What is the correct Nernst equation for M^2+ (aq) + 2e^+ → M (s) at 45°C?(a) E°(M2+/M) + 0.315log10 (1 / [M]^+2)(b) E° (M2+/M) + 0.0425log10 (1 / [M]^+2)(c) E° (M2+/M) + 0.0315log10 (1 / [M]^+2)(d) E° (M2+/M) + 0.0326log10 (1 / [M]^+2)I have been asked this question by my college professor while I was bunking the class.This interesting question is from Electrochemistry in division Electrochemistry of Chemistry – Class 12

What is the correct Nernst equation for M^2+ (aq) + 2e^+ → M (s) at

1.	What is the correct Nernst equation for M^2+ (aq) + 2e^+ → M (s) at 45°C?(a) E°(M2+/M) + 0.315log10 (1 / [M]^+2)(b) E° (M2+/M) + 0.0425log10 (1 / [M]^+2)(c) E° (M2+/M) + 0.0315log10 (1 / [M]^+2)(d) E° (M2+/M) + 0.0326log10 (1 / [M]^+2)I have been asked this question by my college professor while I was bunking the class.This interesting question is from Electrochemistry in division Electrochemistry of Chemistry – Class 12
Answer» Correct answer is (c) E° (M2+/M) + 0.0315log10 (1 / [M]^+2) The best I can explain: Given, Temperature T = 45°C We know, n (number of electrons transferred) = 2 According to Nernst EQUATION, E(M2+/M) = E° (M2+/M) +2.303\(\frac{RT}{nF}\)log10 (M / [M]^+2) Concentration of [M] is taken to be 1 The equation becomes: E° (M2+/M) +2.303 \(\frac{RT}{nF}\)log10 (1/[M]^+2) E(M2+/M) = E° (M2+/M) +2.303 × \(\frac{8.314 \times 318}{2 \times 96500}\)log10 (1 / [M]^+2) =E° (M2+/M) + 0.0315log10 (1 / [M]^+2).

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