Posted by **Andres** on Tuesday, October 2, 2012 at 8:44pm.

Consider a cell at 255 K:

line notation

Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe

Given the standard reduction potentials calculate the cell potential after the reaction operated long enough for Fe3+ to have changed by 1.432M?

I know that Fe3+ is the cathode which will decrease in concentration and Pb2+ is the anode which increases in concentration. The overall electrons transferred is 6 and Q is (Pb2+)^3 over (Fe3+)^2. After calculating the potential i keep getting 0.08195V but the answer is 0.0794V, i don't know where im going wrong, can anyone help.

## Answer This Question

## Related Questions

- Chemistry - Consider a cell at 255 K: line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-...
- Chemistry - Consider a cell at 255 K: line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-...
- Chemistry - Consider a cell at 255 K: line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-...
- Chemistry - Consider a cell at 255 K: line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-...
- Chemistry/Electrochem - Consider a cell at 255 K: line notation Pb-Pb2+(1.27M)--...
- Chemistry (Electrochemistry) - The standard potential for the following galvanic...
- Chemistry - Calculate the cell potential for the following reaction as written ...
- Chemistry - calculate cell potential at 25C for: 3Fe3+(aq) + AL(s)= 3Fe2+(aq)+ ...
- chemistry - Could I get some help with these please? Thanks Calculate the ...
- ap chemistry - I have a lot of questions actually. it's electrochemistry (which ...

More Related Questions