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 relates the cell potential to the concentrations of the reactants and products.
The Nernst equation is given as:
Ecell = E°cell - (0.0592V/n) * log10(Q)
Where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of electrons transferred
Q is the reaction quotient
In this case, you correctly identified that Fe3+ is the cathode (where reduction occurs) and Pb2+ is the anode (where oxidation occurs). The number of electrons transferred (n) is determined by the balanced equation for the overall reaction.
The reaction is:
3 Pb2+ + 2 Fe ↔ 3 Pb + 2 Fe3+
So, n = 2 (since 2 electrons are transferred in the overall reaction).
To calculate Q, you are correct that it is (Pb2+)^3 / (Fe3+)^2.
Given that Fe3+ changes by 1.432M, the new concentration of Fe3+ is 2.29M - 1.432M = 0.858M.
Plugging in the values:
Ecell = E°cell - (0.0592V/2) * log10((Pb2+)^3 / (0.858M)^2)
Now, you need to refer to the standard reduction potentials to find the value of E°cell for this particular half-cell reaction. Make sure you have the correct values for the reduction potentials of Fe3+/Fe and Pb2+/Pb. These values are usually given in a table with respect to the standard hydrogen electrode (SHE).
Once you have determined the standard reduction potentials, you can substitute the values into the equation to calculate the cell potential (Ecell).
It seems that you have correctly applied the above steps, but if you are getting a slightly different answer compared to the provided answer, it could be due to rounding errors in calculations or a slight variation in the standard reduction potentials used.
Double-check your calculations and ensure that you have used the correct values for E°cell. Also, check if there are any rounding errors in your intermediate calculations.