Map A solution containing the following was prepared: 0.14 M Pb2 1.5 x106 M Pb 1

Question

Map A solution containing the following was prepared: 0.14 M Pb2 1.5 x106 M Pb 1.5 x 106 M Mn2 o.14 M Mno4 and 0.80 M HNO3. For this solution, the following balanced reduction half reactions and overall net reaction can occur. 4 2+ 5 Pb 2e Pb E 1.690 v Mn 4H,O E 1.507 V 4+ 8H,O A) Determine E cell. AG, and K for this reaction. Number Number Number cel B) Calculate the value for the cell potential. Eoel, and the free energy, AG, for the given conditions. Number Number AG Scroll down for the rest of the question.

Explanation / Answer

A) Eocell = Ecathode - Eanode

                = 1.690 - 1.507 = 0.183 V

dGo = -nFEocell

        = -10 x 96500 x 0.183 = -176595 kJ

K = [Pb2+]^5.[MnO4^2-]^2.[H+]^16/[Pb4+]^5.[Mn2+]^2

   = (0.14)^7.(0.8)^16/(1.5 x 10^-6)^7

   = 1.76 x 10^33

B) Ecell = Eocell - 0.0592/n logK

              = 0.183 - 0.0592/10 log(1.76 x 10^33)

              = -0.014 V

dG = dGo + RTlnK

      = -176595 + 8.314 x 298 ln(1.76 x 10^33)

      = 13064.68 kJ

C) Ecell at equilibrium = 0 V

D) at equilibrium,

Eocell = 0.0592/n logK

0.183 = 0.0592/10 log[(0.14)^7[H+]^16/(1.5 x 10^-6)^7]

[H+] = 0.715 M

pH = -log[H+] = 0.146

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