Standard Reference Electrode
Standard Hydrogen Electrode (SHE)
SHE:
• Assigned 0.000 V
• Can be anode or cathode
• Pt does not take part in reaction
• Difficult to operate
Standard Conditions:
1 atm for gases, 1.0M for solutions, 25 oC for all (298 K)
Alternative reference electrodes
• Ag/AgCl electrode
AgCl(s) + e-  Cl- + Ag(s)
Ecell = +0.22 V vs. SHE
• Calomel electrode
Hg2Cl2(s) + 2e-  2Cl- + 2Hg(l)
Ecell = + 0.24 V vs.SHE
Standard Reduction Potentials in Aqueous Solution at 25° C
How can we determine which substance is being oxidized and which
is being reduced? …..The MORE POSITIVE reduction
potential gets to be reduced.
Reading the reduction potential chart
 Elements that have the most positive reduction potentials are easily
reduced.
 Elements that have the least positive reduction potentials are easily
oxidized.
 The table can also be used to tell the strength of various oxidizing and
reducing agents.
 It can also be used as an activity series.
Metals having less positive reduction potentials are more active and
will replace metals with more positive potentials.
Zn
1 hour
CuSO4
Calculating Standard Cell Potential
E
o
Cell
E
o
Cathode
E
o
Anode
E
o
Re d
E
o
Ox
IMPORTANT: Write both equations AS IS from the table in the
reduction form with their voltages.
Example: Calculate E0 for the cell shown in the Figure below:
Cu2+ + 2e-  Cu(s)
Eo = 0.337 V
Zn2+ + 2e-  Zn(s)
Eo = -0.763 V
e-
V
e-
o
o
o
o
o
ECell
 ECathode
 E Anode
 ERe

E
d
Ox
Eo = (0.337) – (-0.763) = +1.10 V
Zn
Zn2
+
Cu
NO3-
Cu2+
NO3
-
Remember:
-Cu should be the cathode (it has higher Eo).
- Oxidation occurs at the anode (may show mass decrease).
- Reduction occurs at the cathode (may show mass increase).
- In terms of electrode charge
Electrolytic cell
Anode (+)
Cathode (-)
Galvanic cell
Anode (-)
Cathode (+)
- (°) means standard conditions: 1atm, 1M, 25C.
- Negative Eo implies non-spontaneous
- Positive Eo implies spontaneous (would be a good battery!).
Example: Consider a galvanic cell based on the reaction:
Ag+(aq) + Sn(s) → Ag(s) + Sn2+(aq)
Give the balanced cell reaction and calculate E° for the cell.
Ag+ + e-  Ag(s)
Sn2+ + 2e-  Sn(s)
Eo = 0.80 V
o
o
o
ECell
 ECathode
 E Anode
Eo = -0.14 V
o
o
 EAg
 ESn
2Ag+ + Sn  2Ag(s) + Sn2+
Eo = 0.80 – (-0.14) = 0.94 V
Example: Will the following mixture react spontaneously at standard state?
Zn2+(aq) + Fe2+(aq)  ???
Given:
(a) Zn2+ + 2e- 
(b) Fe3+ + e- 
Zn
Fe2+
(a) -2(b): Zn2+ + Fe2+  Zn(s)
Eo= -0.76 V
Eo = 0.77 V
+ 2Fe3+
E0 for the overall reaction = -0.76 – (0.77) = -1.53 V
The forward reaction is not spontaneous.
Example:
Using the table of standard reduction potentials,
predict whether 1 M HNO3 will dissolve gold metal to form a 1 M
Au3+ solution. …… ….. No!
Au3+ + 3e-  Au(s)
Eo = 1.50 V
NO3- +4H3O+ + 3e-  NO(g) + 6H2O(l)
Eo = 0.96 V
To dissolve Au, it should be as Au3+
NO3- +4H3O+ + Au 
Au3+ + NO(g) + 6H2O(l)
o
o
o
ECell
 ECathode
 E Anode
 0.96  1.50  0.54V
Eo = ?? V
Dependence of Cell Potential on Concentration
Voltaic cells at non-standard conditions -- LeChatlier’s principle can
be applied.
