GATE 2025 CY – Question 63

Physical Chemistry · Electrochemistry: Standard electrode potential and electrochemical cells. Nernst Equation and its application, relationship between electrode potential and thermodynamic quantities, Potentiometric and conductometric titrations. Ionic mobility and conductivity. Kohlrausch’s law, Debye-Hückel limiting law. Debye-Hückel- Onsager equation. · 2 marks · Numerical answer

For the cell reaction, $$\mathrm{Hg_2Cl_2(s)+H_2(1\ atm)\to2Hg(l)+2H^+(a=1)+2Cl^-(a=1)}$$ The standard cell potential is $\mathcal E^0=0.2676$ V, and $(\partial\mathcal E^0/\partial T)_P=-3.19\times10^{-4}\ \mathrm{V\,K^{-1}}$. The standard enthalpy change of the reaction $(\Delta_rH^0)$ at 298 K is $-x\ \mathrm{kJ\,mol^{-1}}$. The value of $x$ is _____ (rounded off to two decimal places). [Given: Faraday constant $F=96500\ \mathrm{C\,mol^{-1}}$]

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Correct answer: 69.64 to 70.34

Explanation

**Thermodynamics of an electrochemical cell.** The Gibbs energy change is $\Delta G=-nF\mathcal E$, and the entropy change is
$$\Delta S=nF\left(\frac{\partial\mathcal E}{\partial T}\right)_P .$$

Since $\Delta H=\Delta G+T\Delta S$,
$$\Delta H=-nF\mathcal E+nFT\left(\frac{\partial\mathcal E}{\partial T}\right)_P=-nF\left[\mathcal E-T\left(\frac{\partial\mathcal E}{\partial T}\right)_P\right].$$

**Numbers.** $n=2$ electrons, $\mathcal E^0=0.2676$ V, $(\partial\mathcal E^0/\partial T)_P=-3.19\times10^{-4}$ V/K, $T=298$ K:
$$T\frac{\partial\mathcal E}{\partial T}=298\times(-3.19\times10^{-4})=-0.09506\text{ V}$$
$$\Delta H^0=-2\times96\,500\times\left[0.2676-(-0.09506)\right]=-193\,000\times0.36266=-69\,994\text{ J/mol}.$$

So $\Delta_rH^0=-69.99$ kJ/mol, and $x=\mathbf{69.99}$.