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, you need to use the Nernst equation, which takes into account the concentrations of the species in the electrochemical cell.
The Nernst equation is as follows:
E = E° - (0.0592V / n) * log(Q)
Where:
- E is the cell potential
- E° is the standard cell potential
- n is the number of electrons transferred in the balanced redox reaction
- Q is the reaction quotient, which is the ratio of the concentrations of the products over the concentrations of the reactants raised to their stoichiometric coefficients
In this case, Fe3+ is reduced to Fe, which involves the gain of 3 electrons (n=3). Pb is oxidized to Pb2+, which involves the loss of 2 electrons (n=2).
Given that the initial concentration of Fe3+ is 2.29M and it changes by 1.432M, the final concentration will be 2.29M - 1.432M = 0.858M.
Now, let's calculate the reaction quotient (Q):
Q = ([Pb2+]/[Fe3+]^2)^2/3
= [(1.27M)/(0.858M)^2]^2/3
Before plugging the values into the Nernst equation, we need to find the standard cell potential (E°) for the given redox reaction. Since the reduction potential for Pb2+ to Pb is not provided, we can use the standard reduction potentials of the individual half-reactions to calculate the cell potential.
The standard reduction potential for Fe3+ to Fe is given, which is 0.77V. The standard reduction potential for Pb2+ to Pb can be obtained from standard reduction potential tables, which is -0.13V.
The overall cell potential (E°) can be calculated as follows:
E° = E°cathode - E°anode
= 0.77V - (-0.13V)
= 0.90V
Now, we can substitute the values into the Nernst equation to calculate the cell potential (E):
E = 0.90V - (0.0592V/6) * log(Q)
Remember to convert the concentrations to their respective powers of 10 before taking the logarithm.
After performing the calculations, the cell potential (E) is approximately 0.0794V.
Therefore, the correct answer is indeed 0.0794V.