Select the correct statements if `9.65 A` current is passed for 1 hour through the cell `:`
`Ag|Ag^(o+)(1M)||Cu^(2+)(1M)|Cu`.

Options

A. `Ag` will oxidize to `Ag^(o+)` and new `[Ag^(o+)]=1.36M`.
B. `Ag^(o+) ` will reduce to `Ag` and new `[Ag^(o+)]=0.64M`
C. `Cu^(2+)` will reduce to `Cu` and new `[Cu^(2+)]=0.82M`.
D. `Cu` will oxidize to `Cu^(2+)` and new `[Cu^(2+)]=0.82M`.

Explanation

To solve the problem, we need to analyze the electrochemical cell and the changes that occur when a current is passed through it. The cell in question is:

\[ \text{Ag} | \text{Ag}^+ (1M) || \text{Cu}^{2+} (1M) | \text{Cu} \]

### Step-by-Step Solution:

1. **Identify the Reactions**:
- In the anode compartment (left side), silver (Ag) is oxidized to silver ions (Ag⁺):
\[ \text{Ag} \rightarrow \text{Ag}^+ + e^- \]
- In the cathode compartment (right side), copper ions (Cu²⁺) are reduced to copper (Cu):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

2. **Calculate Total Charge Passed**:
- Current (I) = 9.65 A
- Time (t) = 1 hour = 3600 seconds
- Total charge (Q) can be calculated using the formula:
\[ Q = I \times t = 9.65 \, \text{A} \times 3600 \, \text{s} = 34740 \, \text{C} \]

3. **Convert Charge to Faraday**:
- 1 Faraday (F) = 96500 C
- The total charge in Faraday is:
\[ \text{Charge in Faraday} = \frac{Q}{F} = \frac{34740 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.36 \, \text{mol} \]

4. **Determine Moles of Ag⁺ Produced**:
- From the oxidation of silver, 1 Faraday produces 1 mole of Ag⁺.
- Therefore, 0.36 Faraday will produce:
\[ \text{Moles of Ag}^+ = 0.36 \, \text{mol} \]

5. **Calculate New Concentration of Ag⁺**:
- Initial concentration of Ag⁺ = 1 M
- Volume of the solution is assumed to be 1 L for simplicity.
- New concentration of Ag⁺:
\[ \text{New concentration of Ag}^+ = 1 + 0.36 = 1.36 \, \text{M} \]

6. **Determine Moles of Cu²⁺ Reduced**:
- For copper, 2 Faraday are required to produce 1 mole of Cu.
- Therefore, 0.36 Faraday will produce:
\[ \text{Moles of Cu}^{2+} \text{ reduced} = \frac{0.36}{2} = 0.18 \, \text{mol} \]

7. **Calculate New Concentration of Cu²⁺**:
- Initial concentration of Cu²⁺ = 1 M
- New concentration of Cu²⁺:
\[ \text{New concentration of Cu}^{2+} = 1 - 0.18 = 0.82 \, \text{M} \]

### Final Statements:
- **Correct Statements**:
- Silver will oxidize to Ag⁺ and new Ag⁺ concentration is 1.36 M.
- Cu²⁺ will reduce to Cu and new Cu²⁺ concentration is 0.82 M.

### Summary of Correct Options:
- Option A: Ag will oxidize to Ag⁺ and new Ag⁺ concentration is 1.36 M. (Correct)
- Option C: Cu²⁺ will reduce to Cu and new Cu²⁺ concentration is 0.82 M. (Correct)