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JEE Main 2022 Chemistry question

ChemistryElectrochemistryJEE Main 2022

Cu(s) + Sn^{2+}(0.001M) \to Cu^{2+}(0.01M) + Sn(s). The Gibbs free energy change for the above reaction at 298 K is kJ . The value of x is . [nearest integer] [Given V; = −0.14 V; F = 96500 C ]

Answer: 983

Solution

E°cell = E^{\circ}(Sn^{2+}/Sn) - E^{\circ}(Cu^{2+}/Cu) = −0.14 − 0.34 = −0.48 V. Then E = E° − = −0.48 − 0.0295 log 10

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Question ID VK-025250