JEE Main 2024 — Electrochemistry Question with Solution

The electromotive force (emf) of a hydrogen electrode can be derived from the Nernst equation, which for the hydrogen half-cell reaction $${\mathrm{H}}_2(\mathrm{g})\to 2{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}$$ is given by:

$$E=E^{\circ }+\frac{RT}{2F}\ln \left(\frac{[{\mathrm{H}}^{+}{]}^2}{P_{{\mathrm{H}}_2}}\right)$$

where:

• E is the electrode potential.
• E^∘ is the standard electrode potential, which is 0 V for the hydrogen electrode.
• R is the gas constant, $$8.314\mathrm{J}\cdot \mathrm{mo}{\mathrm{l}}^{-1}\cdot {\mathrm{K}}^{-1}$$.
• T is the temperature in Kelvin: $$25^{\circ }\mathrm{C}=298\mathrm{K}$$.
• F is the Faraday constant, $$96485\mathrm{C}\cdot \mathrm{mo}{\mathrm{l}}^{-1}$$.
• $$[{\mathrm{H}}^{+}]$$ is the concentration of hydrogen ions.
• $$P_{{\mathrm{H}}_2}$$ is the pressure of hydrogen gas.

In pure water at $$25^{\circ }\mathrm{C}$$, $$[{\mathrm{H}}^{+}]=10^{-7}\mathrm{M}$$. To make the emf zero, we set E to 0 in the Nernst equation:

$$0=0+\frac{8.314\times 298}{2\times 96485}\ln \left(\frac{(10^{-7}{)}^2}{P_{{\mathrm{H}}_2}}\right)$$

Simplifying the constants:

$$\frac{8.314\times 298}{2\times 96485}\approx 0.0128$$

Thus, the equation becomes:

$$0=0.0128\ln \left(\frac{(10^{-7}{)}^2}{P_{{\mathrm{H}}_2}}\right)$$

Since ln term must be zero for this equation to hold true (as 0 divided by any number is still 0), the expression inside the logarithm must equal 1:

$$\frac{(10^{-7}{)}^2}{P_{{\mathrm{H}}_2}}=1$$

Simplifying this gives:

$$(10^{-7}{)}^2=P_{{\mathrm{H}}_2}$$

$$10^{-14}=P_{{\mathrm{H}}_2}$$

Therefore, the required pressure of $${\mathrm{H}}_2$$ to make the emf of the hydrogen electrode zero in pure water at $$25^{\circ }\mathrm{C}$$ is $$10^{-14}$$ bar.

The correct answer is Option B: $$10^{-14}$$.