Asked by DAN
A voltaic cell is constructed that is based on the following reaction.
Sn2+(aq) + Pb(s) -> Sn(s) + Pb2+(aq)
(a) If the concentration of Sn2+ in the cathode half-cell is 1.80 M and the cell generates an emf of +0.219 V, what is the concentration of Pb2+ in the anode half-cell?
(b) If the anode half-cell contains [SO42ā] = 19.08 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?
Sn2+(aq) + Pb(s) -> Sn(s) + Pb2+(aq)
(a) If the concentration of Sn2+ in the cathode half-cell is 1.80 M and the cell generates an emf of +0.219 V, what is the concentration of Pb2+ in the anode half-cell?
(b) If the anode half-cell contains [SO42ā] = 19.08 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?
Answers
Answered by
DrBob222
a.
Ecell = Eocell - 0.0592/2*log Q where
Q = [products]/[reactants]
Substitute into the equation and solve for the one unknown.
b. After you determine (Pb^2+) in the half cell, then (Pb^2+)(SO4^2-) = Ksp
Ecell = Eocell - 0.0592/2*log Q where
Q = [products]/[reactants]
Substitute into the equation and solve for the one unknown.
b. After you determine (Pb^2+) in the half cell, then (Pb^2+)(SO4^2-) = Ksp
Answered by
karonavyress
this question is wack
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.