Question

When the Pb2+ concentration is 3.76x10-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.430V. What is the AP+ concentration? 3Pb2+(aq) + 2Al(s)—>3Pb(s) + 2Al3+(aq) Answer: M

Answer 1

Concentration of Pb+2 = 3.76x10^-1 M = 0.376V

Cell potential = E= 1.430V

3 Pb+2(aq) + 2 Al(s) ------------- 3 Pb(s) + 2 Al+3(aq)

according to given equation

oxidation reaction at anode       [Al(s) ---------------------- Al+3(aq) + 3e-]x2

reduction reaction at cathode      [ Pb+2(aq) +2e- ------------ Pb(s) ] x 3

                                   -------------------------------------------------------------------------------

                                        3 Pb+2(aq) + 2 Al(s) ------------- 3 Pb(s) + 2 Al+3(aq)

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E0 Pb+2/Pb = - 0.13V                                         E0Al+3/Al = - 1.66V

E0cell = E0 cathode - E0anode

E0cell = - 0.13 - ( - 1.66)

E0cell = 1.53V

number of electrons transferred = n= 6e-

Ecell = E0cell - 0.0591/n log[Al+3]^2/[Pb+2]^3

1.430 = 1.53 - 0.0591/6 log[Al+3]^2/(0.376)^3

-0.10 = - 0.00985 log[Al+3]^2/0.05316

10.15 = log[Al+3]^2/0.05316

[Al+3]^2/0.05316= 10^10.15

[Al+3] = 27402.6M

Concentration of Al+3 = 2.74 x10^4 M