Q6 / 180 · KCET 2018 · Chemistry
For a cell reaction involving two electron changes, \( E_{\text {cell }}^{\circ}=0.3 \mathrm{~V} \) at \( 25^{\circ} \mathrm{C} \). The cell equilibrium
constant of the reaction is
constant of the reaction is
- A\( 10^{-10} \)
- B\( 3 \times 10^{-2} \)
- C\( 10 \)
- Correct answerD\( 10^{10} \)
Solution
Using the formula,
\[ \begin{array}{l} E_{\text {cell }}{ }^{0}=\frac{0.0591}{n} \log K \\ 0.3=\frac{0.0591}{2} \log K \\ 10=\log K \\ K=10^{10} \end{array} \]
\[ \begin{array}{l} E_{\text {cell }}{ }^{0}=\frac{0.0591}{n} \log K \\ 0.3=\frac{0.0591}{2} \log K \\ 10=\log K \\ K=10^{10} \end{array} \]
Hence, the correct answer is option: 'D'.