Discussion
The EMF of the following cell at 298K is
Mg(s)|Mg2+(aq) (0.10M) || Ag+(aq)(0.001M)| Ag(s)
(Given: E0cell=3.17V and 2.303RT/F = 0.06 V)
Answer: C
The EMF is 3.02 V.
The cell reaction is Mg + 2Ag+ → Mg2+ + 2Ag, with n = 2. Apply the Nernst equation.
\[ \begin{aligned} E &= E^{0} - \frac{0.06}{2}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^{+}}]^{2}} \\ &= 3.17 - 0.03\log\frac{0.10}{(10^{-3})^{2}} \\ &= 3.17 - 0.03 \times 5 = 3.02\ \text{V} \end{aligned} \]
3.32 V adds the correction (3.17 + 0.15) instead of subtracting it, and 3.17 V is the standard EMF, which would apply only at unit concentrations. Forgetting to square [Ag+] gives log 100 = 2 and 3.11 V, which is not offered.
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