Question

Calculate the cell voltage of the voltaic cell which is set up by joining the following half-cells at 25°C:
Al/Al³⁺ (0.001 M) and Ni/Ni²⁺ (0.001 M)
Given: \( E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V}, \, E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66 \, \text{V} \)

Hint:

E\textsubscript{cell} is always cathode minus anode. The more positive reduction potential is the cathode.

Answer 1

Cell notation: \( \text{Al} | \text{Al}^{3+} (0.001 \, \text{M}) || \text{Ni}^{2+} (0.001 \, \text{M}) | \text{Ni} \)

Cell potential (E_cell):

\( E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.25 \, \text{V}) - (-1.66 \, \text{V}) = 1.41 \, \text{V} \)