The `EMF` of the following cellis `1.05V` at `25^(@)C:`
`Pt,H_(2)(g)(1.0 atm)|NaOH(0.1m),NaCl(0.1M)|AgCl(s),Ag(s)`
Write the cell reaction.

Explanation

To write the cell reaction for the given electrochemical cell, we need to identify the half-reactions occurring at the anode and cathode.

### Step-by-Step Solution:

1. **Identify the Anode and Cathode:**
- The left side of the cell (Pt, H₂(g)(1.0 atm)) is the anode where oxidation occurs.
- The right side of the cell (AgCl(s), Ag(s)) is the cathode where reduction occurs.

2. **Write the Half-Reaction at the Anode:**
- At the anode, hydrogen gas (H₂) is oxidized to hydrogen ions (H⁺) and releases electrons (e⁻).
- The half-reaction can be written as:
\[ \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \]

3. **Write the Half-Reaction at the Cathode:**
- At the cathode, silver chloride (AgCl) is reduced to silver (Ag) and chloride ions (Cl⁻) by gaining electrons.
- The half-reaction can be written as:
\[ \text{AgCl}(s) + e^- \rightarrow \text{Ag}(s) + \text{Cl}^-(aq) \]

4. **Balance the Electrons:**
- The anode reaction produces 2 electrons while the cathode reaction consumes 1 electron. To balance the electrons, we can multiply the cathode reaction by 2:
\[ 2\text{AgCl}(s) + 2e^- \rightarrow 2\text{Ag}(s) + 2\text{Cl}^-(aq) \]

5. **Combine the Half-Reactions:**
- Now, we can add the balanced half-reactions together:
\[ \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \quad (anode) \]
\[ 2\text{AgCl}(s) + 2e^- \rightarrow 2\text{Ag}(s) + 2\text{Cl}^-(aq) \quad (cathode) \]
- Adding these gives the overall cell reaction:
\[ \text{H}_2(g) + 2\text{AgCl}(s) \rightarrow 2\text{Ag}(s) + 2\text{Cl}^-(aq) + 2\text{H}^+(aq) \]

### Final Cell Reaction:
\[ \text{H}_2(g) + 2\text{AgCl}(s) \rightarrow 2\text{Ag}(s) + 2\text{Cl}^-(aq) + 2\text{H}^+(aq) \]