The standard Gibbs free energy change of the reaction shown below is -2.7 kJ mol⁻¹. Sn(s) + Pb²⁺ = Sn²⁺ +… · ABC26PH0608

Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Free Energy Equilibria · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

The standard Gibbs free energy change of the reaction shown below is $-2.7\ \mathrm{kJ\,mol^{-1}}$.

$\mathrm{Sn(s) + Pb^{2+} = Sn^{2+} + Pb(s)}$

Given that $E^0(\mathrm{Pb^{2+}/Pb})$ is $-0.126$ V, the value of $E^0(\mathrm{Sn^{2+}/Sn})$ in V is ____ (up to two decimal places)

Answer

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Explanation

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Study loop for Thermodynamics