A solution containing the following was prepared: 0.19 M Pb2 , 1.5 × 10-6 M Pb4

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, G, 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

Explanation / Answer

A) Calculation of E0 Cell
E0cell = Ecathode – Eanode

We know oxidation takes place on anode , reduction takes place on cathode.

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

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

These are the reductions potentials of both half reactions.

Now we are also given ;

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

From this reaction we can write oxidation and reduction half reaction .

Oxidation half:

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

Reduction half

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

E0cell = (1.693 – 1.507 ) V = 0.186 V

Delta G = - nEF

Number of electrons are shown by n = 10

And F is faraday’s constant.

Delta G = - 10 x 0.186 x 96500 = - 179490 J

Delta G = - RTlnK

- 179490 J = - (8.314 J / K mol ) x 298 K x ln K

K = 2.90 E 31

C) Cell potential

Nernst equation :

Ecell = E0cell – 0.0591 / n   x log ( [oxidant]/[reductant] )

Here n = number of electrons involved in overall cell reaction,

Lets plug the value

Ecell = 0.186– 0.0591 / 10 x log ([ MnO4-] / [Pb2+] )

Ecell = 0.186 – 0.0591 / 10 x log (0.19 / 0.19 )

Ecell = 0.186 V

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