GATE 2022 MT – Question 49
The cell Au–Pb(liquid) | PbCl$_2$–KCl(liquid) | Cl$_2$(gas, 0.5 atm), C(graphite) has emf 1.2327 V at 873 K. Pb activity in the alloy is 0.72; PbCl$_2$ activity in the electrolyte is 0.18. The standard formation Gibbs energy of liquid PbCl$_2$ at 873 K is ______ kJ/mol (one decimal place). Given $R=8.314$ J K$^{-1}$ mol$^{-1}$ and $F=96500$ C/mol.
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Correct answer: -233.5 to -232.5
Explanation
The formation reaction is $Pb+Cl_2\to PbCl_2$, with $Q=0.18/(0.72\times0.5)=0.5$ and two transferred electrons. Use $\Delta G=-2FE=\Delta G^\circ+RT\ln Q$.
Thus $\Delta G^\circ=-2(96500)(1.2327)-8.314(873)\ln0.5=-232880$ J/mol. Answer: **−232.9 kJ/mol**.