Question

Calculate emf of the given cell: Mg(s) $\mid$ Mg$^{2+}$(0.1 M) $\parallel$ Cu$^{2+}$(1.0 × 10$^{-3}$ M) $\mid$ Cu(s)
Given:
$E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34$ V, \quad $E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37$ V
(log 100 = 2)

Hint:

Use:
- \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
- Nernst equation for non-standard conditions: \(E = E^\circ - \frac{0.0591}{n} \log Q\)

Answer 1

Step 1: Identify anode and cathode.
- Mg has lower (more negative) standard reduction potential → acts as anode (oxidation).
- Cu has higher (more positive) reduction potential → acts as cathode (reduction).
Step 2: Standard EMF of the cell:
\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-2.37) = 2.71 \, \text{V}\] Step 3: Apply Nernst equation:
\[E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{products}]}{[\text{reactants}]}\] Step 4: For the reaction:
Mg(s) + Cu$^{2+}$(aq) → Mg$^{2+}$(aq) + Cu(s)
Step 5: Substitute values:
\[E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log \left( \frac{0.1}{1.0 \times 10^{-3}} \right)\] Step 6: Simplify:
\(\frac{0.1}{1.0 \times 10^{-3}} = 100\), so \(\log 100 = 2\)
\[E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \times 2 = 2.71 - 0.0591 = 2.6509 \approx 2.61 \, \text{V}\]