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Standard reduction potentials: Cd²+(aq) + 2e⁻ -> Cd(s), E* = -0.40 V Ag^+(aq) + e⁻ -> Ag(s), E* = +0.80 V Calculate the standard Gibbs free energy change for: 2Ag^+(aq) + Cd(s) -> 2Ag(s) + Cd²+(aq)

1. 115.8 kJ
2. -115.8 kJ
3. -231.6 kJ
4. 231.6 kJ

Correct answer: -231.6 kJ

Solution

E*_cell = 0.80 - (-0.40) = 1.20 V. n=2. delta_G* = -2 * 96500 * 1.20 = -231600 J = -231.6 kJ.

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