# Galvanic Cell Example Problem

## Constructing Galvanic Cells using Standard Reduction Potentials

Galvanic cells are electrochemical cells which use the transfer of electrons in redox reactions to supply an electric current. This example problem illustrates how to form a galvanic cell from two reduction reactions and calculate the cell EMF.

### Galvanic Cell Problem

Given the following reduction half-reactions:

O2 + 4 H+ + 4 e- → 2 H2O
Ni2+ + 2 e- → Ni

Construct a galvanic cell using these reactions. Find:

a) Which half-reaction is the cathode.
b) Which half-reaction is the anode.
c) Write and balance the total cell redox reaction.
d) Calculate E0cell of the galvanic cell.

### How to Find the Solution

To be galvanic, the electrochemical cell must have a total E0cell > 0.

From the Table of Common Standard Reduction Potentials:

O2 + 4 H+ + 4 e- → 2 H2O E0 = 1.229 V
Ni2+ + 2 e- → Ni E0 = -0.257 V

To construct a cell, one of the half-reactions must be an oxidation reaction. To make a reduction half-reaction into an oxidation half-reaction, the half-reaction is reversed. The cell will be galvanic if the nickel half-reaction is reversed.

E0Oxidation = - E0Reduction
E0Oxidation = -(-0.257 V) = 0.257 V

Cell EMF = E0cell = E0Reduction + E0Oxidation
E0cell = 1.229 V + 0.257 V
E0cell = 1.486 V

**Note: If the oxygen reaction had been reversed, the E0cell would not have been positive and the cell would not be galvanic.** In galvanic cells, the cathode is the location of the reduction half-reaction and the anode is where the oxidation half-reaction takes place.

Cathode: O2 + 4 H+ + 4 e- → 2 H2O
Anode: Ni → Ni2+ + 2 e-

To find the total reaction, the two half-reactions must be combined.

O2 + 4 H+ + 4 e- → 2 H2O
+ Ni → Ni2+ + 2 e-

To balance the total number of electrons on both sides, the nickel half-reaction must be doubled.

O2 + 4 H+ + 4 e- → 2 H2O
+ 2 Ni → 2 Ni2+ + 4 e-

Combine the reactions:

O2(g) + 4 H+(aq) + 2 Ni(s) → 2 H2(ℓ) + 2 Ni2+(aq)