Question

If three faradays of electricity is passed through the solutions of \(AgNO_3\), \(CuSO_4\) and \(AuCl_3\), the molar ratio of cations deposited at the cathodes will be:

• A. \(1:1:1\)
• B. \(1:2:3\)
• C. \(3:2:1\)
• D. \(6:3:2\)

Hint:

For electrolysis problems: \[ \text{Moles deposited}=\frac{\text{Faradays passed}}{\text{Valency}} \] Lower the valency, greater will be the amount of substance deposited for the same quantity of electricity.

Answer 1

The Correct Option is D

Concept: According to Faraday’s laws of electrolysis: \[ 1 \text{ mole of electrons } = 1 \text{ Faraday} \] The number of moles of metal deposited depends on the number of electrons required for reduction. General relation: :contentReference[oaicite:0]{index=0} where: \[ n=\text{number of electrons required for reduction} \]

Step 1: Write the cathode reactions. For silver nitrate: \[ Ag^+ + e^- \rightarrow Ag \] Here, \[ n=1 \] For copper sulphate: \[ Cu^{2+}+2e^- \rightarrow Cu \] Here, \[ n=2 \] For gold chloride: \[ Au^{3+}+3e^- \rightarrow Au \] Here, \[ n=3 \]

Step 2: Calculate moles deposited when 3 faradays are passed. For \(Ag^+\): \[ \text{Moles deposited}=\frac{3}{1}=3 \] For \(Cu^{2+}\): \[ \text{Moles deposited}=\frac{3}{2} \] For \(Au^{3+}\): \[ \text{Moles deposited}=\frac{3}{3}=1 \] Thus, the molar amounts are: \[ 3:\frac{3}{2}:1 \]

Step 3: Convert into simplest whole number ratio. Multiply all terms by \(2\): \[ 3\times2 : \frac{3}{2}\times2 : 1\times2 \] \[ 6:3:2 \] But the question asks the ratio of cations deposited based on valency relation. Equivalent deposition ratio becomes: \[ \frac{1}{1}:\frac{1}{2}:\frac{1}{3} \] Taking LCM \(=6\): \[ 6:3:2 \] However, comparing actual moles deposited directly: \[ 3:\frac32:1 \] Multiplying by \(2\): \[ 6:3:2 \] Hence, the correct option is: \[ \boxed{(D)\ 6:3:2} \]