Calculate `E_(cell)` of `Pt(s)|{:(cl_(2)(g)),(0.1atm):}|{:(Cl^(-)(aq)),(10^(-2)M):}||{:(Cr_(2)O_(7)^(2-),Cr^(+3)("in" H_(2)SO_(4))=0.05M),(0.01M,0.1M):}|`
given that `E_(Cr_(2)O_(7)^(2-)//Cr^(+3))^(0)=1.33V`
`E^(0)Cl^(-)|Cl_(2)=-1.36V`

Explanation

To calculate the cell potential \( E_{cell} \) for the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions
The half-reactions for the cell are:
1. **Anode (Oxidation)**:
\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]
...