# chemistry

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).

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1. got this on google unknown half-cell(anode). ... 2Ag+(aq)+Pb(s)→2Ag(s)+Pb2+(aq) E∘cell.

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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 ...

if it does not work feel free to tell me here or on

h t t p s : / / padlet.com/Rae sha wolf/brsu3rnmu1is6shu

just remove all spaces

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2. This is a concentration cell.
Pb ==> Pb^2+ + 2e.......................Eox = ?
Pb^2+ + 2e ==> Pb...................... Ered = - 0.126
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Pb + Pb^2+ ==> Pb^2+ + Pb........ Ecell = 0.1776
Eox + Ered = Ecell
Eox + (-0.126) = 0.1776 which I will round to 0.178
Eox = 0.304 and written as a reduction it is -0.304
-0.304 = Eocell -(0.0592/2)*log (Pb)/(Pb^2+)
-0.304 = - 0.126 - (0.0592/2)*log (1)/(Pb^2+).
Solve for (Pb^2+) in moles/L.

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DrBob222

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