Step 1: Understanding the Concept:
Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction.
Faraday's laws of electrolysis define the relationship between the amount of electricity passed and the quantity of matter deposited.
One Faraday (\(1 F\)) of electricity is the charge carried by one mole of electrons (\(\approx 96485\) Coulombs).
The amount of a metal deposited depends on its oxidation state (valency).
For example, it takes twice as much electricity to deposit one mole of a divalent ion (\(M^{2+}\)) as it does to deposit one mole of a monovalent ion (\(M^+\)).
Step 2: Key Formula or Approach:
1. Faraday's Relation: \(\text{Number of moles deposited} = \frac{\text{Faradays passed}}{\text{Valency of cation } (n)}\).
Step 3: Detailed Explanation:
The amount of electricity passed is \(3\) Faradays.
1. For \(AgNO_3\): The cation is \(Ag^+\). Reaction: \(Ag^+ + e^- \to Ag\). Here \(n = 1\).
\[ \text{Moles of Ag} = \frac{3 F}{1} = 3 \text{ moles} \]
2. For \(CuSO_4\): The cation is \(Cu^{2+}\). Reaction: \(Cu^{2+} + 2e^- \to Cu\). Here \(n = 2\).
\[ \text{Moles of Cu} = \frac{3 F}{2} = 1.5 \text{ moles} \]
3. For \(AuCl_3\): The cation is \(Au^{3+}\). Reaction: \(Au^{3+} + 3e^- \to Au\). Here \(n = 3\).
\[ \text{Moles of Au} = \frac{3 F}{3} = 1 \text{ mole} \]
The molar ratio of Ag : Cu : Au is \(3 : 1.5 : 1\).
To convert this to the simplest whole number ratio, multiply all terms by \(2\):
\[ 3 \times 2 : 1.5 \times 2 : 1 \times 2 = 6 : 3 : 2 \]
Step 4: Final Answer:
The molar ratio of the deposited cations is \(6 : 3 : 2\).