Explain how the Nernst equation reduces to E = E° − (0.05916/n) log Q at 25°C.

• A. At T = 0 K.
• B. Only if Q = 1.
• C. At T = 298 K, E = E° − (RT/nF) ln Q = E° − (0.05916/n) log10 Q.
• D. It never reduces to this form.

Answer: (C)

The Nernst equation links the cell potential to temperature, the number of electrons transferred, and the reaction quotient: E = E° − (RT/nF) ln Q. To use a base-10 logarithm, convert the natural log: ln Q = 2.303 log10 Q. This gives E = E° − (2.303 RT/nF) log10 Q. At 25°C (298 K), the factor RT/F is about 0.025693 V, so 2.303 RT/F ≈ 0.05916 V. Therefore E = E° − (0.05916/n) log10 Q. This shows exactly why the equation takes that simplified form at 25°C. The exact constant would differ at other temperatures, in which case you’d keep the RT/nF ln Q form or use the corresponding (2.303 RT/F) value for that temperature.