Question

Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution:

Reduction Half-Reaction	E˚ (v)
	1.54
	1.16
	0.84

Calculate the standard free-energy change, ∆G˚, in units of kJ/mol at 298 K for the following reaction

Answer 1

Oxidation reaction : 2 Z (s) $\rightarrow$ 2 Z ⁺ (aq) + 2 e ^-

Reduction reaction : Y ²⁺ (aq) + 2 e ^-$\rightarrow$ Y (s)  

Overall reaction : 2 Z (s) + Y ²⁺ (aq) $\rightarrow$ 2 Z ⁺ (aq) + Y (s)  

In overall reaction , two electrons are transferred.

The standard emf of the cell is calculated as E ⁰ cell = E ⁰ cathode - E ⁰ anode

$\therefore$ E ⁰ cell = E ⁰ Y ²⁺ I Y - E ⁰ Z ⁺ I Z

$\therefore$ E ⁰ cell = 1.16 V - 0.84 V = 0.32 V

We have relation between standard Gibbs energy change and emf of the cell as $\Delta$ G ⁰ = - n F E ⁰ cell

$\therefore$$\Delta$ G ⁰ = - 2 $\times$ 96487 C/ mol $\times$ 0.32 V

$\therefore$$\Delta$ G ⁰ = - 61751.7 J /mol

$\therefore$$\Delta$ G ⁰ = - 61.75 kJ / mol

ANSWER : $\Delta$ G ⁰ = - 61.75 kJ / mol