Calculate the standard potential, E ∘ , E°, for this reaction from its Δ G ∘ ΔG° value. X ( s ) + Y 4 + ( aq ) ⟶ X 4 + ( aq ) + Y ( s ) Δ G ∘ = − 48.0 kJ

The standard cell potential is 0.12 V.

We know that;

ΔG° = -nFE° cell

Where;

ΔG° = Standard free energy change

n = Number of moles of electrons = 4

F = Faraday's constant = 96500 C

E° cell = Standard cell potential

Making E° cell the subject of the formula;

E° cell = ΔG°/-nF

E° cell = − 48.0 × 10^3J/-(4  × 96500)

E° cell = 0.12 V

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0.124 V

Explanation:

To calculate standard Gibbs free energy, we use the equation:

Where,

n = number of electrons transferred = 4

F = Faradays constant = 96500 C

= standard cell potential = ?

Putting values in above equation, we get:

Thus the standard potential E°, for this reaction is 0.124 V

12 is it do you think so