Given the standard reduction potentials for the following half-reactions: Cr₂O₇²⁻(aq) + 14H^+(aq) + 6e^- 2Cr³⁺(aq) + 7H₂O(l) E^ = +1.33 V I₂(s) + 2e^- 2I^-(aq) E^ = +0.54 V What is the standard Gibbs free energy change ( G^ ) in kJ mol ⁻¹ for the spontaneous overall cell reaction? (Given: 1 F = 96500 C mol ⁻¹ )

Step-by-step solution

For a spontaneous reaction, the half-cell with the higher reduction potential acts as the cathode, and the one with the lower reduction potential acts as the anode. Cathode (reduction): Cr₂O₇²⁻(aq) + 14H^+(aq) + 6e^- 2Cr³⁺(aq) + 7H₂O(l) E^ = +1.33 V Anode (oxidation): 2I^-(aq) I₂(s) + 2e^- E^ _ anode = +0.54 V The standard cell potential is: E^ _ cell = E^ _ cathode - E^ _ anode = 1.33 - 0.54 = +0.79 V To obtain the balanced overall cell reaction, the oxidation half-reaction must be multiplied by 3 to balance the 6