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Unformatted text preview: PbSO4/Pb b) EE (V) -0.355 Write the line notation for this cell.
Pb(s) c) Since the Cu2+ reduction is
spontaneous (EE>0), but the Pb2+
reduction is not, oxidation of the
Pb must occur. Electrode 2 is the
anode. PbSO4(s) SO42-(0.115 M) Cu2+ (0.0300 M) Cu(s) Calculate the voltage (E) for this cell.
Ecell = Eox + Ered anode: Pb(s) + SO42-(aq) X PbSO4(s) + 2e- cathode: Cu2+(aq) + 2e- X Cu (s) Ecell = 0.327 V + 0.294 V = 0.621 V 0.621 V...
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This document was uploaded on 03/06/2014 for the course CHEM 321 at CSU Northridge.
- Fall '13