A solution containing the following was prepared: 0.14 M Pb^2+, 1.5 times 10^-6

Question

A solution containing the following was prepared: 0.14 M Pb^2+, 1.5 times 10^-6 M Pb^4+, 1.5 times 10^-6 M Mn^2+, 0.14 M MnO_46-, and 0.89 M HNO_3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur. 5[pb^4+ + 2e Pb^2+] E_+^degree = 1.690 V 2[MnO_4^-+8H^+ + 5e' Mn^2+ + 4H_2O] E degree = 1.507 V 5Pb^4+ + 21Mn^2 + + 8H_2O 5Pb^2+ + 2MnO_4^- + H^+ Determine E degree cell Delta G degree, and K for this reaction. Calculate the value for the cell potential, E_cell and the free energy, Delta G, for the given conditions. Determine E degree _cell, Delta G degree, and K for this reaction. Calculate the value for the cell potential, E_cell, and the free energy, Delta_G, for the given conditions. Determine E degree _cell, Delta G degree, and K for this reaction. Calculate the value of E_cell for this system at equilibrium. Delta G, for the given conditions. Calculate the value of E_cell for this system at equilibrium. Determine the pH at which the given concentrations of Pb^2+, Pb^4+, Mn^2+, and MnO_4^- would be at equilibrium.

Explanation / Answer

For the given reaction,

A) Eocell = 1.690 - 1.507 = 0.183 V

dGo = -nFEocell

        = -10 x 96485 x 0.183/1000

        = -176.57 kJ/mol

dGo = -RTlnK

-176567 = -8.314 x 298 lnK

K = 8.92 x 10^30

B) Ecell = Eo - 0.0592/n logK

              = 0.183 - 0.0592/10 log((0.14)^5 x (0.14)^2 x (0.89)^16)/((1.5 x 10^-6)^5 x (1.5 x 10^-6)^2)

              = -0.02 V

dG = dGo + RTlnK

      = 176567 + 8.314 x 298 ln(8.90 x 10^30)

      = -15 kJ

C) Ecell = 0 V at equilibrium

D) Eocell = 0.0592/n logQ

     0.183 = 0.0592/10 log 6.17 x 10^34 [H+]^16

[H+] = 0.572 M at equilibrium

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