Chapter 21 Electrochemistry: Fundamentals
Key Points About Redox Reactions
1. Oxidation (electron loss) always accompanies
reduction (electron gain).
2. The oxidizing agent is reduced, and the reducing
agent is oxidized.
3. The number of electrons gained by the oxidizing
agent always equals the number lost by the
reducing agent.
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 0 to +
2.
REDUCTION
Other reactant gains electrons.
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.
Electrochemical cell
General characteristics of voltaic and electrolytic cells.
VOLTAIC / GALVANIC CELL
System
Energydoes
is released
work on
from
its
spontaneous
surroundings
redox reaction
Oxidation half-reaction
X
X+ + e-
ELECTROLYTIC CELL
Surroundings(power
Energy is absorbed tosupply)
drive a
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+
X+ + Y; DG < 0
Overall (cell) reaction
A- + B+
A + B; DG > 0
A voltaic cell based on the zinc-copper reaction.
Oxidation half-reaction
Zn(s)
Zn2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2eCu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
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)
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq)
2Mn2+(aq) + 5I2(s) + 8H2O(l)
Notation for a Voltaic Cell
components of anode
compartment
components of cathode
compartment
(oxidation half-cell)
(reduction half-cell)
phase of lower
oxidation state
phase of higher
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
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 half-reaction.
Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with
the cathode (reduction).
SOLUTION:
Oxidation half-reaction
Cr(s)
Cr3+(aq) + 3eReduction half-reaction
Ag+(aq) + eAg(s)
Overall (cell) reaction
Cr(s) + Ag+(aq)
Cr3+(aq) + Ag(s)
Voltmeter
e-
salt bridge
Cr
Ag
K+
NO3Cr3+
Ag+
Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)
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)
Reduction half-reaction
2H3O+(aq) + 2eH2(g) + 2H2O(l)
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
Selected Standard Electrode Potentials (298K)
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
strength of oxidizing agent
Half-Reaction
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
+
Cu2+(aq)
stronger
oxidizing agent
Zn2+(aq)
weaker
oxidizing agent
+
Cu(s)
weaker
reducing agent
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.
Writing Spontaneous Redox Reactions and Ranking
Oxidizing and Reducing Agents by Strength
continued (2 of 4)
SOLUTION:
(a)
(1) NO3-(aq) + 4H+(aq) + 3e-
Rev (2) N2H5+(aq)
(1) NO3
-(aq)
+
4H+(aq)
(2) N2H5+(aq)
(A)
N2(g) + 5H+(aq) + 4e+
3e-
E0 = +0.23V
X4
4NO(g) + 3N2(g) + 8H2O(l)
(B) 2NO(g) + 3MnO2(s) + 4H+(aq)
E0 = -0.96V
Mn2+(aq) + 2H2O(l)
NO3-(aq) + 4H+(aq) + 3e-
(3) MnO2(s) +4H+(aq) + 2e-
E0cell = 1.19V
X3
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)
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)
Mn2+(aq) + 2H2O(l)
N2(g) + 5H+(aq) + 4e-
(3) MnO2(s) +4H+(aq) + 2e(C) N2H5+(aq) + 2MnO2(s) + 3H+(aq)
E0 = +0.23V
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:
(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+
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+
Summary
•
•
•
•
•
•
•
•
•
•
•
A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a
salt bridge.
The salt bridge provides ions to maintain the charge balance when the cell operates.
Electrons move from anode to cathode while cation moves from salt bridge to the cathode
half cell.
The output of a cell is called cell potential (Ecell) and is measured in volts.
When all substances are in standard states, the cell potential is the standard cell potential
(Eocell).
Ecell equals Ecathode minus Eanode, Ecell = Ecathode - Eanode.
Conventionally, the half cell potential refers to its reduction half-reaction.
Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking
the oxidizing agent or reducing agent.
Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker
ones.
Spontaneous reaction is indicated negative ∆G and positive ∆E,
∆G = - nF∆E.
We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.
Relative Reactivities (Activities) of Metals
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
strength as reducing agents
1. Metals that can displace H
from acid
Li
K
Ba
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Cd
Co
Ni
Sn
Pb
H2
Cu
Hg
Ag
Au
can displace H
from water
can displace H
from steam
can displace H
from acid
cannot displace H
from any source
The interrelationship of DG0, E0, and K.
