Nernst Equation (CIE A Level Chemistry)

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^–  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:
  • 
  • 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^–  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:
  • 
  • 
  • E  = (+ 0.34) + (– 0.089)
  • E = + 0.25 V