Q19 / 110 · VITEEE 2022 · Chemistry

The standard e.m.f. of a galvanic cell involving cell reaction with \(n=2\) is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction would be (Given \(\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\); \(\mathrm{R}=8.314 \mathrm{~J}^{-1} \mathrm{~mol}^{-1}\))

  • A
    \(2.0 \times 10^{11}\)
  • B
    \(4.0 \times 10^{12}\)
  • C
    \(1.0 \times 10^2\)
  • Correct answer
    D
    \(1.0 \times 10^{10}\)

Solution

\(\mathrm{E}^{\circ}=\frac{0.0591}{\mathrm{n}} \log _{10} \mathrm{~K}\) Here, \(\mathrm{n}=2, \mathrm{E}^{\circ}=0.295\) \(\therefore \log _{10} \mathrm{~K}=\frac{2 \times 0.295}{0.0591}=10 \text { or } \mathrm{K}=10^{10}\)

Hence, the correct answer is option: 'D'.

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