Select the correct cell reaction of the cell `Ag(s)|Ag^+(aq)"||"cu^(2+)(aq)"|"cu(s)`:

Options

A. `2Ag(s)+cu(s)tocu^(+2)(aq)+2Ag^+(aq)`
B. `cu(s)+2Ag^+(aq)tocu^(2+)(aq)+2Ag(s)`
C. `2Ag(s)+cu^(2+)(aq)tocu(s)(aq)+2Ag^+(aq)`
D. `cu^(2+)(aq)+2Ag^+(aq)to2Ag(s)+cu(s)`

Answer

To determine the correct cell reaction for the electrochemical cell represented as `Ag(s)|Ag^+(aq)"||"Cu^(2+)(aq)"|"Cu(s)`, we need to analyze the oxidation and reduction processes occurring in the cell.

### Step-by-Step Solution:

1. **Identify the Anode and Cathode:**
- In the given cell representation, the left side (Ag|Ag⁺) represents the anode where oxidation occurs.
- The right side (Cu²⁺|Cu) represents the cathode where reduction occurs.

2. **Write the Oxidation Half-Reaction:**
- At the anode, silver (Ag) is oxidized to silver ions (Ag⁺). The half-reaction can be written as:
\[ \text{Ag(s)} \rightarrow \text{Ag}^+(aq) + e^- \]

3. **Write the Reduction Half-Reaction:**
- At the cathode, copper ions (Cu²⁺) are reduced to solid copper (Cu). The half-reaction can be written as:
\[ \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu(s)} \]

4. **Balance the Electrons:**
- To combine the half-reactions, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. The oxidation half-reaction produces 1 electron, while the reduction half-reaction consumes 2 electrons. Therefore, we multiply the oxidation half-reaction by 2:
\[ 2\text{Ag(s)} \rightarrow 2\text{Ag}^+(aq) + 2e^- \]

5. **Combine the Half-Reactions:**
- Now, we can add the balanced half-reactions:
\[ 2\text{Ag(s)} + \text{Cu}^{2+}(aq) \rightarrow 2\text{Ag}^+(aq) + \text{Cu(s)} \]

6. **Final Cell Reaction:**
- The overall cell reaction is:
\[ 2\text{Ag(s)} + \text{Cu}^{2+}(aq) \rightarrow 2\text{Ag}^+(aq) + \text{Cu(s)} \]

### Conclusion:
The correct cell reaction for the electrochemical cell `Ag(s)|Ag^+(aq)"||"Cu^(2+)(aq)"|"Cu(s)` is:
\[ 2\text{Ag(s)} + \text{Cu}^{2+}(aq) \rightarrow 2\text{Ag}^+(aq) + \text{Cu(s)} \]