InterviewSolution
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InterviewSolution
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In Nernst equation, the constant 0.0592 at 298 K represents the value ofA. `(RT)/(nF)`B. `(RT)/(F)`C. `(2.303 RT)/(nF)`D. `(2.303RT)/(F)` |
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Answer» Correct option is: (D) \(\frac{2.303 \ RT}{F}\) According to Nernst equation E = \(E_0 - \frac {RT}{nF}\) ln Q or E = \(E_0 - \frac {8.314 \times 298 \times 2.303}{n\times 96500}\) log Q. E = \(E_0 - \frac {0.0592}{n}\) log Q. Therefore, \(\frac{2.303 \ RT}{F}\) = 0.0592 Correct Answer - B::C::D |
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