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.
To calculate the cell potential after the reaction, you need to use the Nernst equation, which takes into account the changes in concentrations of the species involved.
The Nernst equation is given by:
E = E° - (0.0592/n) * log(Q)
Where:
E is the cell potential after the reaction,
E° is the standard cell potential (given in the question),
n is the number of electrons transferred (given in the question),
Q is the reaction quotient based on the concentrations of the species involved.
In this case, the reaction is:
Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe
Given that you correctly mentioned that Fe3+ is the cathode and decreases in concentration, and Pb2+ is the anode and increases in concentration, the reaction quotient (Q) can be calculated as:
Q = ([Pb2+]^3) / ([Fe3+]^2)
Now, let's substitute the given values into the Nernst equation:
E = E° - (0.0592/6) * log(Q)
Given that E° = 0.08195 V (as you previously calculated), and Q = (1.27^3) / (2.29^2), let's calculate the cell potential (E):
E = 0.08195 - (0.0592/6) * log((1.27^3) / (2.29^2))
Calculating this expression, you should arrive at an answer of E = 0.0794 V, which is the correct answer.
If you obtained a different value, please ensure that you have used the correct concentrations in the calculation and check any potential errors in the arithmetic/logarithmic calculations.