The electrode potential of the following half cell at 298 K X| X^2+(0.001 M) \| Y^2+(0.01 M)| Y is _______ x 10^-2 V (Nearest integer) Given: E^0 _ X^2+ mid X=-2.36 V E_ Y^2+ mid Y^0=+0.36 V frac2.303 RT F=0.06 V

X+Y^2+ -> Y+X^2+ E_ Cell ^0=0.36-(-2.36)=2.72 V E_ Cell =2.72-(0.06 / 2) log (0.001 / 0.01) =2.72+0.03=2.75 V =275 x 10^-2 V

The electrode potential of the following half cell at 2... | (Chemistry) | DivineJEE

Exam

Chemistry

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}

$X+Y^{2+} \rightarrow Y+X^{2+} \\$

$E_{\text {Cell }}^0=0.36-(-2.36)=2.72 \mathrm{~V} \\$

$E_{\text {Cell }}=2.72-\frac{0.06}{2} \log \frac{0.001}{0.01} \\$

$=2.72+0.03=2.75 \mathrm{~V} \\$

$=275 \times 10^{-2} \mathrm{~V}$

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