidized.
Zn is the reducing
agent and becomes
oxidized.
•The oxidizing agent is reduced, and the reducing agent is
oxidized.
Oxidation number increases.
The oxidation number
of Zn increases from x
to +2.
•The number of electrons gained by the oxidizing agent
always equals the number lost by the reducing agent.
Other reactant gains
electrons.
REDUCTION
Hydrogen ion gains
electrons.
Oxidizing agent is reduced.
Hydrogen ion is the oxidizing agent
and becomes reduced.
Oxidation number decreases. The oxidation number of H
decreases from +1 to 0.
21-3
21-4
3
4
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Half-Reaction Method for Balancing Redox Reactions
Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq)
Summary: This method divides the overall redox reaction into
oxidation and reduction half-reactions.
1. Divide the reaction into half-reactions -
•Each reaction is balanced for mass (atoms) and charge.
Determine the O.N.s for the species undergoing redox.
•One or both are multiplied by some integer to make the number of
electrons gained and lost equal.
+6
-1
Cr2O72-(aq) + I-(aq)
•The half-reactions are then recombined to give the balanced redox
equation.
Cr2O72I
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.
-
Cr3+
I2
+3
0
Cr3+(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-
21-5
5
Cr3+(aq) + I2(aq)
2 Cr3+
net: +6
2 Cr3+
+ 7H2O(l)
Add 6e - to left.
+ 7H2O(l)
21-6
6
1
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Balancing Redox Reactions in Acidic Solution
6e- + 14H+(aq) + Cr2O722 I
continued
Balancing Redox Reactions in Basic Solution
+ 7H2O(l)
2 Cr3+
-
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Balance the reaction in acid and then add OH- so as to neutralize
the H+ ions.
+ 2e
I2
-
14H+(aq) + Cr2O72-(aq) + 6 I-(aq)
Cr(+6) is the oxidizing agent and I(-1) is the reducing agent.
+ 14OH (aq)
3. Multiply each half-reaction by an integer, if necessary 2 I-
I2 + 2e-
2 Cr3+
2-
7H2O + Cr2O72- + 6 I-
+ 7H2O(l)
3 I2 + 6e-
14H (aq) + Cr2O7 (aq) + 6 I (aq)
+
-
2Cr3+ + 3I2 + 7H2O + 14OH-
Reconcile the number of water molecules.
4. Add the half-reactions together 6 I-
+ 14OH-(aq)
14H2O + Cr2O72- + 6 I-
X3
6e- + 14H+ + Cr2O72-
2Cr3+(aq) + 3I2(s) + 7H2O(l)
-
2Cr3+ + 3I2 + 14OH-
Do a final check on atoms and charges.
2Cr (aq) + 3I2(s) + 7H2O(l)
3+
Do a final check on atoms and charges.
21-7
21-8
7
8
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Figure 21.2
Sample Problem 21.1:
The redox reaction between dichromate ion and iodide ion.
PROBLEM:
Cr2O7
2-
I-
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 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:
Cr3+ + I2
Proceed in acidic solution and then neutralize with base.
SOLUTION:
MnO4+7
MnO44H+ + MnO4-
21-9
9
MnO2(s) + CO32-(aq)
MnO2
+4
MnO2
C2O42+3
C2O42-
MnO2 + 2H2O
C2O42- + 2H2O
+3e-
+2e-
CO32+4
2 CO322CO32- + 4H+
21-10
10
2
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Figure 21.3
Sample Problem 21.1:
Balancing Redox Reactions by the Half-Reaction
Method
VOLTAIC CELL
System
Energydoes
is released
work on
from
its
spontaneous
surroundings
redox reaction
continued:
4H+ + MnO4- +3e-
MnO2+ 2H2O
C2O42- + 2H2O
2CO32- + 4H+ + 2e-
4H+ + MnO4- +3e-
MnO2+ 2H2O
C2O42- + 2H2O
2CO32- + 4H+ + 2e-
8H+ + 2MnO4- +6e-
2MnO2+ 4H2O
8H+ + 2MnO4- +6e3C2O42- + 6H2O
2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l)
3C2O42- + 6H2O
2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq)
6CO32- + 12H+ + 6e-
2MnO2+ 4H2O
6CO32- + 12H+ + 6e-
Oxidation half-reaction
X
X+ + e -
2MnO2(s) + 6CO32-(aq) + 4H+(aq)
+ 4OH-
+ 4OH2MnO2(s) + 6CO32-(aq) + 2H2O(l)
21-11
21-12
11
Reduction half-reaction
B++ eB
Overall (cell) reaction
X + Y+
X+ + Y; DG < 0
Overall (cell) reaction
A- + B+
A + B; DG > 0
12
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The spontaneous reaction between zinc and copper(II) ion.
Figure 21.5
A voltaic cell based on the zinc-copper reaction.
