An aqueous solution containing `1M` each of `Au^(3+),Cu^(2+),Ag^(+),Li^(+)` is being electrolysed by using inert electrodes. The value of standard potentials are `:`
`E_(Ag^(+)//Ag)^(@)=0.80 V,E_(Cu^(+)//Cu)^(@)=0.34V` and `E_(Au^(+3)//Au)^(@)=1.50,E_(Li^(+)//Li)^(@)=-3.03V`
will increasing voltage, the sequence of deposition of metals on the cathode will be `:`

Options

A. `Li,Cu,Ag,Au`
B. `Cu,Ag,Au`
C. `Au,Ag,Cu`
D. `Au,Ag,Cu,Li`

Answer

To determine the sequence of deposition of metals on the cathode during the electrolysis of an aqueous solution containing `1M` each of `Au^(3+)`, `Cu^(2+)`, `Ag^(+)`, and `Li^(+)`, we will follow these steps:

### Step 1: Identify the Standard Reduction Potentials
We have the following standard reduction potentials:
- \( E^\circ (Au^{3+}/Au) = 1.50 \, V \)
- \( E^\circ (Ag^{+}/Ag) = 0.80 \, V \)
- \( E^\circ (Cu^{2+}/Cu) = 0.34 \, V \)
- \( E^\circ (Li^{+}/Li) = -3.03 \, V \)

### Step 2: Determine the Order of Deposition
The order of deposition of metals at the cathode is determined by the standard reduction potentials. The species with the highest (most positive) reduction potential will be reduced first.

1. **Gold (Au)**: \( E^\circ = 1.50 \, V \)
2. **Silver (Ag)**: \( E^\circ = 0.80 \, V \)
3. **Copper (Cu)**: \( E^\circ = 0.34 \, V \)
4. **Lithium (Li)**: \( E^\circ = -3.03 \, V \) (not reduced, will oxidize instead)

### Step 3: Sequence of Deposition
Based on the standard reduction potentials, the sequence of deposition on the cathode will be:
1. **Gold (Au)** will deposit first because it has the highest reduction potential.
2. **Silver (Ag)** will deposit next.
3. **Copper (Cu)** will deposit last.

### Step 4: Conclusion
The sequence of deposition of metals on the cathode will be:
- **Au → Ag → Cu**

Thus, the correct answer is: **Gold (Au), Silver (Ag), Copper (Cu)**.

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