The maximum electrical work obtainable from a cell equals the decrease in Gibbs free energy, expressed as ΔG° = -nFE°_cell, where n is the number of electrons transferred, F is the Faraday constant, and E°_cell is the standard EMF. Substituting n = 3, F = 96500 C/mol, and E°_cell = 0.50 V gives ΔG° = -(3)(96500)(0.50) = -144750 J = -144.75 kJ. The negative sign confirms the reaction is spontaneous, which is consistent with a positive cell potential. Option -48.25 kJ wrongly uses n = 1. Option -96.50 kJ corresponds to n = 2 and a different potential. Option +144.75 kJ has the wrong sign and would imply a non-spontaneous reaction. This relationship is the bridge between electrochemistry and thermodynamics emphasised in NCERT. It is worth emphasising that this is not a special case but a representative example of how gibbs energy and emf operates throughout redox reactions and electrochemistry. A common JEE pitfall is to ignore the role of Gibbs free energy, yet it is exactly this factor that distinguishes the correct answer from the tempting alternatives. Plausibility check: a positive E°_cell must yield a negative ΔG°, and tripling the electron count appropriately scales the magnitude, confirming -144.75 kJ.