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The Nernst equation for a half-cell reaction at 298 K is:

A$E = E^{\circ} - n \cdot 0.059 \log Q$
B$E = E^{\circ} + \dfrac{0.059}{n} \log Q$
C$E = E^{\circ} - \dfrac{0.059}{n} \log Q$
D$E = E^{\circ} - 0.059 \log [\mathrm{H^+}]$
Answer & Solution
Correct answer: C. $E = E^{\circ} - \dfrac{0.059}{n} \log Q$
Full form: $E = E^{\circ} - \dfrac{RT}{nF} \ln Q$. At 298 K with $\log_{10}$, the constants collapse to $\dfrac{0.0592}{n}$, often rounded to $\dfrac{0.059}{n}$. Here $n$ is the number of electrons exchanged in the half-cell, and $Q$ is the reaction quotient. The negative sign matters: when $Q > 1$ (products dominate), $E$ is below $E^{\circ}$. Option B has the wrong sign; option D scrambles where $n$ goes.
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