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The Nernst equation for a cell reaction at $298\,\text{K}$ with $n$ electrons exchanged is:
A$E = E^\circ - \dfrac{0.0591}{n}\log Q$
B$E = E^\circ - \dfrac{2.303 RT}{F}\log Q$
C$E = E^\circ - n \log Q$
D$E = E^\circ + \dfrac{0.0591}{n}\log Q$
Answer & Solution
Correct answer: A. $E = E^\circ - \dfrac{0.0591}{n}\log Q$
The Nernst equation is $E = E^\circ - \dfrac{2.303 RT}{nF}\log Q$. Substituting $T = 298\,\text{K}$, $R = 8.314\,\text{J K}^{-1}\,\text{mol}^{-1}$, $F = 96500\,\text{C mol}^{-1}$ gives $\dfrac{2.303 RT}{F} \approx 0.0591\,\text{V}$, hence the familiar form $E = E^\circ - \dfrac{0.0591}{n}\log Q$. Note the minus sign: a higher $Q$ (more products) lowers the cell EMF.
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