Exams › JEE Main › Chemistry

For the cell reaction Cu²+ (aq) + Fe (s) -> Fe²+ (aq) + Cu (s), determine the standard Gibbs free energy change. Take E°(Cu²+/Cu) = +0.34 V and E°(Fe²+/Fe) = -0.44 V.

1. -150540 J
2. +150540 J
3. -15054 J
4. +15054 J

Correct answer: -150540 J

Solution

Fe is oxidized (anode) and Cu²+ is reduced (cathode). E°cell = 0.34 - (-0.44) = 0.78 V. With n = 2, ΔG° = -2 * 96500 * 0.78 = -150540 J, which is negative, confirming a spontaneous reaction.

Related JEE Main Chemistry questions

• During electrolysis of an aqueous solution of dipotassium succinate, which gas is evolved at the electrodes?
• Which expression gives the equivalent conductance at infinite dilution of Al2(SO4)3, if Λ°Al3+ and Λ°SO4²− denote the equivalent conductances at infinite dilution of the corresponding ions?
• The standard emf of a Daniell cell is 1.10 V. What is the greatest amount of electrical work that can be extracted from this cell?
• During the recharging of a lead storage battery, which reaction takes place at the cathode?
• The limiting molar ionic conductivities of the ions X2+ and Y2− are 57 and 73, respectively. What will be the molar conductivity of the electrolyte formed by these ions?
• According to Kohlrausch’s law, at what condition does each ion contribute a fixed amount to the conductance of an electrolyte, irrespective of the other ion present?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →