A solution of \(AgSO_4\) is electrolyzed for 30 minutes with a current of 5 amperes. What is the mass of the silver deposited?

[Atomic mass of Ag = 108 g/mol]

5.04 g

The correct answer is option 3. 5.04 g.

The total charge (\(Q\)) passed through the solution is given by:

\(Q = I \times t \)

where:

\( I \) is the current (5 amperes),

\( t \) is the time (30 minutes).

\(t = 30 \text{ minutes} = 30 \times 60 \text{ seconds} = 1800 \text{ seconds} \)

Now the total charge is

\(Q = 5 \text{ A} \times 1800 \text{ s} = 9000 \text{ C}\)

The electrolysis of \(Ag^{2+}\) ions will deposit silver metal (Ag) at the cathode. The reaction at the cathode is:

\(Ag^{2+}(aq) + 2e^- \rightarrow Ag (s)\)

The molar mass of silver (\(Ag\)) is 108 g/mol. The equivalent weight of silver is the molar mass divided by the number of electrons transferred, which is 2 (because \(Ag^{2+}\) to \(Ag\) involves a two electron transfer). Thus, the equivalent weight of silver is also 54 g/mol.

Faraday’s constant (\(F\)) is approximately 96500 C/mol, representing the charge per mole of electrons.

Thus, mass of Ag deposited is given by

\(\frac{\text{Molar mass} \times \text{Charge}}{\text{Electrons transferred} \times \text{Faraday's constant}}\)

\(= \frac{108 \times 9000}{2 \times 96500}\)

\(= 5.04\, \ g\)