A cell is set up with copper and lead electrodes in contact with CuSO4(aq) and Pb(NO3)2(aq), respectively, at 25˚C. The standard reduction potentials are:
Pb2+ + 2e- --> Pb E˚=-0.13 V
Cu2+ + 2e- --> Cu E˚=+.34 V
If sulfuric acid is added to the Pb(NO3)2 solution, forming a precipitate of PbSO4, the cell potential:
See above.
increasses
To determine the cell potential when sulfuric acid is added and PbSO4 precipitates, we need to consider the overall reaction of the cell.
The half-reactions involved in the cell are:
1. Pb2+ + 2e- --> Pb (reduction half-reaction)
(Standard reduction potential: E˚ = -0.13 V)
2. Cu2+ + 2e- --> Cu (oxidation half-reaction)
(Standard reduction potential: E˚ = +0.34 V)
Since both half-reactions occur in an acidic solution, we can balance the overall cell reaction by adding H+ ions and water molecules where necessary.
The balanced overall cell reaction is:
Pb2+ + Cu --> Pb + Cu2+
Now, let's calculate the cell potential using the standard reduction potentials and the Nernst equation, which accounts for non-standard conditions:
Ecell = E˚cell - (0.0592 V/n) * log(Q)
where:
Ecell = Cell potential
E˚cell = Standard cell potential
n = number of electrons transferred in the cell reaction
Q = reaction quotient (concentration of products divided by concentration of reactants)
For the cell reaction given above, no reactants or products are present in the reaction quotient (Q) since both sides have pure solids. Therefore, Q = 1.
Plugging in the values, we get:
Ecell = (0.34 V) - (0.0592 V/2) * log(1)
Ecell = 0.34 V - 0.0296 V
Ecell = 0.3104 V
Accordingly, when sulfuric acid is added and PbSO4 precipitates, the cell potential is approximately 0.3104 V.
To determine the cell potential when sulfuric acid is added to the Pb(NO3)2 solution, you need to consider the effect of the precipitate (PbSO4) on the cell. The presence of the precipitate will cause a shift in the reaction and thus affect the overall cell potential.
To start, let's break down the overall cell reaction into its two half-reactions:
1. Reduction half-reaction: Pb2+ + 2e- --> Pb (E˚ = -0.13 V)
2. Oxidation half-reaction: Cu --> Cu2+ + 2e- (E˚ = +0.34 V)
Next, we need to determine which half-reaction will take place at each electrode based on their respective reduction potentials. The half-reaction with the more positive value will occur as the reduction half-reaction, while the other will be the oxidation half-reaction.
Since Cu2+ + 2e- --> Cu has a more positive reduction potential (+0.34 V) compared to Pb2+ + 2e- --> Pb (-0.13 V), the Cu half-reaction will occur as the reduction half-reaction at the cathode (negative electrode). The Pb half-reaction will be the oxidation half-reaction at the anode (positive electrode).
Now, when sulfuric acid is added to the Pb(NO3)2 solution, it will react with the lead ions (Pb2+) to form lead sulfate (PbSO4), which is a solid precipitate. This precipitation reaction will remove Pb2+ ions from the solution, thereby reducing the concentration of Pb2+ available for the oxidation half-reaction.
The decrease in the concentration of Pb2+ ions will shift the equilibrium of the oxidation half-reaction to the left, reducing its occurrence. This, in turn, will increase the occurrence of the Cu half-reaction as the reduction half-reaction at the cathode.
Therefore, the cell potential will be determined by the Cu half-reaction. The reduction potential for Cu2+ + 2e- --> Cu (E˚ = +0.34 V) is positive, and thus, the cell potential will remain positive.
In conclusion, adding sulfuric acid to the Pb(NO3)2 solution and forming a precipitate of PbSO4 will not change the fact that the cell potential remains positive, as it is determined by the Cu half-reaction. However, the concentration of Pb2+ ions will decrease due to the formation of the precipitate, shifting the equilibrium of the oxidation half-reaction to the left.