Ecell = Eocell + (0.0592/#e)log(Pb^2+)/(Pb^2+)
Ecell = 0.1776
Eocell= I don't have my text handy but I think this is -0.126. You should verify that.
#e = 2
For the log term, the top number there is 1M and the lower number is x. You solve for x
The measured potential of the following cell was 0.1776 volts:
Pb|Pb2+(??M)||Pb2+(1.0M)|Pb
Calculate the electrode potential of the unknown half-cell(anode).
Not sure how to approach this, considering the molarity is unknown. There isn't a lot of space on the page for the question, so I'm guessing it's a relatively simple solution.
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