The `EMF` of a galvanic cell `Pt|H_(2)(1 atm)|HCl(1M)|Cl_(2)(g)|Pt` is `1.29V`. Calculate the partial pressure of` Cl_(2)(g)`. `E^(c-)._(Cl_(2)|Cl^(c-))=1.36V`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the half-reactions
For the given galvanic cell, we identify the half-reactions occurring at the anode and cathode.

- **Anode (oxidation):**
\[
\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-
...