alculate the standard free-energy change at 25 ∘C for the following reaction: Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)?

alculate the standard free-energy change at 25 ∘C for the following reaction:

The half reactions involved of this redox reaction and their respective standard electrode potentials are:
(1) Mg²⁺(aq) + 2 e⁻ ⇄ Mg(s)
E₁ = – 2.372 V
(2) Fe²⁺(aq) + 2 e⁻ ⇄ Fe(s)
E₂ = – 0.44 V
( standard electrode potentials from wikipedia [1] )
In the reaction
Mg(s) + Fe²⁺(aq) ⇄ Mg²⁺(aq) + Fe(s)
the first half reaction is the one proceeding in reverse direction. So the cell potential of this reaction is
E° = E₂ – E₁ = – 0.44 V – (−2.372 V) = 1.932 V
From the half reactions you can see that two electrons are transferred.
Hence,
ΔG° = – n∙F∙E°
= – 2 ∙ 96500 C∙mol⁻¹ ∙ 1.932 V
= – 2 ∙ 96500 J∙V⁻¹∙∙mol⁻¹ ∙ 1.932 V
= – 372876 J∙mol⁻¹
= – 372.876 kJ∙mol⁻¹

actually its just J
-3.71*10^5 J

∆G for the formation of Au and Mg = 0 since these are elements but you need the values of the ions or the values for Eºcell for the cell
reduction potential Au3+ –> Ag = +1.50
reduction potential Mg2+ –> Mg = -2.372…Mg is being oxidized in the reaction so +2.372
Ecell = Ered + Eox = 1.5 + 2.372 = 3.872V
∆G = -nFE…n = moles e-, F = faraday, E = Ecell
∆G = -6(96485)(3.872) = -2.24×10^6J or -2.24×10^3kJ

the units are just kJ, so the answer is -373 kJ

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