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Galvanic Cell Example Problem

Constructing Galvanic Cells using Standard Reduction Potentials

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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.

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.

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)

Answers:

a. The half-reaction O2 + 4 H+ + 4 e- → 2 H2O is the cathode.
b. The half-reaction Ni → Ni2+ + 2 e- is the anode.
c. The balanced cell reaction is:
O2(g) + 4 H+(aq) + 2 Ni(s) → 2 H2(ℓ) + 2 Ni2+(aq)
d. The cell EMF is 1.486 volts.

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