A voltaic cell is set up at 25°C with the following half cells: Al/Al³⁺(0.001M) and Ni/Ni²⁺(0.50M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential. E°Ni²⁺/Ni = -4.25V and E°Al³⁺/Al = -5.66V (log(8×10⁻⁶) = -5.097)

Al³⁺ + 3e⁻ → Al E° = -1.66V
Ni²⁺ + 2e⁻ → Ni E° = -0.25V
E°cell = -0.25 - (-1.66) = 1.41 V
Thus, the aluminum electrode is the anode and the nickel electrode is the cathode.
3Ni²⁺ + 2Al → 3Ni + 2Al³⁺
E° = E°cell
E = E° - (0.05916/n)logQ
E = 1.41 + (0.05916/6)log[(0.001)²/(0.5)³] = 1.41 + (0.05916/6)log(8×10⁻⁶) = 1.41 - 0.05916/6 * 5.097 = 1.41 - 0.0508 = 1.359V