The cell reaction `2Ag^+(aq)+H_2(g)to2H^+(aq)+2Ag(s)`,best represented by :

Options

A. `Ag(s)|Ag^+"(aq)"||"H^+(aq)|H_2(g)|Pt(s)`
B. `Pt(s)|H_2(g)|H^+(aq)"||"Ag^+(aq)|Ag(s)`
C. `Ag(s)|Ag^+"(aq)"||"H_2(g)|H^+(aq)|Pt(s)`
D. `Ag^+"(aq)"|"Ag(s)||H_2(g)|H^+(aq)`

Answer

To represent the cell reaction \(2Ag^+(aq) + H_2(g) \rightarrow 2H^+(aq) + 2Ag(s)\), we can follow these steps:

### Step 1: Identify the Anode and Cathode
In a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode.

- **Oxidation**: In the given reaction, \(H_2(g)\) is oxidized to \(2H^+(aq)\).
- **Reduction**: \(Ag^+(aq)\) is reduced to \(Ag(s)\).

### Step 2: Assign the Half-Reactions
From the overall cell reaction, we can write the half-reactions:

- **Oxidation half-reaction**:
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]
- **Reduction half-reaction**:
\[ 2Ag^+(aq) + 2e^- \rightarrow 2Ag(s) \]

### Step 3: Determine the Electrode Materials
- For the oxidation half-reaction involving \(H_2\), we will use a **platinum electrode** as it is inert and can facilitate the reaction.
- For the reduction half-reaction involving \(Ag^+\), we will use a **silver electrode**.

### Step 4: Write the Cell Representation
The cell representation follows the format:
\[ \text{Anode} | \text{Anode Solution} || \text{Cathode Solution} | \text{Cathode} \]

1. **Anode**: \(Pt(s) | H_2(g) | H^+(aq)\)
2. **Cathode**: \(Ag^+(aq) | Ag(s)\)

Combining these, we get:
\[ Pt(s) | H_2(g) | H^+(aq) || Ag^+(aq) | Ag(s) \]

### Step 5: Finalize the Cell Representation
The complete cell representation is:
\[ Pt(s) | H_2(g) | H^+(aq) || Ag^+(aq) | Ag(s) \]

### Conclusion
The best representation of the cell reaction \(2Ag^+(aq) + H_2(g) \rightarrow 2H^+(aq) + 2Ag(s)\) is:
\[ Pt(s) | H_2(g) | H^+(aq) || Ag^+(aq) | Ag(s) \]