For the cell in which of the following occurs:
\(2Au^{3+} (aq) + 6Br^- (aq) \longrightarrow 2Au(s) + 3Br_2 (l)\) has \(E^o_{cell} = 2.49 V\) at \(298 K\)
The standard Gibbs energy change will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(- 1442 kJ\)
The standard Gibbs energy change of a cell reaction is related to its standard emf by
ΔrG° = −nFE°cell
where n is the number of electrons transferred in the reaction as written. In \( \ce{2Au^{3+}} \) + \( \ce{6Br^{-}} \) → 2Au + \( \ce{3Br2} \), each \( \ce{Au^{3+}} \) gains 3 electrons, so \( \ce{2Au^{3+}} \) gain 6 electrons (and \( \ce{6Br^{-}} \) lose 6): n = 6.
ΔrG° = −6 × 96487 C mol−1 × 2.49 V
= −1441516 J mol−1
≈ −1442 kJ
NCERT (2026-27 edition):
Electrochemical Cell and Gibbs Energy of the Reaction – “… Solution: ΔrG° = − nFE°(cell). n in the above equation is 2, F = 96487 C mol−1 and E°(cell) = 1.1 V. Therefore, ΔrG° = − 2 × 1.1 V × 96487 C mol−1 = − 21227 J mol−1 = − 212.27 kJ mol−1.”