What is standard decrease in Gibbs energy for the following cell? Al _ (s) Al ⁺³_ (1M) Cd ⁺²_ (1M) Cd if E^ , Al ⁺³/ Al = -1.66 V, E^ , Cd ⁺²/ Cd = -0.40 V

Step-by-step solution

The standard cell potential is given by: E^ _ cell = E^ _ cathode - E^ _ anode E^ _ cell = -0.40 - (-1.66) = 1.26 V The half-cell reactions are: Anode: Al Al ⁺³ + 3e^- Cathode: Cd ⁺² + 2e^- Cd To balance the number of electrons, the overall reaction involves the transfer of 6 electrons, so n = 6 . The standard Gibbs free energy change is: G^ = -n F E^ _ cell G^ = -6 96500 1.26 G^ = -729540 J = -729.54 kJ Answer: -729.54 kJ