At 298 , K,the standard reduction potential f | vedclass

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(D) The dissolution equilibrium is $Cu(OH)_{2(s)} ightleftharpoons Cu^{2+}_{(aq)} + 2OH^{-}_{(aq)}$.$K_{sp} = [Cu^{2+}][OH^{-}]^2 = 1 	imes 10^{-20

Vedclass · Accepted Answer

$25$

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(D) The dissolution equilibrium is $Cu(OH)_{2(s)} ightleftharpoons Cu^{2+}_{(aq)} + 2OH^{-}_{(aq)}$.$K_{sp} = [Cu^{2+}][OH^{-}]^2 = 1 	imes 10^{-20}$.At $pH = 14$,$pOH = 14 - 14 = 0$,so $[OH^{-}] = 10^0 = 1 \, M$.$[Cu^{2+}] = \frac{K_{sp}}{[OH^{-}]^2} = \frac{10^{-20}}{1^2} = 10^{-20} \, M$.Using the Nernst equation for $Cu^{2+} + 2e^{-} ightarrow Cu_{(s)}$:$E = E^{\circ} - \frac{0.059}{2} \log_{10} \frac{1}{[Cu^{2+}]}$.$E = 0.34 - \frac{0.059}{2} \log_{10} \frac{1}{10^{-20}}$.$E = 0.34 - \frac{0.059}{2} 	imes 20 = 0.34 - 0.59 = -0.25 \, V$.Given $E = -x 	imes 10^{-2} \, V$,we have $-0.25 = -x 	imes 10^{-2}$,so $x = 25$.

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