Calculate the cell potential of voltaic cells that contain the following pairs of half-cells.

a)A half-cell containing both HgCl2 and Hg2Cl2; lead in a solution of Pb2+ ions.

Hug

To calculate the cell potential of a voltaic cell, you need to consider the half-cell potentials of the two half-cells and their respective reactions. The cell potential is determined by the difference in the standard reduction potentials of the half-cells.

In this case, we have two half-cells:

1) Mercury (II) chloride and mercury(I) chloride (HgCl2 and Hg2Cl2)
2) Lead (Pb) in a solution of lead (II) ions (Pb2+)

First, we need to find the half-cell potentials for each of these reactions. The standard reduction potentials are typically provided in a standard reduction potential table.

Let's find the half-cell potential for each half-cell:

1) Mercury (II) chloride and mercury(I) chloride:

The reaction can be written as follows:

Hg2Cl2 + 2e- → 2Hg(l) + 2Cl-

The standard reduction potential for this half-reaction is typically given as -0.336 V.

2) Lead (Pb) in a solution of lead (II) ions:

The reaction can be written as follows:

Pb2+(aq) + 2e- → Pb(s)

The standard reduction potential for this half-reaction is typically given as -0.126 V.

Now, we can calculate the overall cell potential by subtracting the reduction potential of the anode (the one that gets oxidized) from the reduction potential of the cathode (the one that gets reduced).

Cell potential = Reduction potential of the cathode - Reduction potential of the anode

Cell potential = (-0.126 V) - (-0.336 V)

Cell potential = 0.21 V

Therefore, the cell potential of the voltaic cell containing the given half-cells is 0.21 V.