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  • The electrode potential of the following half cell at 298 K <img alt="\mathrm{X}\left|\mathrm{X}^{2+}(0.001 \mathrm{M}) \| \mathrm{Y}^{2+}(0.01 \mathrm{M})\right| \mathrm{Y}$ is $\times 1

The electrode potential of the following half cell at 298 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}_{\mathrm{x}^{2+}{ }_{\mathrm{X}}}^0=-2.36 \mathrm{~V}
\begin{aligned} & \mathrm{E}_{\mathrm{Y}^{+2} \mid \mathrm{Y}}^0 \\ & \mathrm{E}^0 \mathrm{Y}^{2+1 Y}=+0.36 \mathrm{~V} \\ & \frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V} \end{aligned}

Option: 1

275


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{x}+\mathrm{y}^{+2} \rightarrow \mathrm{y}+\mathrm{x}^{+2} \\ & \mathrm{E}^{\circ} \text { Cell }=\mathrm{E}_{\text {Cathode }}-\mathrm{E}_{\text {Anode }}^o \\ & \mathrm{E}^{\circ} \text { Cell }=0.36-(-2.36)=2.72 \mathrm{~V} \\ & \mathrm{E}_{\text {Cell }}=2.72-\frac{0.06}{2} \log \frac{\mathrm{x}^{+2}}{\mathrm{y}^{+2}} \\ & \mathrm{E}_{\text {Cell }}=2.72-\frac{0.06}{2} \log \frac{0.001}{0.01} \\ & =2.72+0.03=2.75 \mathrm{~V} \\ & =275 \times 10^{-2} \mathrm{~V} \end{aligned}

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