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Electrochemical Cell EMF Example Problem

How to Calculate Cell EMF for an Electrochemical Cell


EMF is the net voltage of half-reactions in an electrochemical cell.

EMF is the net voltage of half-reactions in an electrochemical cell.

Maria Toutoudaki, Getty Images

The cell electromotive force, or cell EMF is the net voltage between the oxidation and reduction half-reactions taking place between two redox half-reactions. Cell EMF is used to determine whether or not the cell is galvanic. This example problem shows how to calculate the cell EMF using standard reduction potentials.

The Table of Standard Reduction Potentials is needed for this example.


Consider the redox reaction:

Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)

a) Calculate the cell EMF for the reaction.
b) Identify if the reaction is galvanic.


Step 1: Break the redox reaction into reduction and oxidation half-reactions.

Hydrogen ions, H+ gain electrons when forming hydrogen gas, H2. The hydrogen atoms are reduced by the half-reaction:

2 H+ + 2 e- → H2

Magnesium loses two electrons and is oxidized by the half-reaction:

Mg → Mg2+ + 2 e-

Step 2: Find the standard reduction potentials for the half-reactions.

Reduction: E0 = 0.0000 V

The table shows reduction half-reactions and standard reduction potentials. To find E0 for an oxidation reaction, reverse the reaction.

Reversed reaction:

Mg2+ + 2 e- → Mg

This reaction has a E0 = -2.372 V.

E0Oxidation = - E0Reduction

E0Oxidation = - (-2.372 V) = + 2.372 V

Step 3: Add the two E0 together to find the total cell EMF, E0cell

E0cell = E0reduction + E0oxidation

E0cell = 0.0000 V + 2.372 V = +2.372 V

Step 4: Determine if reaction is galvanic. Redox reactions with a positive E0cell value are galvanic.
This reaction's E0cell is positive and therefore galvanic.


The cell EMF of the reaction is +2.372 Volts and is galvanic.

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