Question

Use the tabulated half-cell potentials to calculate ΔG° for the following redox reaction.

2 Al(s) + 3 Mg²⁺(aq) → 2 Al³⁺(aq) + 3 Mg(s)

	+6.8 x 102 kJ
	-7.8 x 102 kJ
	+4.1 x 102 kJ
	+1.4 x 102 kJ
	-2.3 x 102 kJ

Answer is 4.1-10^2 kJ

Please explain every details

When I did it, I got the potential energy -2.37+(-1.66)=-4.03

I don't understand why it's -2.37+1.66!!!

Please explain why,

Answer 1

Al3+(aq) + 3e- ? Al(s) -1.66

Mg2+(aq) + 2e- ? Mg(s) -2.37

note that in reality, we have Solid Aluminium in the left side (actually, there should be an oxidation and a reduction, you can'0t have only 2 reductions)

then we need to invert Al+3 reaction

Al(s) ?Al3+(aq) + 3e- +1.66

Mg2+(aq) + 2e- ? Mg(s) -2.37

Note that the sign of E Al+3 changes

Ecell = Ered + Eox = -2.37+1.66 = -0.71 V

dG = -nF*Ecell

note that n = electrons being transferred is 2*3 = 6

F = faraday constant 96500

then

dG = -nF*Ecell = -6*96500*-0.71 = 411090 J/mol = 411.09 kJ/mol

whichi s 4.1*10^2 kJ