\(Cu(S)+Sn^{2+} (0.001M) Cu^{2+} (0.01M) + Sn(s)\)
The Gibbs free energy change for the above reaction at 298 K is x × 10-1 × kJ mol-1. The value of x is ________ .[nearest integer]
[Given \(E^{-}_{cu^{2+} / cu} = 0.34V; E^{-}_{Sn^{2+} / Sn} = - 0.14V; F = 96500 C ∼ mol^{-1}\)]
\(Cu(S)+Sn^{2+} (0.001M) Cu^{2+} (0.01M) + Sn(s)\)
The Gibbs free energy change for the above reaction at 298 K is x × 10-1 × kJ mol-1. The value of x is ________ .[nearest integer]
[Given \(E^{-}_{cu^{2+} / cu} = 0.34V; E^{-}_{Sn^{2+} / Sn} = - 0.14V; F = 96500 C ∼ mol^{-1}\)]
Correct Answer: 983
Solution and Explanation
The correct answer is 983
\(Cu + Sn^{2+} → Cu^{2+} + Sn(s)\)
\(E°_{cell} = E°_{Ox} + E°_{Red }\)
= - 0.34 - 0.14
= 0.48 V
\(E = E° - \frac{0.0591}{2} log\frac{[ Cu^{2+}]}{[ Sn^{2+}]}\)
= - 0.48 - 0.0295 log 10
= - 0.5095 V
\(ΔG = -nFE\)
= -2 × 96500 × 0.5095 J / mol
= 98333.5 × 10-3 kJ / mol
= 983.3 × 10-1 kJ / mol
= 983 × 10-1 kJ / mol
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Concepts Used:
Gibbs Free Energy
The energy associated with a chemical reaction that can be used to do work.It is the sum of its enthalpy plus the product of the temperature and the entropy (S) of the system.
The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system. In completely reversible process maximum enthalpy can be obtained.
ΔG=ΔH−TΔS
The Conditions of Equilibrium
If both it’s intensive properties and extensive properties are constant then thermodynamic system is in equilibrium. Extensive properties imply the U, G, A.
