The standard reduction potential for the half-cell reaction,`Cl_2 + 2e^(-) to 2Cl^(-)` will be `(Pt^(2+)+2Cl^(-)to Pt + Cl_2 , E_"cell"^@=-0.15 V , Pt^(2+) + 2e^(-) to Pt, E^@=1.20 V)`

Options

A. `-1.35 V`
B. `+1.35 V`
C. `1.05 V`
D. `+1.05 V`

Explanation

To solve the problem, we need to determine the standard reduction potential for the half-cell reaction:

\[ \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \]

We are given two half-cell reactions with their standard reduction potentials:

1. \( \text{Pt}^{2+} + 2\text{Cl}^- \rightarrow \text{Pt} + \text{Cl}_2 \), with \( E^\circ_{\text{cell}} = -0.15 \, \text{V} \)
2. \( \text{Pt}^{2+} + 2e^- \rightarrow \text{Pt} \), with \( E^\circ = 1.20 \, \text{V} \)

### Step 1: Reverse the first reaction
To find the standard reduction potential for the reaction involving chlorine, we first reverse the first reaction:

\[ \text{Pt} + \text{Cl}_2 \rightarrow \text{Pt}^{2+} + 2\text{Cl}^- \]

When we reverse a reaction, the sign of the standard potential changes. Therefore, the new potential becomes:

\[ E^\circ = +0.15 \, \text{V} \]

### Step 2: Use the formula for cell potential
The standard cell potential \( E^\circ_{\text{cell}} \) is given by the equation:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

In our case:
- The cathode reaction is the reduction of \( \text{Cl}_2 \) to \( \text{Cl}^- \) (which we are trying to find).
- The anode reaction is the reduction of \( \text{Pt}^{2+} \) to \( \text{Pt} \).

### Step 3: Substitute known values into the equation
From the given data:
- \( E^\circ_{\text{anode}} = 1.20 \, \text{V} \) (for the Pt reaction)
- \( E^\circ_{\text{cell}} = -0.15 \, \text{V} \) (for the overall cell reaction)

Substituting these values into the equation gives:

\[ -0.15 = E^\circ_{\text{cathode}} - 1.20 \]

### Step 4: Solve for \( E^\circ_{\text{cathode}} \)
Rearranging the equation to solve for \( E^\circ_{\text{cathode}} \):

\[ E^\circ_{\text{cathode}} = -0.15 + 1.20 \]
\[ E^\circ_{\text{cathode}} = 1.05 \, \text{V} \]

### Step 5: Conclusion
Thus, the standard reduction potential for the half-cell reaction \( \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \) is:

\[ E^\circ = 1.05 \, \text{V} \]