Question

When [Ag+] = 1.05 M, the observed cell potential at 298K for an electrochemical cell with the reaction shown below is 3.268 V. What is the Mg2+ concentration in this cell?

2Ag+(aq) + Mg(s)-->2Ag(s) + Mg2+(aq)

Answer 1

Given data:

The reactio in the electrochemical cell is

2Ag+(aq) + Mg(s)-->2Ag(s) + Mg2+(aq)

From Nernst equation, we can write

E(cell) = Eo(cell) - (0.059/n)*Log([P]/[R])

Here, Eocell = Eo(ox) + Eo(red); Here Eo(ox) = Eo(Mg/Mg2+) and Eo(red) = Eo(Ag+/Ag)

                   = 2.37 + 0.80, i.e. 3.17 V

E(cell) = 3.267 V, [Ag+] = 1.05 M

And n = 2, [P] = [Mg2+], [R] = [Ag+]^2

Therefore, 3.267 = 3.17 - (0.059/2)*Log{[Mg2+]/(1.05)^2}

                        = 3.17 - 0.0295*Log{[Mg2+]/1.1025}

      3.267 - 3.17 = -0.0295*Log{[Mg2+]/1.1025}

               0.097 = -0.0295*Log{[Mg2+]/1.1025}

Log{[Mg2+]/1.1025} = -0.097/0.0295, i.e. -3.288

       [Mg2+]/1.1025 = 10^( -3.288), i.e. 5*10^-4

Hence         [Mg2+] = 5.68*10^-4 M