An increase in the concentration of a reactant will favor the forward
reaction and the cell potential will increase.
The converse is also true!
Example: For the cell reaction:
2Al + 3Mn2+ → 2Al3+ + 3Mn E°cell = ??
Predict whether Ecell is larger or smaller than E°cell for the following
cases:
a. [Al3+ ] = 2.0 M, [Mn2+ ] = 1.0 M
A: Ecell < E°cell
b. [Al3+ ] = 1.0 M, [Mn2+] = 3.0 M
B: Ecell > E°cell
When cell is not at standard conditions,
use Nernst Equation
• In a chemical reaction such as:
aA + bB  cC + dD
c
C
a
A
d
D
b
B
a .a
o
G  G  2.3RT log
 G  2.3RT log Q
a .a
o
Substitute
G  nFE
G o  nFE o
Where:
Go : Free energy change when all the reactants and products are
in their standard states (unit activity).
R : is the gas constant.
T : is the temperature in the absolute temperature
Q : Reaction Quotient
aCc .aDd
G  G  2.3RT log a b
a A .aB
o
c
d
[
C
]
[
D
]
 nFE  nFE o  2.3RT log
[ A]a [ B]b
c
d
2
.
3
RT
[
C
]
[
D
]
E  Eo 
log
nF
[ A]a [ B]b
Where concentrations are substituted for activities
At 298 K the equation becomes
K
c
d
0
.
0591
[
C
]
[
D
]
E  Eo 
log
…… Nernst Equation
n
[ A]a [ B]b
 At Equilibrium, G = 0, E = 0. Hence
0.0591
0E 
log K
n
o
0.0591
E 
log K
n
o
Concentration Cells
We can construct a cell where both
compartments contain the same
components
BUT
at
different
concentrations.
In
the
picture,
Silver will be deposited on the right electrode, thus lowering the
concentration of Ag+ in the right compartment. In the left
compartment the silver electrode dissolves [producing Ag+ ions]
to raise the concentration of Ag+ in solution.
Example: Using the table of standard reduction potentials,
calculate ∆G° for the reaction:
• Is this reaction spontaneous?
Cu2+ + Fe → Cu + Fe2+
………. Yes!
Example : Determine Eocell and Ecell based on the following halfreactions:
VO2+ + 2H+ + e- → VO2+ + H2O
Zn2+ + 2e- → Zn
E° = 1.00 V
E° = -0.76V
Where: 25°C, [VO2+] = 2.0 M, [H+] = 0.50 M,
[VO2+] = 1.0 x 10-2 M, [Zn2+] = 1.0 x 10-1 M
o
Cell
E
E
o
Cathode
E
o
Anode
= 1.00 – (-0.76) = 1.76 V
2VO2+ + 4H+ + Zn → 2VO2+ + 2H2O + Zn2+
0.0591
[VO 2 ]2 .[ Zn 2 ]
EE 
log
n
[VO2 ]2 .[ H  ]4
o
0.0591
(0.01) 2 .(0.1)
 1.76 
log
 1.89V
2
4
2
(2.0) .(0.5)
Example:
Calculate Kw, the ion-product constant of water, using the given
data.
H 2 O + e- 
H+ + e- 
1/2H2 + OH-
Eo = -0.83 V
Eo = 0.00 V
1/2H2


K w  [ H ].[OH ]
Should be
Anode
Therefore, we need this equation: the ion-product
H2O
 H+ + OH-
E oCell  E ocathode  E oAnode  (0.83)  (0)  0.83V
E oCell  0.83  0.0591* log[ H  ].[OH  ]
0  0.83  0.0591* log Kw
Kw = 1 x 10-14
At equilibrium,
Ecell = 0, and
[H+][OH-] = Kw