How do I calculate the cell potential for PbO2(s)+4H+(aq)+Sn(s)→Pb2+(aq)+2H2O(l)+Sn2+(aq)?

How do I calculate the cell potential for PbO2(s)+4H+(aq)+Sn(s)→Pb2+(aq)+2H2O(l)+Sn2+(aq)?

pbo2(s)+4h+(aq)+sn(s)→pb2+(aq)+2h2o(l)+sn2+(aq)

Answer:

Look up the standard reduction potentials for the two half reactions:

PbO2(s) + 4 H+ + 2 e– –> Pb2+ + 2 H2O(l) Eo = +1.46

Sn2+ + 2e—> Sn(s) Eo = -0.136

Now, in your reaction, the second half-reaction is an oxidation, so change the sign of its Eo, and add the two together:

Ecell = +1.46 V + 0.136 V = +1.60 V

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