Nernst Equation | CIE A Level Chemistry Revision Notes 2025

The Nernst Equation

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

$E = E^{Θ} + \frac{R T}{z F} l n \frac{[o x i d i s e d s p e c i e s]}{[r e d u c e d s p e c i e s]}$

$E = E^{Θ} + \frac{0.059}{z} \log_{10} \frac{[o x i d i s e d s p e c i e s]}{[r e d u c e d s p e c i e s]}$

- 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

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)

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:
- $E = 0.77 + \frac{0.059}{1} \log_{10} \frac{[0.034]}{[0.64]}$
- 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)

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:
- $E = E^{Θ} + \frac{0.059}{z} \log_{10} \frac{[o x i d i s e d s p e c i e s]}{[r e d u c e d s p e c i e s]}$
- $E = 0.34 + \frac{0.059}{2} \log_{10} \frac{[0.0010]}{[1.0]}$
- E = (+ 0.34) + (– 0.089)
- E = + 0.25 V

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 to:

$E = E^{Θ} + \frac{0.059}{z} \log_{10} \frac{[o x i d i s e d s p e c i e s]}{[r e d u c e d s p e c i e s]}$

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