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:increase, decrease or remain the same
When sulfuric acid is added to the Pb(NO3)2 solution, it reacts with the lead (Pb) ions, resulting in the formation of a precipitate of PbSO4. This reaction can be represented as follows:
Pb2+ + SO4^2- → PbSO4
Since PbSO4 is insoluble in water, it forms a solid precipitate. As a result, the concentration of Pb2+ ions in the solution decreases.
The cell potential (or voltage) of a cell depends on the difference in the standard reduction potentials of the half-cell reactions that occur at the electrodes.
Before adding sulfuric acid, the reduction potential of the lead electrode is -0.13 V, and the reduction potential of the copper electrode is +0.34 V. The overall cell potential is calculated by subtracting the reduction potential of the anode (the electrode undergoing oxidation) from the reduction potential of the cathode (the electrode undergoing reduction).
E(cell) = E(cathode) - E(anode)
Thus, E(cell) = +0.34 V - (-0.13 V) = +0.47 V
Now, when sulfuric acid is added to the Pb(NO3)2 solution, the concentration of Pb2+ ions decreases. The reduction potential of the Pb2+ half-cell reaction remains the same (-0.13 V), but the concentration of Pb2+ ions affects the overall cell potential.
Since the concentration of Pb2+ ions decreases, it will affect the reduction potential of the half-cell reaction at the anode (Pb electrode). This decrease in concentration will result in a slightly more positive reduction potential for the Pb electrode. Therefore, the overall cell potential will increase, and the cell potential will be larger than 0.47 V.
Thus, when sulfuric acid is added to the Pb(NO3)2 solution, forming a precipitate of PbSO4, the cell potential will increase.
To determine how the cell potential will be affected when sulfuric acid is added to the Pb(NO3)2 solution, we need to consider the half-reactions involved in the cell setup.
The half-reactions are as follows:
1. Pb2+ + 2e- → Pb (reduction on the lead electrode)
2. Cu2+ + 2e- → Cu (reduction on the copper electrode)
The cell potential is calculated by taking the difference between the reduction potentials of the two half-reactions. Since both half-reactions involve the gain of electrons (reduction) and we use the reduction potentials, we need to reverse the sign of the reduction potential for the half-reaction of copper (Cu) to align it with the overall reaction.
Therefore, the cell potential (Ecell) is given by:
Ecell = E°(reduction potential of Pb) - E°(reduction potential of Cu)
= -0.13 V - (+0.34 V)
= -0.47 V
Now, when sulfuric acid is added to the Pb(NO3)2 solution, it reacts with the lead (Pb) ions to form lead sulfate (PbSO4):
Pb2+ + SO4^2- → PbSO4
Since the lead ions (Pb2+) are consumed in the formation of lead sulfate, the concentration of Pb2+ decreases. According to the Nernst equation, which accounts for changes in concentration, the cell potential (Ecell) would decrease as the concentration of Pb2+ decreases. This is because the reduction potential depends on the concentration of the ions involved in the half-reaction.
Therefore, when sulfuric acid is added to the Pb(NO3)2 solution, forming a precipitate of PbSO4, the cell potential will decrease.
E = Eo -0.06/2*log(1/(Pb^+2)
If Pb^+2 + SO4^= -->PbSO4.
Therefore, the addition of sulfate causes the PbSO4 to ppt thus decreasing the (Pb^+2). What does that do to the E value for Pb and how will that affect the cell voltage?