Question

If each of these ions were reduced to metal with one coulomb, which would yield the greatest mass?

(A) Cu²⁺(aq)
(B) Ag⁺(aq)
(C) Hg₂²⁺(aq)
(D) Cu⁺(aq)

Answer 1

The ion that would yield the greatest mass when reduced with one coulomb of charge is Hg₂⁺²(aq), because it has the highest formula mass among the ions that require only a single electron to reduce and will give the greatest mass per mole of electrons (option C).

Step-by-step explanation:

The student has asked which ion would yield the greatest mass when reduced with one coulomb of charge. For each metal ion, the number of coulombs required to reduce it to metal needs to be considered in relation to its molar mass. As specified, 1 mol of electrons, equivalent to 1 faraday (96,485 C), is needed to reduce 1 mol of Ag⁺ ion to solid Ag, but 1 mol of electrons reduces only 0.5 mol of Cu⁺² ion to solid Cu. For Cu⁺ and Hg₂⁺², the reduction would involve either 1 or 2 electrons, respectively, to be fully reduced to their solid form.

Knowing that 1 coulomb is much less than a faraday, we can deduce that the greatest mass of metal would come from the ion that requires the fewest electrons to be reduced. Since Ag⁺ Cu⁺, and Hg₂⁺² ions all require only a single electron to be reduced (Ag⁺ to Ag, Cu⁺ to Cu, and Hg₂⁺² to Hg2), the ion with the greatest atomic or formula mass will give the greatest mass of metal per mole of electrons. Among these, Hg2 has the greatest formula mass, therefore Hg₂⁺² would yield the greatest mass when reduced by one coulomb.

Thus, the correct answer is option (C) Hg₂⁺²(aq).

Answer 2

Cu⁺(aq) would yield the greatest mass. So, Option D is correct.

Step-by-step explanation:

The mass of the metal deposited during electrolysis is directly proportional to the quantity of charge (in coulombs) passed through the electrolyte. The relationship is given by Faraday's laws of electrolysis. The equation for the mass deposited (m) is given by:

m = (Q . M)/(n . F)

where:

Q is the charge in coulombs,

M is the molar mass of the substance deposited,

n is the number of moles of electrons exchanged during the reaction, and

F is Faraday's constant.

Comparing the given ions, Cu⁺ has a lower charge than Cu²⁺ and Hg₂²⁺, and it is equal to Ag⁺. Since the charge (Q) is the same for all ions (one coulomb), the lower charge of Cu⁺ implies that a greater quantity of moles of Cu⁺ will be deposited during the reduction reaction. Additionally, the molar mass of copper (Cu) is lower than that of silver (Ag), making Cu⁺ the ion that would yield the greatest mass of metal.

In summary, Cu⁺(aq) would yield the greatest mass because of its lower charge and lower molar mass compared to the other ions. This prediction is based on the fundamental principles of electrolysis, specifically Faraday's laws, which govern the relationship between the quantity of charge passed and the mass of the substance deposited during electrolysis.