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

You need to know the Nernst equation, so make sure you learn it
- CIE specifically ask students to learn use this version:

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

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)