Question

A solution containing the following was prepared: 0.19 M Pb2 , 1.5 × 10-6 M Pb4 , 1.5 × 10-6 M Mn2 , 0.19 M MnO4–, and 0.81 M HNO3. For this solution, the following balanced reduction half-reactions and overall net reaction can occur.

5[Pb^4+ + 2e^- > Pb^2+] E^o=1.690

2[MnO4^-+8H^+ + 5e^-> Mn^2+ + 4H2O] E^o=1.507V

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5Pb^4+ + 2Mn^2+ + 8 H2O> 5Pb^2+ + 2MnO4^- + 16H^+

A) Determine E^ocell, deltaG, and K for the Reaction.

B) Calculate the value for the cell potential, Ecell, and the free energy, ΔG, for the given conditions.

C) Calculate the value of Ecell for this system at equilibrium.

D) Determine the pH at which the given concentrations of Pb2 , Pb4 , Mn2 , and MnO4– would be at equilibrium. THANKS

Answer 1

A) Determine E°cell, ΔG°, and K for this reaction. (V,J,K)

Here, the reaction is

5 Pb⁴⁺ + 2 Mn²⁺ + 8H₂O       5Pb²⁺ + 2MnO₄^- + 16H⁺

G^ocell = -nFE^ocell = -10 x 96485 x 0.183 = -176567 J/mol

G^o = -RTlnK = -8.314 x 298 K x lnK

lnK = 71.27

K = 8.96 x 10³⁰

B) Calculate the value for the cell potential, Ecell, and the free energy, ΔG, for the given conditions.(V,J)

We have Nernst equation,

Ecell = E^ocell – (0.0591/n)log[Pb²⁺]⁵[MnO₄^-]²[H⁺]¹⁶/[Pb⁴⁺]⁵[Mn²⁺]²

         = 0.183 V – (0.0591/10)log[0.19]⁵[0.19]²[0.86]¹⁶/[1.5 x 10^-6]⁵[1.5 x 10^-6]² = -0.022 V

G = -nFE = -10 x 96485 x 0.022 = 21226.7 J/mol

C) Calculate the value of Ecell for this system at equilibrium.(V)

At equilibrium G^o = 0

Therefore,

Ecell = 0

D) Determine the pH at which the given concentrations of Pb²⁺, Pb⁴⁺, Mn²⁺, and MnO₄- would be at equilibrium.(pH)

Ecell = E^ocell – (0.0591/n)log[Pb²⁺]⁵[MnO₄^-]²[H⁺]¹⁶/[Pb⁴⁺]⁵[Mn²⁺]²

    0 = 0.183 V – (0.0591/10)log[0.19]⁵[0.19]²[H⁺]¹⁶/[1.5 x 10^-6]⁵[1.5 x 10^-6]² = -0.022 V

30.91 = log(5.23 x 10³⁵) + 16 log[H⁺]

pH = 0.3