Nernst Equation (CIE A Level Chemistry)

• Under non-standard conditions, the cell potential of the half-cells is shown by the symbol E_cell
• The effect of changes in temperature and ion concentration on the E_cell can be deduced using the Nernst equation

E = electrode potential under nonstandard conditions

E^θ = standard electrode potential

R = gas constant (8.31 J K^-1 mol^-1)

T = temperature (kelvin, K)

z = number of electrons transferred in the reaction

F = Faraday constant (96 500 C mol^-1)

ln = natural logarithm

• This equation can be simplified to

• • At standard temperature, R, T and F are constant
  • ln x = 2.303 log₁₀ x

• The Nernst equation only depends on aqueous ions and not solids or gases
• The concentrations of solids and gases are therefore set to 1.0 mol dm^-3

Applying Nernst Equation

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

Answer

• From the question, the concentrations of ions for the Fe³⁺ / Fe²⁺ half-cell are as follows:
  • [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:

      E = (+0.77) + (-0.075)

      E = +0.69 V

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

Answer

• From the question, the concentrations of ions for the Fe³⁺ / Fe²⁺ half-cell are as follows:
  • [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 as 1.0
• z is 2 as 2 electrons are transferred in this reaction
• The Nernst equation for this half-reaction is, therefore:

      E  = (+ 0.34) + (– 0.089)

      E = + 0.25 V