QJEEM24Ch6713: JEE Main 2024 Chemistry - Class 12 JEEM24Ch6713 Solution: What pressure (bar) of would be required to make emf of…

The electromotive force (emf) of a hydrogen electrode can be derived from the Nernst equation, which for the hydrogen half-cell reaction $\mathrm{H_2(g)} \rightarrow 2\mathrm{H^+(aq)} + 2\mathrm{e^-}$ 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, $8.314 \, \mathrm{J \cdot mol^{-1} \cdot K^{-1}}$ .
• T is the temperature in Kelvin: $25^{\circ} \mathrm{C} = 298 \, \mathrm{K}$ .
• F is the Faraday constant, $96485 \, \mathrm{C \cdot mol^{-1}}$ .
• $[\mathrm{H^+}]$ 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: $10^{-14}$ .