Posted by **Andres** on Tuesday, October 2, 2012 at 8:46pm.

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.