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Nernst Equation Example Problem

Calculate Cell Potential in Nonstandard Conditions

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The Nernst equation is used to calculate cell potential.

The Nernst equation is used to calculate cell potential.

Maria Toutoudaki, Getty Images

Standard cell potentials are calculated in standard conditions. The temperature and pressure are at standard temperature and pressure and the concentrations are all 1 M aqueous solutions. In non-standard conditions, the Nernst equation is used to calculate cell potentials. It modifies the standard cell potential to account for temperature and concentrations of the reaction participants. This example problem shows how to use the Nernst equation to calculate a cell potential.

Problem:

Find the cell potential of a galvanic cell based on the following reduction half-reactions at 25 °C

Cd2+ + 2 e- → Cd    E0 = -0.403 V
Pb2+ + 2 e- → Pb    E0 = -0.126 V

where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M.

Solution:

The first step is to determine the cell reaction and total cell potential.

In order for the cell to be galvanic, E0cell > 0.

**Review Galvanic Cell Example Problem for the method to find cell potential of a galvanic cell.

For this reaction to be galvanic, the cadmium reaction must be the oxidation reaction. Cd → Cd2+ + 2 e- E0 = +0.403 V
Pb2+ + 2 e- → Pb E0 = -0.126 V

The total cell reaction is:

Pb2+(aq) + Cd(s) → Cd2+(aq) + Pb(s)

and E0cell = 0.403 V + -0.126 V = 0.277 V

The Nernst equation is:

Ecell = E0cell - (RT/nF) x lnQ

where
Ecell is the cell potential
E0cell refers to standard cell potential
R is the gas constant (8.3145 J/mol·K)
T is the absolute temperature
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant 96485.337 C/mol )
Q is the reaction quotient, where

Q = [C]c·[D]d / [A]a·[B]b

where A, B, C, and D are chemical species; and a, b, c, and d are coefficients in the balanced equation:

a A + b B → c C + d D

In this example, the temperature is 25 °C or 300 K and 2 moles of electrons were transferred in the reaction.

RT/nF = (8.3145 J/mol·K)(300 K)/(2)(96485.337 C/mol)
RT/nF = 0.013 J/C = 0.013 V

The only thing remaining is to find the reaction quotient, Q.

Q = [products]/[reactants]

**For reaction quotient calculations, pure liquid and pure solid reactants or products are omitted.**

Q = [Cd2+]/[Pb2+]
Q = 0.020 M / 0.200 M
Q = 0.100

Combine into the Nernst equation:

Ecell = E0cell - (RT/nF) x lnQ
Ecell = 0.277 V - 0.013 V x ln(0.100)
Ecell = 0.277 V - 0.013 V x -2.303
Ecell = 0.277 V + 0.023 V
Ecell = 0.300 V

Answer:

The cell potential for the two reactions at 25 °C and [Cd2+] = 0.020 M and [Pb2+] = 0.200 M is 0.300 volts.

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