Question

Calculate the emf of the following cell at 25°C : \[ \text{Zn(s)} | \text{Zn}^{2+} (0.1 \, \text{M}) \parallel \text{H}^+ (0.01 \, \text{M}) | \text{Hg}(l) | \text{Hg}_2^{2+} (1 \, \text{bar}), \, \text{Pt(s)} \] \[ \left[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V}, \, E^\circ_{\text{H}_2/\text{H}^+} = 0.00 \, \text{V}, \, \log 10 = 1 \right] \]

Hint:

To calculate emf from the Nernst equation, always remember to substitute correct values for concentrations, partial pressures, and standard electrode potentials.

Answer 1

The Nernst equation is used to calculate the emf of the cell: \[ E = E^\circ - \dfrac{0.0591}{n} \log Q \] For the given reaction, n = 2 (since 2 electrons are involved) and the reaction quotient \(Q\) is given by: \[ Q = \dfrac{[\text{Zn}^{2+}] [\text{Hg}_2^{2+}]}{[\text{H}^+]^2 [\text{Zn}] [\text{Hg}]} \] Substituting the given concentrations and partial pressures: \[ Q = \dfrac{(0.1) (1)}{(0.01)^2 (1) (1)} = 10^4 \] Now, we can calculate the emf: \[ E = 0.00 \, \text{V} - \dfrac{0.0591}{2} \log (10^4) = 0.00 - \dfrac{0.0591}{2} \times 4 = 0.00 - 0.1182 = -0.1182 \, \text{V} \]