JEE Main 2023 — Electrochemistry Question with Solution

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:

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

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.