The reaction
`(1)/(2)H_(2)(g)+AgCl(s)hArrH^(+)(aq)+Cl^(-)(aq)+Ag(s)`
occurs in the galvanic cell

Options

A. `Ag|AgCl(s)|KCl("soln")|AgNO_(3)|Ag`
B. `Pt|H_(2)(g)|HCl|("soln")|AgNO_(3)("soln")|Ag`
C. `Pt|H_(2)(g)HCL("soln")|AgCl(s)|Ag`
D. `Pt|H_(2)(g)|KCl("soln")|AgCl(s)|Ag`

Answer

To solve the problem regarding the galvanic cell reaction, we will break down the steps involved in identifying the anode and cathode reactions, as well as the overall cell representation.

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
The given reaction is:
\[ \frac{1}{2} H_2(g) + AgCl(s) \rightleftharpoons H^+(aq) + Cl^-(aq) + Ag(s) \]
Here, we have hydrogen gas (H₂), silver chloride (AgCl), hydrogen ions (H⁺), chloride ions (Cl⁻), and solid silver (Ag).

2. **Determine Oxidation and Reduction**:
- **Oxidation**: This involves the loss of electrons. In the reaction, H₂ is oxidized to H⁺:
\[ \frac{1}{2} H_2 \rightarrow H^+ + e^- \]
- **Reduction**: This involves the gain of electrons. In the reaction, Ag⁺ (from AgCl) is reduced to Ag:
\[ Ag^+ + e^- \rightarrow Ag \]

3. **Assign Anode and Cathode**:
- The **anode** is where oxidation occurs. Therefore, the reaction at the anode is:
\[ \frac{1}{2} H_2 \rightarrow H^+ + e^- \]
- The **cathode** is where reduction occurs. Therefore, the reaction at the cathode is:
\[ Ag^+ + e^- \rightarrow Ag \]

4. **Write the Cell Representation**:
The galvanic cell can be represented as:
\[ \text{Anode} \, | \, \text{Electrolyte} \, | \, \text{Cathode} \]
In this case:
\[ \frac{1}{2} H_2(g) | H^+(aq), Cl^-(aq) | Ag(s), AgCl(s) \]

5. **Identify the Type of Cell**:
The overall cell reaction indicates a galvanic cell where chemical energy is converted to electrical energy. The cell representation shows the flow of electrons from the anode (H₂ oxidation) to the cathode (Ag⁺ reduction).

6. **Final Answer**:
The reaction occurs in a galvanic cell, and the correct representation is:
\[ \text{Option C} \]