JEE MAIN - Chemistry (2023 - 30th January Evening Shift - No. 20)
The electrode potential of the following half cell at $298 \mathrm{~K}$
$\mathrm{X}\left|\mathrm{X}^{2+}(0.001 \mathrm{M}) \| \mathrm{Y}^{2+}(0.01 \mathrm{M})\right| \mathrm{Y}$ is _______ $\times 10^{-2} \mathrm{~V}$ (Nearest integer)
Given: $\mathrm{E}^{0} _ {\mathrm{X}^{2+} \mid \mathrm{X}}=-2.36 \mathrm{~V}$
$\mathrm{E}_{\mathrm{Y}^{2+} \mid \mathrm{Y}}^{0}=+0.36 \mathrm{~V}$
$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$
$\mathrm{X}\left|\mathrm{X}^{2+}(0.001 \mathrm{M}) \| \mathrm{Y}^{2+}(0.01 \mathrm{M})\right| \mathrm{Y}$ is _______ $\times 10^{-2} \mathrm{~V}$ (Nearest integer)
Given: $\mathrm{E}^{0} _ {\mathrm{X}^{2+} \mid \mathrm{X}}=-2.36 \mathrm{~V}$
$\mathrm{E}_{\mathrm{Y}^{2+} \mid \mathrm{Y}}^{0}=+0.36 \mathrm{~V}$
$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$
Answer
275
Explanation
$$
\begin{aligned}
& \text { Cell reaction : } X+Y^{2+}(0.01 \mathrm{M}) \longrightarrow X^{2+}(0.001 \mathrm{M})+Y \\\\
& E_{\text {cell }}^o=0.36-(-2.36)=2.72 \mathrm{~V} \\\\
& E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log \frac{\left[X^{2+}\right]}{\left[Y^{2+}\right]} \\\\
& =2.72-\frac{0.06}{2} \log \frac{(0.001)}{0.01} \quad[n=2 \text {, as two electrons are involved in the reaction] } \\\\
& =2.72-(0.03) \times(-1) \\\\
& =2.72+0.03 \\\\
& =2.75 \mathrm{~V}=275 \times 10^{-2} \mathrm{~V} \\\\
&
\end{aligned}
$$
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