calculate the standard free energy change for the following

calculate the standard free energy change for the following reaction at 25C.

2Au3+(aq) + 3Mg(s) <---> 2Au(s) + 3Mg2+ (aq)

Solution

The cell is galvanic which means that a positive voltage is produced. The only way you could produce a voltage of 3.98 V is if the two half-reactions are as follows:

Reduction: Au3+ + 3e- ==> Au . . .Eo = +1.50 V
Oxidation: Mg ==> Mg2+ + 2e- . .Eo = +2.37 V

To add the two together: the number of electrons must cancel, and the lowest multiple of 2 and 3 is 6, So multiply the first reaction by 2 to get 6e- on the left and multiply the second reaction by 3 to get 6e-.on the right.

2Au3+ + 6e- ==> 2Au . . .Eo = +1.50 V
3Mg ==> 3Mg2+ + 6e- . ..Eo = +2.37 V
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2Au3+ + 3Mg ==> 2Au + 3Mg2+ . . .Eo = +3.87 V

Remember that when we multiply the equations,

delta G= -nFE = -6*96500*3.87 = 2240.73KJ