$\mathrm{E}^{\circ}$ cell is $1 \cdot 049 \mathrm{~V}$ and involves transfer of 2 electrons, calculate…

$\mathrm{E}^{\circ}$ cell is $1 \cdot 049 \mathrm{~V}$ and involves transfer of 2 electrons, calculate equilibrium constant of cell?
  1. $2.75 \times 10^{35}$
  2. $2.75 \times 10^{10}$
  3. $0 \cdot 524 \times 10^{35}$
  4. $2 \cdot 098 \times 10^{10}$

Solution

$\mathrm{E}_{\text {cell }}^{0}=1.049 \mathrm{~V}, \quad \mathrm{n}=2, \quad \mathrm{~K}=?$ $\mathrm{E}_{\text {cell }}^{0}=\frac{0.0592}{\mathrm{n}} \log _{10} \mathrm{~K}$ $\therefore \log _{10} \mathrm{~K}=\frac{\mathrm{E}_{\text {cell }}^{0} \times \mathrm{n}}{0.0592}=\frac{1.049 \times 2}{0.0592}$ $\therefore \log _{10} \mathrm{~K}=35.439$ $\therefore \mathrm{K} =$ Antilog $(35.439)$ $\therefore \mathrm{K}=2.75 \times 10^{35}$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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