If ${E}_{{Cu}^{2+} / {Cu}}^{\circ}=+0.34 \,{~V}$ then reduction potential of ${Cu}^{2+} / {Cu}$ electrode at ${pH}=14$ is $\left({K}_{{sp}}\right.$ of ${Cu}({OH})_2=1$ $\times 10^{-19}$ )

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Easy

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Application +4 | -1

{E}_{{Cu}^{2+} / {Cu}}^{\circ}=+0.34 \,{~V}

then reduction potential of

{Cu}^{2+} / {Cu}

electrode at

{pH}=14

\left({K}_{{sp}}\right.

{Cu}({OH})_2=1

\times 10^{-19}

)

View answer and explanation

Correct option: B

-0.22 \,{~V}

Option B is correct because it uses the Nernst equation and the Ksp of Cu(OH)2 to calculate the reduction potential at pH=14.

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