The Nernst equation describes the relationship between the concentration of aqueous ions in a half-cell and the electrode potential.
\( E = E^\ominus + \left(\frac{0.0592}{z}\right) \ln \left(\frac{[\text{oxidised}]}{[\text{reduced}]}\right) \)
E is the electrode potential under non-standard conditions
E⦵ is the standard electrode potential
z is the number of electrons transferred in the reaction
When the concentration of aqueous ions changes from 1 moldm-3, it will cause the electrode value of the half-cell equation to increase or decrease. Therefore, the electrode potential changes. Let's try to understand this better by solving a question.
[Question]
Calculate the electrode potential of a Cu(s) / Cu2+(aq) electrode at 298 K when the concentration of Cu2+(aq) ions is 0.001 mol dm-3.
Step 1: Identify the half-reaction and the number of electrons transferred (z).
Cu2+(aq) + 2e− ⇌ Cu(s).
z=2 as 2 electrons are transferred
Step 2: Determine the standard electrode potential (E⦵) for the half-cell.
By referring to a table of half-cell electrode values, we find that E⦵ for Cu(s)/Cu2+ is +0.34V.
Step 3: Identify the concentrations of the oxidised and reduced forms of the species in the half-cell.
The oxidised form is the species with a higher oxidation number. Cu2+ has an oxidation number of +2 and Cu has an oxidation number of 0.
Thus, Cu2+ is the oxidised form and Cu is the reduced form.
Step 4: Substitute the values into the Nernst equation and calculate E.
Cu is a solid, so its concentration is set to 1moldm-3. Concentration of solids in half-cells is always considered to be 1moldm-3.
\( E = +0.34 + \left(\frac{0.0592}{2}\right) \ln \left(\frac{\text{0.001}}{\text{1}}\right) = +0.34 + (-0.089) = +0.25V\)
[Important note]
If you are trying to find the change in the voltage of a galvanic cell containing two types of metals (such as a Zn/Cu cell), you first calculate the half-cell electrode potentials for both Zn and Cu individually. Then, you plug them into the equation \( E_{cell} = E_{cathode} - E_{anode} \).