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e-
Zn(s) + Cu2+(aq)
13
Oxidation half-reaction
AA + e-
Reduction half-reaction
Y++ e- Y
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21-13
ELECTROLYTIC CELL
Surroundings(power
Energy is absorbed tosupply)
drive a
nonspontaneous
redox reaction
do work on system(cell)
X3
X2
Figure 21.4
General characteristics of voltaic and electrolytic cells.
Reduction half-reaction
Cu2+(aq) + 2eCu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Zn2+(aq) + Cu(s)
21-14
14
3
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Figure 21.6
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
A voltaic cell using inactive electrodes.
Oxidation half-reaction
2I-(aq)
I2(s) + 2e-
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5eMn2+(aq) + 4H2O(l)
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
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq)
2Mn2+(aq) + 5I2(s) + 8H2O(l)
inert electrode
21-15
21-16
15
16
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Sample Problem 21.2:
PROBLEM:
PLAN:
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Diagramming Voltaic Cells
Why Does a Voltaic Cell Work?
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.
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.
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).
e-
SOLUTION:
Oxidation half-reaction
Cr(s)
Cr3+(aq) + 3e-
Ecell > 0 for a spontaneous reaction
Voltmeter
salt bridge
Cr
1 Volt (V) = 1 Joule (J)/ Coulomb (C)
Ag
K+
NO3-
Reduction half-reaction
Ag+(aq) + eAg(s)
Cr3+
Ag
+
Overall (cell) reaction
Cr(s) + Ag+(aq)
Cr3+(aq) + Ag(s)
21-17
17
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
21-18
18
4
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Figure 21.7
Table 21.1 Voltages of Some Voltaic Cells
Voltaic Cell
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 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)
21-19
21-20
19
20
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PROBLEM:
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Calculating an Unknown E0half-cell from E0cell
Sample Problem 21.3:
Table 21.2 Selected Standard Electrode Potentials (298K)
Half-Reaction
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
SOLUTION:
E0Zn
E0cell
=
E0cathode
E0bromine
21-21
Zn2+(aq) + 2e
anode: Zn(s)
as
-
Zn2+(aq)
E0anode
+
2e-
= 1.83 =
-
strength of oxidizing agent
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.
E = +0.76
Zn(s) is -0.76V
E0bromine
- (-0.76)
= 1.86 - 0.76 = 1.07 V
F2(g) + 2e2F-(aq)
Cl2(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)
2H+(aq) + 2eH2(g)
N2(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
PLAN:
21
Reduction half-reaction
2H3O+(aq) + 2eH2(g) + 2H2O(l)
21-22
22
5
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Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
Writing Spontaneous Redox Reactions
•By convention, electrode potentials are written as reductions.
PROBLEM: (a) Combine the following three half-reactions into three balanced
equations (A, B, and C) for spontaneous reactions, and
calculate E0cell for each.
•When pairing two half-cells, you must reverse one reduction half-cell to
produce an oxidation half-cell. Reverse the sign of the potential.
(b) Rank the relative strengths of the oxidizing and reducing agents:
(1) NO3-(aq) + 4H+(aq) + 3e-
•The reduction half-cell potential and the oxidation half-cell potential are
added to obtain the E0cell.
(2) N2(g) + 5H+(aq) + 4e(3) MnO2(s)
•When writing a spontaneous redox reaction, the left side (reactants)
must contain the stronger oxidizing and reducing agents.
Zn(s)
Example:
Cu2+(aq)
+
stronger
reducing agent
Zn2+(aq)
stronger
oxidizing agent
+
weaker
oxidizing agent
PLAN:
Cu(s)
weaker
reducing agent
21-23
+
E0 = 0.96V
2e-
Mn2+(aq)
E0 = -0.23V
+ 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
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.
21-24
23
24
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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:
(a)
(1)
(1) NO3-(aq) + 4H+(aq) + 3e-
Rev (2)
N2H5+(aq)
NO3-(aq)
4H+(aq)
+
(2) N2H5+(aq)
(A)
N2(g) +
+
3e-
N2(g) + 5H+(aq) + 4e-
(1) NO(g) + 2H2O(l)
E0
(B) 2NO(g) + 3MnO2(s) + 4H+(aq)
Rev (2) N2H5+(aq)
X4
= +0.23V
E0cell = 1.19V
(2) N2H5+(aq)
(C) N2H5+(aq) + 2MnO2(s) + 3H+(aq)
+
+
3e-
Mn2+(aq) + 2H2O(l)
X2
E0cell
E0 = 1.23V
E0cell = 1.46V
Mn2+(aq) + 2H2O(l)
X2
N2(g) + 2Mn2+(aq) + 4H2O(l)
(b) Ranking oxidizing and reducing agents within each equation:
E0 = -0.96V
E0
E0 = +0.23V
Mn2+(aq) + 2H2O(l)
N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e-
X3
Mn2+(aq) + 2H2O(l)
4H+(aq)
N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e-
4NO(g) + 3N2(g) + 8H2O(l)
NO3-(aq)
(3) MnO2(s) +4H+(aq) + 2e-
21-25
+
4e-
NO3-(aq) + 4H+(aq) + 3e-
(3) MnO2(s) +4H+(aq) + 2e-
Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (3 of 4)
NO(g) + 2H2O(l) E0 = 0.96V
5H+(aq)
NO(g) + 2H2O(l)
4NO3-(aq) + 3N2H5+(aq) + H+(aq)
Rev (1) NO(g) + 2H2O(l)
25
+4H+(aq)
NO(g) + 2H2O(l)
N2H5+(aq)
= 1.23V
(A): oxidizing agents: NO3- > N2
reducing agents: N2H5+ > NO
= 0.27V
(B): oxidizing agents: MnO2 > NO3-
reducing agents: NO > Mn2+
(C): oxidizing agents: MnO2 > N2
reducing agents: N2H5+ > Mn2+
X3
2NO3-(aq) + 3Mn3+(aq) + 2H2O(l)
21-26
26
6
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Sample Problem 21.4: Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (4 of 4)
Relative Reactivities (Activities) of Metals
1. Metals that can displace H from acid
A comparison of the relative strengths of oxidizing and reducing
agents produces the overall ranking of
2. Metals that cannot displace H from acid
Oxidizing agents: MnO2 > NO3- > N2
Reducing agents: N2H5+ > NO > Mn2+
3. Metals that can displace H from water
4. Metals that can displace other metals from
solution
21-27
21-28
27
28
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Figure 21.8
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The reaction of calcium in water.
