Find the emf of the cell described by the cell diagram Ni | Ni²⁺ (0.750 M) || Cu²⁺ (0.0500 M) | Cu.

Question

Quizplus · Accepted Answer

The emf of the cell can be calculated using the Nernst equation: $$E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log \frac{[Cu^{2+}]}{[Ni^{2+}]}$$. Given that the standard reduction potentials (E°) for Cu²⁺/Cu and Ni²⁺/Ni are +0.34 V and -0.25 V respectively, the standard emf (E°cell) is the difference between the two, which is 0.34 V - (-0.25 V) = 0.59 V. Since both reactions involve a 2-electron transfer (n = 2), and the concentrations of Cu²⁺ and Ni²⁺ are 0.0500 M and 0.750 M respectively, the equation simplifies to: $$E_{cell} = 0.59 V - \frac{0.0592}{2} \log \frac{0.0500}{0.750}$$. Calculating this gives an emf close to 0.56 V, making option E the correct answer.

Find the Emf of the Cell Described by the Cell