At $298 \mathrm{~K}$, the standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ electrode is $0.34 \mathrm{~V}$.

Given : $\mathrm{K}_{\mathrm{sp}} \mathrm{Cu}(\mathrm{OH})_{2}=1 \times 10^{-20}$

Take $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$

The reduction potential at $\mathrm{pH}=14$ for the above couple is $(-) x \times 10^{-2} \mathrm{~V}$. The value of $x$ is ___________

Given: <br/><br/> Standard reduction potential for Cu²⁺/Cu, E° = 0.34 V<br/><br/> Ksp of Cu(OH)₂ = 1 × 10⁻²⁰<br/><br/> 2.303RT/F = 0.059 V<br/><br/> pH = 14 <br/><br/> First, we have the solubility equilibrium for Cu(OH)₂: <br/><br/> $$\mathrm{Cu}(\mathrm{OH})_2(\mathrm{~s}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$$ <br/><br/> The Ksp expression for this reaction is: <br/><br/> $\mathrm{Ksp}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2$ <br/><br/> At pH 14, the concentration of OH⁻ ions is 1 M: <br/><br/> $\left[\mathrm{OH}^{-}\right] = 1 \mathrm{M}$ <br/><br/> Now we can find the concentration of Cu²⁺: <br/><br/> $$\left[\mathrm{Cu}^{2+}\right]=\frac{\mathrm{Ksp}}{\left[\mathrm{OH}^{-}\right]^2}=\frac{1 \times 10^{-20}}{1^2}=10^{-20} \mathrm{M}$$ <br/><br/> The half-cell reaction for the reduction of Cu²⁺ is: <br/><br/> $$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$$ <br/><br/> Now we can use the Nernst equation to calculate the reduction potential at pH 14: <br/><br/> $E = E° - \frac{0.059}{n} \log_{10} \frac{1}{\left[\mathrm{Cu}^{2+}\right]}$ <br/><br/> Here, n = 2 (number of electrons transferred in the Cu²⁺/Cu couple). <br/><br/> $E = 0.34 - \frac{0.059}{2} \log_{10} \frac{1}{10^{-20}}$<br/><br/> $E = 0.34 - \frac{0.059}{2} \times 20$<br/><br/> $E = 0.34 - 0.59$<br/><br/> $E = -0.25 \mathrm{~V}$ <br/><br/> Thus, the reduction potential at pH 14 for the Cu²⁺/Cu couple is -0.25 V. In terms of x × 10⁻² V: <br/><br/> $(-) x \times 10^{-2} \mathrm{~V} = -0.25 \mathrm{~V}$ <br/><br/> The value of x is 25.