Question

Use tabulated half-cell potentials to calculate ΔG∘rxn for each of the following reactions at 25 ∘C.

Answer 1

Answer:- $\Delta G^0$ for the reaction is -58 kJ/mol.

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

$3Sn(s)\rightarrow 3Sn^2^+(aq) + 6e^-$ $E^0$ = 0.14 V

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

$2Fe^3^+(aq) + 6e^-\rightarrow 2Fe(s)$ $E^0$ = -0.04V

$E^0$ for the cell = $E_(reduction) + E_(oxidation)$

$E^0$ for the cell = -0.04V + 0.14V = 0.10V

Now we could easily calculate $\Delta G^0$ by using the below formula--

$\Delta G^0$ = -nF$E^0$

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...

$\Delta G^0$ = -(6)(96500)(0.10)

$\Delta G^0$ = -57900 J

Since, 1000 J = 1 kJ

So, $\Delta G^0$ = -58 kJ/mol