Question

Calculate the e.m.f. of the following cell: \[ \text{Mg(s)} | \text{Mg}^{2+} (0.1 M) || \text{Cu}^{2+} (0.001 M) | \text{Cu(s)} \] Given: \[ E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34V, \quad E^0_{\text{Mg}^{2+}/\text{Mg}} = -2.37V \]

Hint:

The Nernst equation helps calculate the actual cell potential under non-standard conditions.

Answer 1

Step 1: Cell Reaction: \[ \text{Mg} + \text{Cu}^{2+} \rightarrow \text{Mg}^{2+} + \text{Cu} \] Step 2: Standard E.M.F. Calculation: \[ E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0 \] \[ = 0.34 - (-2.37) = 2.71V \] Step 3: Nernst Equation: \[ E_{\text{cell}} = E_{\text{cell}}^0 - \frac{0.0591}{n} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \] \[ E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log \frac{0.1}{0.001} \] \[ E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \times 2 \] \[ E_{\text{cell}} = 2.71 - 0.0591 = 2.6509V \] Thus, the e.m.f. of the cell is 2.65V.