Free Energy and Electrical Work
Oxidation half-reaction
Ca(s)
Ca2+(aq) + 2e-
Reduction half-reaction
2H2O(l) + 2eH2(g) + 2OH-(aq)
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-
29
DG0 = - RT ln K
E0cell = - (RT/n F) ln K
1V = 1J/C
F = 9.65x104J/V*mol e-
Overall (cell) reaction
Ca(s) + 2H2O(l)
Ca2+(aq) + H2(g) + 2OH-(aq)
21-29
DG0 = -n F E0cell
21-30
30
7
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Figure 21.9 The interrelationship of DG0, E0, and K.
DG
DG
0
DG0 = -nFEocell
0
K
E
0
cell
PROBLEM:
Reaction at
standard-state
conditions
<0
>1
>0
0
1
0
at equilibrium
>0
<1
<0
nonspontaneous
spontaneous
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:
DG0 = -RT lnK
0
cell
2X
E0cell =
E0cell = (0.0592V/n) log K (at 298 K)
K
log K =
E0cell = -RT lnK
nF
21-31
Pb2+(aq) + 2eAg+(aq) + e-
Pb(s)
Ag(s)
E0 = -0.13V
E0 = 0.80V
E0 = 0.13V
E0 = 0.80V
Pb(s)
Pb2+(aq) + 2eAg+(aq) + eAg(s)
0.592V
n
n x E0cell
0.592V
log K
=
E0cell = 0.93V
DG0 = -nFE0cell = -(2)(96.5kJ/mol*V)(0.93V)
(2)(0.93V)
0.592V
K = 2.6x1031
DG0 = -1.8x102kJ
21-32
31
32
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Sample Problem 21.6:
The Effect of Concentration on Cell Potential
[Zn2+] = 0.010M
-nF Ecell = -nF Ecell + RT ln Q
Ecell = E0cell -
RT
nF
ln Q
SOLUTION: Determining E0cell :
E0cell
21-33
0.0592
2H+(aq) + 2e-
H2(g)
E0 = 0.00V
Zn2+(aq) + 2e-
Zn(s)
E0
Zn(s)
•When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell
-
PH 2 = 0.30atm
PLAN: Find E0cell and Q in order to use the Nernst equation.
•When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell =
Ecell =
[H+] = 2.5M
Calculate Ecell at 298 K.
•When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell
E0cell
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:
DG = DG0 + RT ln Q
33
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 DG0 at 298 K for this reaction.
By substituting standard state
values into E0cell, we get
E
Calculating K and DG0 from E0cell
Sample Problem 21.5:
Ecell = E0cell -
log Q
n
Ecell = 0.76 -
Zn2+(aq)
0.0592V
n
+
2e-
log Q
P x [Zn2+]
H2
[H+]2
= -0.76V
E0 = +0.76V
(0.0592/2)log(4.8x10-4)
Q=
Q=
(0.30)(0.010)
(2.5)2
Q = 4.8x10-4
= 0.86V
21-34
34
8
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Figure 21.10
Figure 21.11
A concentration cell based on the Cu/Cu2+ half-reaction.
The relation between Ecell and log Q for the zinc-copper cell.
Oxidation half-reaction
Cu(s)
Cu2+(aq, 0.1M) + 2e-
Cu
Reduction half-reaction
1.0M) + 2eCu(s)
2+(aq,
Overall (cell) reaction
Cu2+(aq,1.0M)
Cu2+(aq, 0.1M)
21-35
21-36
35
36
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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 0.0100M 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?
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
Ecell = E0cell -
0.0592V
1
log
[Ag+]
Ag+(aq, 4.0x10-4M) half-cell B
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.
21-37
37
9