Answer:

To determine the amount of copper deposited by one Faraday of current in each solution, we need to consider the stoichiometry of the electrochemical reactions involved.

a) 1M Cu2Cl2:
The balanced half-reaction for the reduction of Cu2+ ions in acidic solution is:

Cu2+ + 2e- -> Cu

From the reaction, we can see that 2 moles of electrons are required to deposit 1 mole of copper ions. Therefore, 1 Faraday of charge (96485 Coulombs) will deposit 0.5 moles of copper.

b) 2M Cu(NO3)2:
The balanced half-reaction for the reduction of Cu2+ ions in acidic solution is the same as in option a.

Similarly, 2 moles of electrons are required to deposit 1 mole of copper ions. Therefore, 1 Faraday of charge will deposit 1 mole of copper.

c) 5M CuSO4:
The balanced half-reaction for the reduction of Cu2+ ions in acidic solution is the same as in option a.

Again, 2 moles of electrons are required to deposit 1 mole of copper ions. Therefore, 1 Faraday of charge will deposit 2.5 moles of copper.

d) 5M Cu3(PO4)2:
The balanced half-reaction for the reduction of Cu2+ ions in acidic solution is the same as in option a.

Similarly, 2 moles of electrons are required to deposit 1 mole of copper ions. Therefore, 1 Faraday of charge will deposit 2.5 moles of copper.

Summary:
The amount of copper deposited by one Faraday current will be maximum in option b) 2M Cu(NO3)2, where 1 mole of copper is deposited. In options a), c), and d), only 0.5 moles of copper are deposited by one Faraday of current.

Note:
It is important to note that the concentration of the solution does not affect the amount of copper deposited per Faraday, as the stoichiometry of the electrochemical reaction remains the same. The concentration only affects the total amount of copper that can be deposited in a given volume of the solution.