Question

What is the cell potential for the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 43 ∘C when [Fe2+]= 3.10 M and [Mg2+]= 0.110 M .

E=____V

Answer 1

We are going to use the nerst equation

E= E⁰ - RT/nF lnQ

where E is the cell potential, E⁰ is the standard potential cell, R ideal gas constant, T temperature, n electrons transfered, F faraday`s constant, Q is the reaction cocient

First we have to calculate E⁰

E⁰ = E⁰redcat - E⁰red anode so the value of the standard potential of each reaction we found it in any table of standard potential, one of the species reduces that is the Fe because it gain electrons and the Mg oxidice because it loses electrons.

Where the reduction occurs is the cathode and where the oxidation occurs is the anode.

Because in the tables of standard potential we only have reduction potentials we change the sign of the Mg reaction to indicate that the reaction occuring is the oxidation and in the Fe we leave it the same to indicate that the reduction is taking place.

For Mg+2 + 2e ----------Mg0 E= -2,38V and for Fe+2 + 2e------Fe0 E= -0,44V

E⁰= (-0,44) - (+2,38V) = -2,82V

E= -2,82V - (0,082latm/Kmol * 316K)/2*96500C ln[Mg+2]/[Fe+2]

E=  -2,82V - (25,912/193000)ln (0,110/3,10)

E=  -2,82V - (1,34*10^-4) ln 0,035

E=  -2,82V - (1,34*10^-4) -3,34

E= -2,82V

So the concentration doesn´t affect the potential of this reaction.