Question

What is the cell potential for the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 85 ∘C when [Fe2+]= 3.50 M and [Mg2+]= 0.110 M .

Answer 1

Answer –

We are given, redox reaction –

Mg(s) + Fe²⁺(aq) ----> Mg²⁺(aq) + Fe(s)

[Fe²⁺]= 3.50 M and [Mg²⁺]= 0.110 M.

First we need to calculate E^ocell from the standard reduction potential

We know in the reaction Mg(s) converted to Mg²⁺ means it gets oxidized and Fe²⁺(aq) to Fe(s) so Fe²⁺ (aq) gets reduced

So half reactions with E^o value

Mg(s) ---> Mg²⁺(aq) + 2e^- E^ooxd = +2.37 V

Fe²⁺(aq) + 2e^- ------> Fe(s) E^oredn = -0.44 V

Mg(s) + Fe²⁺(aq) ----> Mg²⁺(aq) + Fe(s) , E^ocell = 1.93 V

We know the Nernst equation –

Ecell = E^ocell – 0.0592/n * log [Mg²⁺(aq)] / [Fe²⁺(aq)]

         = 1.93 V – 0.0592/2 * log 0.110 M /3.50 M

         = 1.93 V – (-0.0445 V)

         = 1.97 V