Standard Reduction
Potentials
Electrochem III
Balancing Oxidation-Reduction Equations
Perhaps the
easiest way to
balance the
equation of an
oxidationreduction
reaction is via
the half-reaction
method.
Balancing Oxidation-Reduction Equations
This involves
treating the
oxidation and
reduction as two
separate
processes,
balancing these
half reactions, and
then combining
them to attain the
balanced equation
for the overall
reaction.
The Half-Reaction Method
1.Assign oxidation
numbers to determine
what is oxidized and
what is reduced.
2.Write the oxidation
and reduction halfreactions.
The Half-Reaction Method
3.Balance each halfreaction.
•
•
•
•
Balance elements other
than H and O.
Balance O by adding
H2O.
Balance H by adding H+.
Balance charge by
adding electrons.
4.Multiply the half-
reactions by integers
so that the electrons
gained and lost are the
same.
The Half-Reaction Method
5.Add the half-
reactions,
subtracting things
that appear on both
sides.
Make sure the
equation is balanced
according to mass.
Make sure the
equation is balanced
according to charge.
6.
7.
The Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4− (aq) + C2O42− (aq)  Mn2+ (aq) + CO2 (aq)
The Half-Reaction Method
• First, we assign oxidation numbers.
+7
+3
+2
+4
MnO4− + C2O42-  Mn2+ + CO2
• Manganese is oxidized.
• Carbon is reduced.
The Half-Reaction Method
C2O42−  CO2
To balance the carbon, we add a
coefficient of 2:
C2O42−  2 CO2
The Half-Reaction Method
C2O42−  2 CO2
The oxygen is now balanced as well. To
balance the charge, we must add 2
electrons to the right side.
C2O42−  2 CO2 + 2 e−
The Half-Reaction Method
MnO4−  Mn2+
The manganese is balanced; to balance
the oxygen, we must add 4 waters to the
right side.
MnO4−  Mn2+ + 4 H2O
The Half-Reaction Method
MnO4−  Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+
to the left side.
8 H+ + MnO4−  Mn2+ + 4 H2O
The Half-Reaction Method
8 H+ + MnO4−  Mn2+ + 4 H2O
To balance the charge, we add 5 e− to
the left side.
5 e− + 8 H+ + MnO4− 
Mn2+ + 4 H2O
The Half-Reaction Method
Now we evaluate the two half-reactions together:
C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
To attain the same number of electrons on each
side, we will multiply the first reaction by 5 and
the second by 2.
The Half-Reaction Method
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The Half-Reaction Method
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides
are the electrons. Subtracting them, we
are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2++ 8 H2O + 10 CO2
Half Reactions in Basic Solutions
If a reaction occurs in basic solution,
one can balance it as if it occurred in
acid.
Once the equation is balanced, add
OH− to each side to “neutralize” the
H+ in the equation and create water in
its place.
If this produces water on both sides,
you might have to subtract water
from each side.
•
•
•
•
•
Standard Cell Potentials
A galvanic cell is always a redox reaction that
has been separated into 2 half reactions.
Each half reaction is assigned a potential from
which the cell potential can be derived.
• For example:
• Redox rxn: 2H (aq) + Zn(s)  Zn
• The Cell Potential is 0.76 V
+
Anode
Zn (s)  Zn2+ (aq) +2e-
2+
(aq) + H2 (g)
Cathode
2H+(aq) + 2e- H2 (g)
Standard Hydrogen Electrode
Definition:
When the platinum cathode is this half reaction has a
concentration of 1 M [H+] with H2 gas at 1 atm.
We can measure Ecell of the total cell at 0.76 V, but
there is no way to measure each half reaction
directly.
To deal with this, the standard hydrogen electrode
has been assigned a cell potential of 0.0V.
Since the Standard Hydrogen Electrode is assigned
0.0V and the total Ecell = 0.76V,
then Zn(s)  Zn2+(aq) +2e- has a potential of 0.76 V
Why???
Because E°cell= E°cathode +E°Anode
E°cell= E°H+H2+ E°Zn  Zn2+
The ° indicates standard states are employed
with 2H+(aq) + 2e- H2 (g) as 0.0V
We can assign values to all other half reactions.
For example:
Zn (s) + Cu 2+  Zn 2+ + Cu (s)
What are the half reactions?
Anode Zn(s) Zn2+ (aq) + 2eCathode Cu (aq) + 2e-  Cu (s)
LEO
GER
Zn (s) + Cu
2+
 Zn
2+
+ Cu (s)
E°cell= E°cathode +E°Anode
The E°cell is measured at 1.10 V,
And E°Anode is determined that Zn(s) Zn2+ (aq)
+ 2e- is 0.76V,
Then E°cathode Cu (aq) + 2e-  Cu (s) must be
0.34 V
Because 1.10V = 0.76 V + 0.34 V
measured known calculated
The value of the standard
hydrogen electrode.
The value of the
standard hydrogen
electrode of 0.0V is
accepted by the
scientific
community.
Standard Reduction Potentials
When the cell potentials of half reactions are given,
they are written as:
Standard Reduction Potentials: All half reactions
have solutions at 1 M and gases at 1 atm.
Zn2+ (aq) + 2e-  Zn (s)
Cu+ (aq) + e-  Cu (s)
2H+(aq) + 2e- H2 (g)
They are all written as a gain of electrons,
reduction. GER
Combining 2 half reactions to obtain a
balanced oxidation-reduction reaction
require 2 manipulations:
1.One of the reactions must be reversed, which
means the sign of its E° must be reversed.
2.Since the number of electrons lost by the anode
must equal the number of electrons gained by
the cathode. The half reactions are multiplied to
find the lowest common value.
But the value of the E° does not change!!
Note: Standard reduction potentials are intensive
properties. It does not matter how many times
the reaction occurs, only that it occurs.
Consider a Galvanic Cells
Based on the unbalanced Reaction
Fe 3+ (aq) + Cu (s)  Cu 2+ (aq) + Fe 2+
(aq)
The two half reactions are as standard
reduction potentials:
Fe 3+ (aq) + 1e-  Fe 2+ (aq)
E° = 0.77V
Cu 2+ (aq) + 2e-  Cu (s)
E° = 0.34V
E°cell continued
To balance the cell reaction and
calculate the E° equation number 2
must be reversed and the sign of the
E° must be reversed.
Cu (s)  Cu 2+ (aq) + 2eE°= -0.34V
And one must double the first equation to
balance the electrons.
2Fe 3+(aq) + 2e-  2Fe 2+(aq) E°=0.77V
NOTE: The value of E°is not multiplied
Sum up the two half reactions
Cu (s)  Cu 2+ (aq) + 2eE°= -0.34V
2Fe 3+(aq) + 2e-  2Fe 2+(aq) E°=0.77V
2Fe3+(aq)+Cu(s) 2Fe2+(aq)+Cu2+(aq)
E°=0.43V
Other Examples
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