Question

What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 79 ∘C when [Fe2+]= 3.20 M...

What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 79 ∘C when [Fe2+]= 3.20 M and [Mg2+]= 0.310 M .

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Answer #1

anode reaction: oxidation takes place

Mg(s) -------------------------> Mg+2 (aq) + 2e-   ,   E0Mg+2/Mg = - 2.38 V

cathode reaction : reduction takes palce

Fe+2(aq) + 2e- -----------------------------> Fe(s) , E0Fe+2/Fe = -0.44V

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net reaction: Mg(s) +Fe+2(aq) -------------------------> Mg+2 (aq) + Fe(s)

E0cell= E0cathode- E0anode

E0cell= E0Fe+2/Fe - E0Mg+2/Mg

          = -0.44 - (-2.38)

          = 1.94 V

nernest equation

Ecell = E0cell -(0.0591/n)* log [Mg+2]/[Fe+2]

Ecell = 1.94 - (0.059 x1/2) *log (0.310 / 3.20 }

Ecell   = 1.97 V

cell potential =Ecell   = 1.97 V

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