Question

Calculate the cell voltage of the voltaic cell which is set up by joining the following half-cells at 25°C:
Al/Al³⁺ (0.002 M) and Ni/Ni²⁺ (0.002 M)
Given: \( E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \)
\( E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66 \, \text{V} \)
\( \log 5 = 0.6990 \)

Hint:

Always identify the anode and cathode correctly before applying the Nernst equation.

Answer 1

Cell Reaction: \( 2\text{Al} + 3\text{Ni}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Ni} \)

Cell potential calculation:
\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.25 - (-1.66) = 1.41 \, \text{V} \)

Using Nernst Equation:

\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Al}^{3+}]^2}{[\text{Ni}^{2+}]^3} \right) \)

Given:
\( n = 6 \)

\( E_{\text{cell}} = 1.41 - \frac{0.0591}{6} \log \left( \frac{(0.002)^2}{(0.002)^3} \right) \)

\( = 1.41 - \frac{0.0591}{6} \log(0.002) \)

\( = 1.41 - \frac{0.0591}{6} \cdot (-2.699) \)

\( = 1.41 + 0.0266 \)

\( = 1.4366 \, \text{V} \) (approx)