Three Faradays of electricity was passed through an aqueous solution…

The correct answer is option 2. 84 g.

To determine the mass of iron deposited at the cathode when 3 Faradays of electricity are passed through an aqueous solution of iron (II) bromide (\(\text{FeBr}_2\)), follow these steps:

Faraday's constant (\(F\)) is approximately 96,485 C/mol.

The reduction reaction for iron (II) bromide is:

\(\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}\)

This indicates that 2 moles of electrons are required to reduce 1 mole of iron ions (\(\text{Fe}^{2+}\)) to iron metal (\(\text{Fe}\)).

To deposit 1 mole of iron, you need 2 Faradays of charge (because 2 moles of electrons are needed).

With 3 Faradays of charge, the number of moles of iron deposited is:

\(\text{Moles of Fe} = \frac{\text{Total Faradays}}{\text{Faradays per mole of Fe}} = \frac{3}{2} = 1.5 \text{ moles}\)

The atomic mass of iron is 56 g/mol. Therefore, the mass of iron deposited is:

\( \text{Mass} = \text{Moles} \times \text{Atomic mass} = 1.5 \times 56 = 84 \text{ grams}\)

Conclusion

The mass of iron metal deposited at the cathode is 84 grams.