Answer:
a) Ecell = 0.5123 V
b) Ecell = 0.4695 V
c) [Pb2 +] = 4.75 M
Explanation:
a)
The reaction at the cathode is represented as follows:
Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V
The reaction at the anode is equal to:
Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V
The number of moles of the electrons that are involved is equal to n = 2
Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V
The initial cell potential can be calculated with the following formula:
Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V
b)
The reaction in the cell is equal to:
Cu2 + + Pb (s) -> Cu (s) + Pb2 +
The concentration of Cu2 that gives the exercise is equal 0.2 M
Therefore, the change in concentration for Cu2 + is equal to:
Cu2 + = 1.4 M - 0.2 M = 1.2 M
We use the formula from part a)
Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V
c)
To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:
Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))
0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)
Clearing Pb2 +:
[Pb2 +] = 4.75 M
