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.
To calculate the cell potential of a voltaic cell, you need to know the standard reduction potentials of the half-reactions involved.
First, let's write the two half-reactions for the given half-cell pair:
1. Hg2Cl2(s) + 2e- ↔ 2Hg(l) + 2Cl- (aq) (Reduction half-reaction)
2. Pb2+ (aq) + 2e- ↔ Pb(s) (Reduction half-reaction)
The standard reduction potentials for these reactions can be found in a reference table, such as a standard reduction potentials table.
Once you have the reduction potentials, you can calculate the standard cell potential, E°cell, using the formula:
E°cell = E°cathode - E°anode
Now, let's find the standard reduction potentials for the given half-reactions:
1. The standard reduction potential for the reduction of Hg2Cl2 to Hg is given as +0.28V.
2. The standard reduction potential for the reduction of Pb2+ to Pb is given as -0.13V.
Now, we can calculate the cell potential:
E°cell = E°cathode - E°anode
E°cell = (+0.28V) - (-0.13V)
E°cell = +0.28V + 0.13V
E°cell = +0.41V
So, the cell potential for the given voltaic cell is +0.41V.