Discussion
In the following cell reaction,
Zn(s) + Cu2+ (0.1 M) ⇋ Zn2+ (0.001 M) + Cu(s) , at 298 K,
Calculate the Ecell at 298 K if E0cell at this temperature is 1.1V. (2.303 RT/F = 0.059 V at 298 K)
Answer: C
\(E_{cell}\) is 1.159 V.
Two electrons are transferred. By the Nernst equation,
\[ \begin{aligned} E &= E^{0} - \frac{0.059}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \\ &= 1.1 - 0.0295\log\frac{0.001}{0.1} \\ &= 1.1 + 0.059 = 1.159\ \text{V} \end{aligned} \]
The low Zn2+ concentration pushes the reaction forward, so E is above \(E^{0}\). 1.041 V gets the sign of the log term wrong, and 1.218 V uses n = 1.
Note the half in \(\frac{0.059}{n}\): with n = 2 the log term changes E by 0.0295 V per factor of ten in the concentration ratio, and the ratio here is a hundredfold.
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