Three moles of electrons are passed through three solutions in succession containing `AgNO_3`, `CuSO_4` and `AuCl_3` respectively the molar ratio of amounts of cations reduced at cathode will be

Options

A. `1:2:3`
B. `3:2:1`
C. `2:1:3`
D. `6:3:2`

Answer

To solve the problem, we need to determine the molar ratio of the amounts of cations reduced at the cathode when 3 moles of electrons are passed through solutions of AgNO3, CuSO4, and AuCl3.

### Step-by-Step Solution:

1. **Identify the Reduction Reactions:**
- For AgNO3: The reduction reaction is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (1 \text{ mole of } e^- \text{ reduces } 1 \text{ mole of } Ag^+) \]
- For CuSO4: The reduction reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (2 \text{ moles of } e^- \text{ reduce } 1 \text{ mole of } Cu^{2+}) \]
- For AuCl3: The reduction reaction is:
\[ \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \quad (3 \text{ moles of } e^- \text{ reduce } 1 \text{ mole of } Au^{3+}) \]

2. **Calculate the Moles of Cations Reduced:**
- **For AgNO3:**
- 1 mole of electrons reduces 1 mole of Ag^+.
- Therefore, 3 moles of electrons will reduce:
\[ 3 \text{ moles of } e^- \rightarrow 3 \text{ moles of } Ag^+ \]

- **For CuSO4:**
- 2 moles of electrons reduce 1 mole of Cu^2+.
- Therefore, 3 moles of electrons will reduce:
\[ \frac{3 \text{ moles of } e^-}{2} = 1.5 \text{ moles of } Cu^{2+} \]

- **For AuCl3:**
- 3 moles of electrons reduce 1 mole of Au^3+.
- Therefore, 3 moles of electrons will reduce:
\[ \frac{3 \text{ moles of } e^-}{3} = 1 \text{ mole of } Au^{3+} \]

3. **Establish the Molar Ratio:**
- Now we have the amounts reduced:
- Ag: 3 moles
- Cu: 1.5 moles
- Au: 1 mole
- The molar ratio of Ag : Cu : Au is:
\[ 3 : 1.5 : 1 \]
- To simplify this ratio, we can multiply each term by 2 to eliminate the decimal:
\[ 6 : 3 : 2 \]

4. **Final Molar Ratio:**
- The final molar ratio of the amounts of cations reduced at the cathode is:
\[ 6 : 3 : 2 \]