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  4. The electrode potential of the following half cell at 298 K X | X 2+(0.001 M ) | Y 2+(0.01 M )| Y is × 10-2 V (Nearest integer). Given: E x 2+ mid x 0=-2.36 V E Y 2+|Y 0=+0.36 V (2.303 RT / F )=0.06 V

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Q. The electrode potential of the following half cell at $298\, K$
$X \left| X ^{2+}(0.001 M ) \| Y ^{2+}(0.01 M )\right| Y$ is _____$\times 10^{-2} V$ (Nearest integer).
Given : $E _{ x ^{2+} \mid x }^0=-2.36\, V$
$E _{ Y ^{2+}|Y }^0=+0.36 \,V$
$\frac{2.303 RT }{ F }=0.06 \,V$

JEE MainJEE Main 2023Electrochemistry

Solution:

$ X + Y ^{2+} \rightarrow Y + X ^{2+} $
$ E _{\text{ Cell} }^0=0.36-(-2.36)=2.72 V $
$ E _{\text {Cell }}=2.72-\frac{0.06}{2} \log \frac{0.001}{0.01} $
$ =2.72+0.03=2.75 V$
$ =275 \times 10^{-2} V $