JEE Main 2024 — Electrochemistry Question with Solution

To find the potential of the given half cell at 298 K, we can use the Nernst equation, which in this scenario is simplified due to the standard conditions for the hydrogen ion concentration ($[{\mathrm{H}}^{+}]=1M$). The half-reaction involved is the reduction of hydrogen ions to hydrogen gas:

$$2{\mathrm{H}}_{(\mathrm{aq})}^{+}+2{\mathrm{e}}^{-}⟶{\mathrm{H}}_2(g)$$

Given:

• Pressure of ${\mathrm{H}}_2=2\text{ atm}$
• Hydrogen ion concentration $[{\mathrm{H}}^{+}]=1M$
• $2.303\frac{\mathrm{RT}}{\mathrm{F}}=0.06V$ (a constant that simplifies calculation at 298 K)
• $\log 2=0.3$

The Nernst equation for this reaction under the given conditions becomes:

$$\mathrm{E}={\mathrm{E}}^{\circ }-\frac{0.06}{n}\log \frac{{\mathrm{P}}_{{\mathrm{H}}_2}}{[{\mathrm{H}}^{+}{]}^2}$$

Where ${\mathrm{E}}^{\circ }=0.00V$ is the standard electrode potential for the hydrogen half-cell, $n=2$ is the number of electrons transferred, and the pressure of ${\mathrm{H}}_2$ is given as 2 atm. Inserting the values, we get:

$$\mathrm{E}=0.00-\frac{0.06}{2}\log \frac{2}{1^2}=-\frac{0.06}{2}\times 0.3=-0.03\times 0.3=-0.009=-0.9\times 10^{-2}\text{V}$$

Therefore, the potential of the given half-cell at 298 K is $-0.9\times 10^{-2}\text{V}$.