`Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+)`
`E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(o)` is

Options

A. `-0.10V`
B. `+0.18V`
C. `-0.54V`
D. `0.54V`

Explanation

To solve the problem, we need to find the standard reduction potential \( E^\circ_{I_2} \) for the half-reaction involving iodine. We are given the overall cell reaction and the standard reduction potentials for the other half-reaction. Here are the steps:

### Step 1: Identify the half-reactions
The overall reaction is:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{I}^- \rightarrow \text{I}_2 + \text{Cr}^{3+} \]

From this, we can identify the half-reactions:
1. Reduction half-reaction (at the cathode):
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
This half-reaction has a standard reduction potential \( E^\circ = 1.33 \, \text{V} \).

2. Oxidation half-reaction (at the anode):
\[ 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \]
We need to find the standard reduction potential \( E^\circ_{I_2} \) for the half-reaction:
\[ \text{I}_2 + 2e^- \rightarrow 2\text{I}^- \]

### Step 2: Write the Nernst equation for the cell
The standard cell potential \( E^\circ_{cell} \) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

### Step 3: Substitute known values
We know:
- \( E^\circ_{cell} = 0.79 \, \text{V} \)
- \( E^\circ_{cathode} = E^\circ_{Cr_2O_7^{2-}} = 1.33 \, \text{V} \)
- \( E^\circ_{anode} = E^\circ_{I_2} \)

Substituting these values into the equation, we get:
\[ 0.79 = 1.33 - E^\circ_{I_2} \]

### Step 4: Solve for \( E^\circ_{I_2} \)
Rearranging the equation to isolate \( E^\circ_{I_2} \):
\[ E^\circ_{I_2} = 1.33 - 0.79 \]

Calculating this gives:
\[ E^\circ_{I_2} = 0.54 \, \text{V} \]

### Conclusion
The standard reduction potential for iodine is:
\[ E^\circ_{I_2} = 0.54 \, \text{V} \]