Galvanic Cell Example Problem

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.

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 E⁰_cell of the galvanic cell.

To be galvanic, the electrochemical cell must have a total E⁰_cell > 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.

E⁰_Oxidation = - E⁰_Reduction
E⁰_Oxidation = -(-0.257 V) = 0.257 V

Cell EMF = E⁰_cell = E⁰_Reduction + E⁰_Oxidation
E⁰_cell = 1.229 V + 0.257 V
E⁰_cell = 1.486 V

**Note: If the oxygen reaction had been reversed, the E⁰_cell 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.