Q47 / 200 · Manipal (MET) 2011 · Chemistry

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

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

Solution

$\begin{aligned} \Delta G^{\circ}= & n F E^{\circ} \\ \Delta G^{\circ} & =-2.303 R T \log K_c \\ n F E^{\circ} & =2.303 R T \log K_c \\ \log K_c & =\frac{n F E^{\circ}}{2.303 R T}=\frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298} \\ \log K_c^{\prime} & =9.97 \\ K_c & =1 \times 10^{10}\end{aligned}$

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

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