The chemical reaction
`2AgCl_("(fused)")+H_(2(g))rarr2HCl_((aq))+2Ag_((s))`
taking place in a galvanic cell is represented by the notation

Options

A. `Pt_((s))|H_(2(g)).1"bar"|1MKCl_((aq))|AgCl_((s))|Ag_((s))`
B. `Pt_((s))|H_(2(g)).1"bar"|"1M HCl"_((aq))|"1M Ag^(+)(aq)|Ag_((s))`
C. `Pt_((s))|H_(2(g))."1 bar"|"1M HCl"_((aq))|AgCl_((fused))|Ag`
D. `Pt_((s))|H_(2(g))."1 bar"|"1M HCl"_((aq))|Ag_((s))|AgCl_((s))`

Answer

To represent the given chemical reaction in a galvanic cell notation, we need to follow a systematic approach. Here’s the step-by-step solution:

### Step 1: Identify the Oxidation and Reduction Half-Reactions
In the reaction:
\[ 2AgCl_{(fused)} + H_{2(g)} \rightarrow 2HCl_{(aq)} + 2Ag_{(s)} \]

- **Oxidation**: \( H_{2(g)} \rightarrow 2H^+ + 2e^- \) (Hydrogen gas is oxidized to hydrogen ions)
- **Reduction**: \( 2Ag^+ + 2e^- \rightarrow 2Ag_{(s)} \) (Silver ions from AgCl are reduced to metallic silver)

### Step 2: Write the Anodic and Cathodic Reactions
- **Anodic Reaction** (oxidation occurs here):
\[ H_{2(g)} \rightarrow 2H^+ + 2e^- \]

- **Cathodic Reaction** (reduction occurs here):
\[ 2Ag^+ + 2e^- \rightarrow 2Ag_{(s)} \]

### Step 3: Identify the Components for Cell Notation
- **Anode**: The oxidation half-reaction involves hydrogen gas (H2) and H+ ions from HCl.
- **Cathode**: The reduction half-reaction involves silver ions (Ag+) from AgCl and solid silver (Ag).

### Step 4: Construct the Cell Notation
The general format for cell notation is:
\[ \text{Anode} | \text{Anodic Solution} || \text{Cathodic Solution} | \text{Cathode} \]

For this reaction:
- Anode: \( H_{2(g)} | HCl_{(aq)} \)
- Cathode: \( AgCl_{(fused)} | Ag_{(s)} \)

Since both half-reactions occur in the same compartment, we can represent it as:
\[ H_{2(g)} | HCl_{(aq)} | AgCl_{(fused)} | Ag_{(s)} \]

### Final Cell Notation
Thus, the cell notation for the given reaction is:
\[ H_{2(g)} | HCl_{(aq)} | AgCl_{(fused)} | Ag_{(s)} \]