Q9 / 90 · JEE Main 2019 (11 Jan Shift 2) · Chemistry
Given the equilibrium constant:
$\mathrm{K}_{\mathrm{C}}$ of the reaction:
$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ is
$10 \times 10^{15}$ calculate the $E_{\text {cell }}^{0}$ of this reaction at $298 \mathrm{~K}$ $\left[2.303 \frac{\mathrm{RT}}{\mathrm{F}}\right.$ at $\left.298 \mathrm{~K}=0.059 \mathrm{~V}\right]$
- A$0.04736 \mathrm{mV}$
- B$0.4736 \mathrm{mV}$
- Correct answerC$0.4736 \mathrm{~V}$
- D$0.04736 \mathrm{~V}$
Solution
$E_{\text {cell }}^{0}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{C}}$ or $\mathrm{E}_{\text {cell }}^{0}=\frac{0.059 \mathrm{~V}}{\mathrm{n}} \log \mathrm{K}_{\mathrm{C}}$
$$
=\frac{0.059 \mathrm{~V}}{2} \log 10^{16}=0.4736 \mathrm{~V}
$$
Hence, the correct answer is option: 'C'.