Q31 / 225 · JEE Main 2003 · Chemistry

For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be

  • A
    $29.5 \times 10^{-2}$
  • B
    10
  • C
    $1 \times 10^{10}$
  • Correct answer
    D
    $1 \times 10^{-10}$

Solution

The equilibrium constant is related to the standard emf of cell by the expression $\log K=E_{\text {cell }}^0 \times \frac{n}{0.059}=0.295 \times \frac{2}{0.059}$ $\log \mathrm{K}=\frac{590}{59}=10$ or $\mathrm{K}=1 \times 10^{-10}$

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

Relevant Questions