Consider two cells, the first with Al and Ag electrodes, and the second with Zn and Ni electrodes, each in appropriate 1.00 M solutions of their ions. a) If connected as voltaic cells in series, which two metals are plated, and what is the total potential? If 2.50 g of metal is plated in the voltaic cell, how much metal is plated in the electrolytic cell?

To determine which metals are plated in the voltaic cells and the total potential, we need to compare the reduction potentials of the metals involved. The metal with the higher reduction potential will be plated at the cathode (positive electrode), while the metal with the lower reduction potential will be plated at the anode (negative electrode). Additionally, we can calculate the total potential by summing the reduction potentials of the two cells.

The reduction potentials for some common metals are as follows:
- Al: -1.66 V
- Ag: +0.80 V
- Zn: -0.76 V
- Ni: -0.25 V

a) Since the reduction potential for Ag (+0.80 V) is higher than that of Al (-1.66 V), Ag will be plated at the cathode, and Al will be plated at the anode in the first cell (Al-Ag).

In the second cell (Zn-Ni), the reduction potential for Zn (-0.76 V) is lower than that of Ni (-0.25 V). So, Zn will be plated at the anode, and Ni will be plated at the cathode.

To find the total potential of the series connection, we add up the reduction potentials of the two cells:
Total Potential = Reduction potential of Al-Ag cell + Reduction potential of Zn-Ni cell

Total Potential = (-1.66 V) + (-0.76 V) = -2.42 V

Therefore, the total potential of the series connection is -2.42 volts.

b) To determine the amount of metal plated in the electrolytic cell, we need to know the current and the time of electrolysis. Given that 2.50 g of metal is plated, we need to calculate the number of moles of metal for the corresponding metal and then convert it into grams for the other metal.

Let's assume that the current is expressed in amperes (A) and the time of electrolysis is given in seconds (s).

- Determine the number of moles of the metal that has been plated by Al-Ag cell:
molar mass of Ag = 107.87 g/mol
moles of Ag = mass of Ag plated / molar mass of Ag
= 2.50 g / 107.87 g/mol

- Convert the moles of Ag plated into moles of the other metal (Ni) using the stoichiometry:
From the balanced equation for the cell reaction: Al + Ag⁺ → Al³⁺ + Ag
1 mole of Ag is equivalent to 1 mole of Al (1:1 ratio).

- Convert the moles of Ni into grams of Ni:
molar mass of Ni = 58.69 g/mol
mass of Ni plated = moles of Ni * molar mass of Ni

By following these steps, you can calculate the amount of metal plated in the electrolytic cell.