Nernst Equation (Cambridge (CIE) A Level Chemistry): Revision Note

Calculating the electrode potential of a Fe³⁺ / Fe²⁺ half-cell

Calculate the electrode potential at 298K of a Fe³⁺ / Fe²⁺ half-cell.

Fe³⁺ (aq) + e^– $\rightleftharpoons$ Fe²⁺ (aq)

• [Fe³⁺] = 0.034 mol dm^-3 

• [Fe²⁺] = 0.64 mol dm^-3 

• E^θ = +0.77 V

Answer

• From the question, the relevant values for the Fe³⁺ / Fe²⁺ half-cell are:

  • [Fe³⁺] = 0.034 mol dm^-3

  • [Fe²⁺] = 0.64 mol dm^-3 

  • E^Θ = + 0.77 V

• The oxidised species is Fe³⁺ as it has a higher oxidation number (+3)

• The reduced species is Fe²⁺ as it has a lower oxidation number (+2)

• z is 1 as only one electron is transferred in this reaction

• The Nernst equation for this half-reaction is, therefore:

  • $\mathit{E}\mathit{=}\mathbf{0}\mathbf{.}\mathbf{77}\mathbf{+}\frac{\mathbf{0}\mathbf{.}\mathbf{059}}{\mathbf{1}}{\mathbf{log}}_{\mathbf{10}}\frac{\left[0.034\right]}{\left[0.64\right]}$

  • E = (+0.77) + (-0.075)

  • E = +0.69 V

Calculating the electrode potential of a Cu²⁺ / Cu half-cell

Calculate the electrode potential at 298K of a Cu²⁺ / Cu half-cell.

Cu²⁺ (aq) + 2e^– $\rightleftharpoons$ Cu (s)

• [Cu²⁺] = 0.001 mol dm^-3 

• E^θ = +0.34 V

Answer

• From the question, the relevant values for the Cu²⁺ / Cu half-cell are:

  • [Cu²⁺] = 0.0010 mol dm^-3

  • E^Θ = + 0.34 V

• The oxidised species is Cu²⁺ as it has a higher oxidation number (+2)

• The reduced species is Cu as it has a lower oxidation number (0)

• Cu is solid which means that it is not included in the Nernst equation

  • Its concentration does not change and is, therefore, fixed at 1.0

• z is 2 as 2 electrons are transferred in this reaction

• The Nernst equation for this half-reaction is, therefore:

  • $\mathit{E}\mathit{=}{\mathit{E}}^{\mathit{\Theta }}\mathit{+}\frac{\mathbf{0}\mathbf{.}\mathbf{059}}{\mathit{z}}{\mathbf{log}}_{\mathit{10}}\frac{\left[oxidisedspecies\right]}{\left[reducedspecies\right]}$

  • $\mathit{E}\mathit{=}\mathbf{0}\mathbf{.}\mathbf{34}\mathbf{+}\frac{\mathbf{0}\mathbf{.}\mathbf{059}}{\mathbf{2}}{\mathbf{log}}_{\mathbf{10}}\frac{\left[0.0010\right]}{\left[1.0\right]}$

  • E  = (+ 0.34) + (– 0.089)

  • E = + 0.25 V

• You need to know the Nernst equation, so make sure you learn it

  • CIE specifically ask students to learn use this version:

$\mathit{E}\mathit{=}{\mathit{E}}^{\mathit{\Theta }}\mathit{+}\frac{\mathbf{0}\mathbf{.}\mathbf{059}}{\mathit{z}}{\mathbf{log}}_{\mathit{10}}\frac{\left[oxidisedspecies\right]}{\left[reducedspecies\right]}$

• Make sure you always check what the temperature is

• If the temperature is not 298 K (or 25 ^oC) the full Nernst equation should be used

• You don’t need to know how to simplify the Nernst equation

• You are only expected to use the equation when the temperature is 298 K (or 25 ^oC)