Question

Calculate Delta G° for the following reaction at 25°C:

3Zn²⁺_(aq) + 2Al _(s) = 3Zn _(s) + 2Al ³⁺_(aq)

Delta G° = ? kJ/mol

Answer 1

we will be using the formula

     

In this equation

• n is the number of moles of electrons exchanged in the cell reaction.
• F is 96.485= 96500 kJ volt^-1 mol^-1 (the "Faraday").

first we have to calculate Eo cell:

oxidation half reaction:

Al = Al³⁺ + 3 e

Ereduction = -1.66 V

so for oxidtion, E= 1.66 V

reduction reaction:

Zn(s)Zn2+(aq) + 2e–

E° = 0.763 V

therefore Ereduction= -0.763

after balancing the reactions and adding the half reactions we get the final reaction.

E^o_cell = E^o_reduction + E^o_oxidation

therefore E^o_cell= -0.763 + 1.66 = 0.897 V

total 6 electron are transfered during the reaction.

So,

dG= 6*96500*0.897= 519363 kJ/mol