Question

A solution containing 0.190.19 M Pb2+Pb2+ , 1.5×10−61.5×10−6 M Pb4+Pb4+ , 1.5×10−61.5×10−6 M Mn2+Mn2+ , 0.190.19...

A solution containing 0.190.19 M Pb2+Pb2+ , 1.5×10−61.5×10−6 M Pb4+Pb4+ , 1.5×10−61.5×10−6 M Mn2+Mn2+ , 0.190.19 M MnO−4MnO4− , and 0.900.90 M HNO3HNO3 was prepared. For this solution, the balanced reduction half‑reactions and overall net reaction shown can occur.

5[Pb4++2e−↽−−⇀Pb2+]5[Pb4++2e−↽−−⇀Pb2+] ?∘+=1.690 VE+°=1.690 V
2[MnO−4+8H++5e−↽−−⇀Mn2++4H2O]2[MnO4−+8H++5e−↽−−⇀Mn2++4H2O] ?∘−=1.507 VE−°=1.507 V
5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO−4+16H+5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO4−+16H+

A. Determine ?∘cellEcell∘, Δ?∘ΔG∘, and ?K for this reaction.

?∘cell=Ecell∘=

V

Δ?∘=ΔG∘=

J

?=K=

B. Calculate the value for the cell potential, ?cellEcell, and the free energy, Δ?ΔG, for the given conditions.

?cell=Ecell=

V

Δ?=ΔG=

J

C. Calculate the value of ?cellEcell for this system at equilibrium.

?cell=Ecell=

V

D. Determine the pH at which the given concentrations of Pb2+Pb2+ , Pb4+Pb4+ , Mn2+Mn2+ , and MnO−4MnO4− would be at equilibrium.

pH=pH=

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A solution containing 0.190.19 M Pb2+Pb2+ , 1.5×10−61.5×10−6 M Pb4+Pb4+ , 1.5×10−61.5×10−6 M Mn2+Mn2+ , 0.190.19...
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