What will be the reduction potential of Cu in | vedclass

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(B) $pH$ of the solution $= 12$. $	herefore pOH = 14 - 12 = 2$,so $[OH^-] = 10^{-2} \ M$. For $Cu(OH)_2$,$K_{sp} = [Cu^{+2}][OH^-]^2 = 10^{-19}$. $

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$-0.103 \ V$

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(B) $pH$ of the solution $= 12$. $	herefore pOH = 14 - 12 = 2$,so $[OH^-] = 10^{-2} \ M$. For $Cu(OH)_2$,$K_{sp} = [Cu^{+2}][OH^-]^2 = 10^{-19}$. $	herefore [Cu^{+2}] = \frac{10^{-19}}{(10^{-2})^2} = 10^{-15} \ M$. Using the Nernst equation: $E_{Cu^{+2}/Cu} = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.0591}{2} \log \frac{1}{[Cu^{+2}]}$. $E_{Cu^{+2}/Cu} = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-15}}$. $E_{Cu^{+2}/Cu} = 0.34 - \frac{0.0591 	imes 15}{2} = 0.34 - 0.44325 = -0.10325 \ V \approx -0.103 \ V$.

What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^{\circ}_{Cu^{+2}/Cu} = 0.34 \ V$.

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