Discussion
What is the emf of the cell at 298 K in which the following reaction takes place?
Ni(s) + 2Ag+(0.002M) → Ni2+(0.04M) + 2Ag(s). (E°cell at 298 K = 1.05V)
Answer: B
The emf is 0.93 V.
Two electrons are transferred (n = 2). By the Nernst equation at 298 K:
\[ \begin{aligned} E &= E^{\circ} - \frac{0.059}{2}\log\frac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 1.05 - 0.0295\log\frac{0.04}{(0.002)^{2}} \\ &= 1.05 - 0.0295 \times 4 = 0.932\ \text{V} \end{aligned} \]
The ratio inside the logarithm is \(\frac{0.04}{4 \times 10^{-6}} = 10^{4}\). Forgetting to square \([\mathrm{Ag^+}]\), or adding the correction, leads to the other values.
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