Question

Calculate the cell voltage of the voltaic cell which is set up by joining following half-cells at 25°C:
Al/Al$^{3+}$(0.001 M) and Ni/Ni$^{2+}$(0.1 M)
Given: $E^\circ_{\text{Al}} = -1.66$ V, $E^\circ_{\text{Ni}} = -0.25$ V, $n_{\text{Al}} = 3$, $n_{\text{Ni}} = 2$

Hint:

Use the Nernst equation to calculate the cell potential by considering the concentrations of the ions involved in the reaction.

Answer 1

The cell potential for the voltaic cell can be calculated using the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \] First, calculate the standard cell potential: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.25) - (-1.66) = 1.41 \, \text{V} \] Next, calculate the reaction quotient (Q): \[ Q = \frac{[\text{Ni}^{2+}] \times [\text{Al}^{3+}]}{[\text{Ni}] \times [\text{Al}]} \] Substitute the values: \[ Q = \frac{(0.1) \times (0.001)}{(1) \times (1)} = 0.0001 \] Now, using the Nernst equation: \[ E_{\text{cell}} = 1.41 - \frac{0.0592}{3} \log (0.0001) = 1.41 - \frac{0.0592}{3} \times (-4) = 1.41 + 0.0789 = 1.4889 \, \text{V} \] Thus, the cell voltage is 1.49 V.