Use tabulated half-cell potentials to calculate δg∘rxn for each of the following reactions at 25 ∘c. 2fe3+(aq)+3sn(s)→2fe(s)+3sn2+(aq)

- for the reaction is -58 kJ/mol.

Solution:- Oxidation half reaction taking place at the anode is--

= 0.14 V

Reduction half equation taking place at the cathode is---

= -0.04V

for the cell =

for the cell = -0.04V + 0.14V = 0.10V

Now we could easily calculate by using the below formula--

= -nF

where n is the number of electrons transferred in the over all reaction. Looking at two half equations the value of n is 6.

F is Farady constant and it's value is 96500 C/mol.

Plugging in the values in the formula...

= -(6)(96500)(0.10)

= -57900 J

Since, 1000 J = 1 kJ

So, = -58 kJ/mol