Q81 / 200 · NEET 2007 · Chemistry

The equilibrium constant of the reaction: $\mathrm{Cu}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \rightarrow \mathrm{Cu}_{(a q)}^{2+}+2 \mathrm{Ag}_{(s)}$ $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ at $298 \mathrm{~K}$ is:

  • A
    $2.4 \times 10^{16}$
  • B
    $4.0 \times 10^{10}$
  • Correct answer
    C
    $4.0 \times 10^{15}$
  • D
    $2.4 \times 10^{10}$

Solution

As $\mathrm{E}_{\mathrm{cell}}^0=\frac{0.0591}{n} \log \mathrm{K}_c$ $\begin{aligned} & \therefore 0.46=\frac{0.0591}{2} \log \mathrm{K}_c \\ & \therefore \log \mathrm{K}_c=\frac{2 \times 0.46}{0.0591}=15.57 \\ & \text {or } \mathrm{K}_c=\text { Antilog } 15.57=3.7 \times 10^{15} \end{aligned}$

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

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