The following cell was found to have a potential of 0.416 V. Calculate the k for Pb3(ASO4)2.
Pb3(s)| Pb3(AsO)2(saturated) AsO4^3-(1.00x10^-3 M) || S.H.E.
Pb^2+ +2e- <---> Pb(s) E^o = -0.126 V
H3AsO4 + 2H+ + 2e- <---> H3AsO3 + H2O E^o = 0.559 V© BrainMass Inc. brainmass.com October 10, 2019, 7:04 am ad1c9bdddf
The oxidation half-reaction of the cell is the oxidation of Pb to Pb2+, and the reduction half reaction is the SHE half reaction.
OHR: Pb(s) <=> Pb2+(aq) + 2e- E° = 0.126 V
RHR: 2H+ + 2e- <=> H2O E° = 0.000 V
Overall cell ...
This solution shows the calculation of solubility product constant from electrochemical data.