We have enabled User registrations.
0 votes

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


1 Answer

0 votes
selected by
Best 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

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.