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A galvanic cell consists of two half cells and then with standard reduction potentials shown: Co2 + (aq) + 2 e-Co ≡ (s) .......... -0.28 Volts Cd2 + (aq) E-≡ + 2 CD (s) .......... -0.40 Volt current concentrations are: Co2 + (aq) = 0.00100 M, Cd2 + = 0.100 M What is the energy change freely associated with this particular design? Please explain how you get the answer. Thanks in advance.

The cell reaction is: Cd + Co (2 +) ==> Co + Cd (2 +), E ° = 0.12V nFΔE =- =- ΔG ° + RT Ln nFE Q * = -2 * 0.12 * 96 485 * 8314 298 * ln (0.100/0.001) = - 11.7 KJ