DG0
DG0 = -nFEocell
Reaction at
standard-state
conditions
DG0
K
E0cell
<0
>1
>0
0
1
0
at equilibrium
>0
<1
<0
nonspontaneous
spontaneous
DG0 = -RT lnK
By substituting standard state
values into E0cell, we get
E0
K
cell
E0cell = -RT lnK
nF
E0cell = (0.0592V/n) log K (at 298 K)
The Effect of Concentration on Cell Potential
DG = DG0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Nernst equation
Ecell =
E0
cell
RT
-
nF
ln Q
•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 =
E0
cell
n
log Q
Calculating K and DG0 from E0cell
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 DG0 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:
Pb2+(aq) + 2eAg+(aq) + e-
Ag(s)
2X Ag+(aq) + e-
Ag(s)
E0cell = - (RT/n F) ln K
log K =
Pb(s)
n x E0cell
0.592V
E0cell =
=
0.592V
n
(2)(0.93V)
K = 2.6x1031
0.592V
E0 = -0.13V
Anode
E0 = 0.80V
Cathode
E0cell = E0cathode – E0anode = 0.93V
log K
DG0 = -nFE0cell
= -(2)(96.5kJ/mol*V)(0.93V)
DG0 = -1.8x102kJ
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 -
H2
[H+]2
Q=
(0.30)(0.010)
(2.5)2
0.0592V
n
Q=
log Q
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
Q = 4.8x10-4
Diagramming Voltaic Cells
PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that
consists of one half-cell with a Zn bar in a Zn(NO3)2 solution, another half-cell
with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement
indicates that the Zn electrode is negative relative to the Ag electrode.
PLAN:
Identify the redox reactions
Write each half-reaction.
Associate the (-)(Zn) pole with the anode (oxidation)
and the (+) (Ag) pole with the cathode (reduction).
SOLUTION:
Oxidation half-reaction
Zn2+(aq) + 2eZn(s)
e-
Voltmeter
salt bridge
Zn
Ag
K+
Reduction half-reaction
Ag+(aq) + eAg(s)
Overall (cell) reaction
Zn(s) + 2Ag+(aq)
Zn2+(aq) + 2Ag(s)
NO3Zn2+
Anode
Ag+
Cathode
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
Free Energy and Electrical Work
DG a -Ecell
-Ecell =
-wmax
charge
If there is no
current flows,
the potential
represents the
maximum work
the cell can do.
If there is no current
flows, no energy is lost
to heat the cell
component.
DG = wmax = charge x (-Ecell)
DG = - n F Ecell
charge = n F
e-
n = # mols
F = Faraday constant
In the standard state
DG0
=-nF
E0
cell
All components
are at standard
state.
F = 96,485 C/mol
DG0 = - RT ln K
1V = 1J/C
F = 9.65x104J/V*mol
E0cell = - (RT/n F) ln K
E0cell = - (0.05916/n) log K at RT
The Effect of Concentration on Cell Potential
Cell operates with all
components at standard
states. Most cells are starting
at Non-standard state.
Nernst equation
DG = DG0 + RT ln Q
-nF Ecell = -nF Ecell + RT ln Q
Ecell =
E0
cell
RT
-
nF
ln Q
• When Q < 1 and thus [reactant] > [product], lnQ < 0, so Ecell > E0cell forward reaction
• When Q = 1 and thus [reactant] = [product], lnQ = 0, so Ecell = E0cell Equilibrium
• When Q >1 and thus [reactant] < [product], lnQ > 0, so Ecell < E0cell Reverse reaction
0.0592
Ecell =
E0
cell
n
log Q
Sample Problem
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: 2H+(aq) + Zn (s)
H2(g) + Zn2+ (aq)
[Zn2+] = 0.010M
[H+] = 2.5M
PH = 0.30atm
2
PLAN:
Calculate Ecell at 298 K.
Find E0cell and Q in order to use the Nernst equation.
Ecell =
E0
cell
-
RT
nF
ln Q
0.0592
Ecell =
E0
cell
-
n
SOLUTION: Determining E0cell :
2H+(aq) + 2eZn2+(aq)
+
2e-
H2(g)
E0 = 0.00V cathode
Zn(s)
E0
= -0.76V anode
Ecell0 = E0c-E0a = 0.00-(-0.76)V = 0.76 V
Ecell =
E0
cell
-
Q=
H2
[H+]2
Q=
(0.30)(0.010)
(2.5)2
0.0592
n
P x [Zn2+]
log Q
Ecell = 0.76 - (0.0592/2)log(4.8x10-4) = 0.86V
Q = 4.8x10-4
log Q
Summary
•
•
•
•
•
•
•
•
•
•
•
A voltaic cell contains of oxidation (anode) and reduction (cathode) half-cells, connected by a
salt bridge.
The salt bridge provides ions to maintain the charge balance when the cell operates.
Electrons move from anode to cathode while cation moves from salt bridge to the cathode
half cell.
The output of a cell is called cell potential (Ecell) and is measured in volts.
When all substances are in standard states, the cell potential is the standard cell potential
(Eocell).
Ecell equals Ecathode minus Eanode, Ecell = Ecathode - Eanode.
Conventionally, the half cell potential refers to its reduction half-reaction.
Using standard H2 reference electrode, other Eo half-cell can be measured and used for ranking
the oxidizing agent or reducing agent.
Spontaneous redox reactions combine stronger oxidizing and reducing agent to form weaker
ones.
Spontaneous reaction is indicated negative ∆G and positive ∆E,
∆G = -nF∆E.
We can determine K using ∆E, ∆Go = -nF∆Eo = - RTlnK.