The easiest way to do this is to reverse equation 1 and add in equation 2 for
PbO2 + 4H^+ + SO4^2- + Pb + SO4^2- --> PbSO4 + PbSO4 or
PbO2 + 2H2SO4 --> 2PbSO4 + 2H2O Ecell = 2.041
Then Ecell = Eocell - (0.05916/n)*log(1/(H2SO4). Plug in the numbers and solve for Ecell.
Given:
PbSO4 + 2e− Pb + SO42− E° = -0.356 V
PbO2 + 4H+ + SO42− + 2e− PbSO4 + 2H2O E° = 1.685 V
Determine the cell voltage of the lead storage battery when the concentration of sulfuric acid is 4.00 M.
Where do I start, thank you
2 answers
I'm confused. Why do you use log and why do you do 1/H2SO4? I found this equation in our textbook:
Ecell=Eocell-(RT/nFl)nQ
When I use that equation I get 2.00, when I use the equation you've given I get 2.05. I know it's a small difference but why?
Thank you
Ecell=Eocell-(RT/nFl)nQ
When I use that equation I get 2.00, when I use the equation you've given I get 2.05. I know it's a small difference but why?
Thank you
1 answer