Question

Calculate Ecello. free energy for the following battery. Mg(s) , Mg+2 ; Al+3 , Al(s)

Reduction potentials : Al = -1.66V Mg = -2.37 V

Calculate E cell for the new conditons: How does the voltage change and why? (please explain using Le Chateliers principles)

Mg(s) , Mg+2(0.500M) ; Al+3 (2.500M), Al(s)

Answer 1

1: Oxidation-half cell: Mg(s) ---- > Mg²⁺ + 2e- ; E⁰(oxi) = +2.37 V

Reduction-half cell: Al³⁺ + 3e- ---- > Al(s); E⁰(red) = - 1.66

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Overall cell reaction: 3Mg(s) + 2Al³⁺ ---- > 3Mg²⁺  + 2Al(s);

E⁰(cell) = E⁰(oxi) + E⁰(red) =   2.37 V - 1.66 V = 0.71 V (answer)

standard free energy, G⁰ = - nFxE⁰(cell) = - 6 x 96500 x 0.71 V = 411090 J (answer)

2: The voltage will change according to the concentration of Mg2+ and Al3+ that can be calculated according to Nernst equation.

E(cell) = E⁰(cell) - (0.0591/n) x log[Mg2+] / [Al3+]

=> E(cell) = 0.71 V - (0.0591/ 6) x log(0.500 M / 2.500 M) = 0.717 V (answer)