QJEEM23Ch181d: JEE Main 2023 Chemistry - Class 12 JEEM23Ch181d Solution: At , the standard reduction potential for electrode is . Given…

Given:

Standard reduction potential for Cu²⁺/Cu, E° = 0.34 V

Ksp of Cu(OH)₂ = 1 × 10⁻²⁰

2.303RT/F = 0.059 V

pH = 14

First, we have the solubility equilibrium for Cu(OH)₂:

$\mathrm{Cu}(\mathrm{OH})_2(\mathrm{~s}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$

The Ksp expression for this reaction is:

$\mathrm{Ksp}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2$

At pH 14, the concentration of OH⁻ ions is 1 M:

$\left[\mathrm{OH}^{-}\right] = 1 \mathrm{M}$

Now we can find the concentration of Cu²⁺:

$\left[\mathrm{Cu}^{2+}\right]=\frac{\mathrm{Ksp}}{\left[\mathrm{OH}^{-}\right]^2}=\frac{1 \times 10^{-20}}{1^2}=10^{-20} \mathrm{M}$

The half-cell reaction for the reduction of Cu²⁺ is:

$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$

Now we can use the Nernst equation to calculate the reduction potential at pH 14:

$E = E° - \frac{0.059}{n} \log_{10} \frac{1}{\left[\mathrm{Cu}^{2+}\right]}$

Here, n = 2 (number of electrons transferred in the Cu²⁺/Cu couple).

$E = 0.34 - \frac{0.059}{2} \log_{10} \frac{1}{10^{-20}}$

$E = 0.34 - \frac{0.059}{2} \times 20$

$E = 0.34 - 0.59$

$E = -0.25 \mathrm{~V}$

Thus, the reduction potential at pH 14 for the Cu²⁺/Cu couple is -0.25 V. In terms of x × 10⁻² V:

$(-) x \times 10^{-2} \mathrm{~V} = -0.25 \mathrm{~V}$

The value of x is 25.