Question

Calculate emf of the following cell at 25°C: \[ \text{Zn}(s) \, | \, \text{Zn}^{2+} (0.1M) \, || \, \text{Cd}^{2+} (0.01M) \, | \, \text{Cd}(s) \] Given: \[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.40 \, \text{V}, \quad E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.76 \, \text{V}. \]

Hint:

The Nernst equation is used to calculate the emf of an electrochemical cell under non-standard conditions by considering the concentrations of the ions involved.

Answer 1

The emf of the cell is given by the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \] Where: - \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \). - \( n = 2 \) (the number of electrons transferred). The standard cell potential is: \[ E^\circ_{\text{cell}} = (-0.76) - (-0.40) = -0.36 \, \text{V}. \] Now, substitute into the Nernst equation: \[ E_{\text{cell}} = -0.36 - \frac{0.0592}{2} \log \left( \frac{0.01}{0.1} \right) \] \[ E_{\text{cell}} = -0.36 - \frac{0.0592}{2} \log (0.1) \] \[ E_{\text{cell}} = -0.36 - \frac{0.0592}{2} \times (-1) = -0.36 + 0.0296 = -0.33 \, \text{V}. \] Thus, the emf of the cell at 25°C is \( -0.33 \, \text{V} \).