An aqueous solution containing 1 M each of `Au^(3+),Cu^(2+),Ag^+, Li^(+)` is being electrolysed by using inert electrodes . The value of standard potentials are `E_(Ag^+//Ag)^@=0.80V,E_(cu^+//Cu)^@=0.34V and E_(Au^(3+)//Au)^@ = 1.50 V, E_(Li^+//Li)^@=-3.03V` with 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 the given solution, we need to analyze the standard reduction potentials (SRP) of the ions present in the solution. The ions in the solution are Au³⁺, Cu²⁺, Ag⁺, and Li⁺, and their respective standard reduction potentials are as follows:

- E°(Ag⁺/Ag) = 0.80 V
- E°(Cu²⁺/Cu) = 0.34 V
- E°(Au³⁺/Au) = 1.50 V
- E°(Li⁺/Li) = -3.03 V

### Step-by-Step Solution:

1. **Identify the Standard Reduction Potentials**:
We start by listing the standard reduction potentials for each ion:
- Au³⁺ + 3e⁻ → Au (E° = 1.50 V)
- Ag⁺ + e⁻ → Ag (E° = 0.80 V)
- Cu²⁺ + 2e⁻ → Cu (E° = 0.34 V)
- Li⁺ + e⁻ → Li (E° = -3.03 V)

2. **Determine the Order of Deposition**:
The metal with the highest standard reduction potential will be deposited first because it is the most favorable reduction reaction. Therefore, we will arrange the metals in order of their standard reduction potentials from highest to lowest:
- Au³⁺ (1.50 V)
- Ag⁺ (0.80 V)
- Cu²⁺ (0.34 V)
- Li⁺ (-3.03 V)

3. **Sequence of Deposition**:
Based on the standard reduction potentials, the sequence of deposition on the cathode will be:
- First: Au (from Au³⁺)
- Second: Ag (from Ag⁺)
- Third: Cu (from Cu²⁺)
- Last: Li (from Li⁺)

### Final Answer:
The sequence of deposition of metals on the cathode will be: **Au, Ag, Cu, Li**.