Question

What will be the emf of the following cell at 25°C?
\[ \text{Fe} \, / \, \text{Fe}^{2+}(0.001M) \, || \, \text{H}^{+}(0.01M) \, | \, \text{H}_2(g) \, (1 \, \text{Bar}) \, | \, \text{Pt}(s) \] Given: \[ E_{\text{Fe}^{2+} / \text{Fe}}^\circ = -0.44V, \quad E_{\text{H}^{+} / \text{H}_2}^\circ = 0.00V \]

• A. 0.44V
• B. -0.44V
• C. 0.41V
• D. -0.41V

Hint:

Use the Nernst equation to calculate the emf of electrochemical cells, remembering to adjust for the concentration of ions in solution.

Answer 1

The Correct Option is C

The emf of the cell can be calculated using the Nernst equation: \[ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0592}{n} \log Q \] Where \(n\) is the number of electrons transferred (here \(n = 2\) for the iron half-reaction), and \(Q\) is the reaction quotient. For the given cell: \[ E_{\text{cell}} = E_{\text{Fe}^{2+} / \text{Fe}}^\circ - \frac{0.0592}{2} \log \left( \frac{[\text{Fe}^{2+}]}{[\text{H}^{+}]^2} \right) \] Substitute the given concentrations: \[ E_{\text{cell}} = -0.44V - \frac{0.0592}{2} \log \left( \frac{0.001}{(0.01)^2} \right) \] Simplify the logarithm: \[ E_{\text{cell}} = -0.44V - \frac{0.0592}{2} \log (10) \] \[ E_{\text{cell}} = -0.44V - 0.0296 \times 1 \] \[ E_{\text{cell}} = -0.44V - 0.0296V = -0.41V \] Thus, the emf of the cell is 0.41V.