The standard electrode potential for Cu^{+2}/ | vedclass

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(C) The reduction half-reaction is: $Cu^{+2} + 2e^{-} \longrightarrow Cu$Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{2} \log \frac{1}{[Cu

Vedclass · Accepted Answer

$-0.22$

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(C) The reduction half-reaction is: $Cu^{+2} + 2e^{-} \longrightarrow Cu$Using the Nernst equation: $E = E^{\circ} - \frac{0.059}{2} \log \frac{1}{[Cu^{+2}]}$. Taking the approximation $\frac{0.06}{2} = 0.03$:For the solubility equilibrium: $Cu(OH)_{2} ightleftharpoons Cu^{+2} + 2OH^{-}$$K_{sp} = [Cu^{+2}][OH^{-}]^{2} = 1 	imes 10^{-19}$Given $pH = 14$,$pOH = 14 - 14 = 0$,so $[OH^{-}] = 10^{0} = 1 \ M$.Substituting into $K_{sp}$: $[Cu^{+2}](1)^{2} = 10^{-19} \implies [Cu^{+2}] = 10^{-19} \ M$.Now,calculate the reduction potential:$E = 0.34 - 0.03 \log \left( \frac{1}{10^{-19}} ight)$$E = 0.34 - 0.03 \log (10^{19})$$E = 0.34 - 0.03 	imes 19$$E = 0.34 - 0.57 = -0.22 \ V$.

The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

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