A solution contains `Cr^(+2),Cr^(+3)` and `I^(-)` ions, This solution was treated with iodine at `35^(@)C.E^(@)` for `cr^(+3)//Cr^(+2)` is `-0.41V` and `E^(@)` for `I_(2)//2I^(-)=0.536V`, The favourable redox reaction is :-

Options

A. `Cr^(+2)` will be oxidised to `Cr^(+3)`
B. `Cr^(+3)` will reduced to `Cr^(+2)`
C. There will be no redox reaction
D. `I^(-)` will oxidised to `I_(2)`

Explanation

To solve the problem, we need to determine the favorable redox reaction in a solution containing Cr²⁺, Cr³⁺, and I⁻ ions when treated with iodine at 35°C. We have the standard reduction potentials for the relevant half-reactions.

### Step-by-Step Solution:

1. **Identify the half-reactions and their standard reduction potentials:**
- For the reduction of chromium:
\[ \text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \quad E^\circ = -0.41 \, \text{V} \]
- For the reduction of iodine:
\[ \text{I}_2 + 2e^- \rightarrow 2\text{I}^- \quad E^\circ = 0.536 \, \text{V} \]

2. **Determine the oxidation and reduction reactions:**
- In this scenario, Cr³⁺ will be reduced to Cr²⁺, and I₂ will be reduced to I⁻. However, since Cr²⁺ can also be oxidized to Cr³⁺, we need to consider the oxidation reaction:
\[ \text{Cr}^{2+} \rightarrow \text{Cr}^{3+} + e^- \quad E^\circ = +0.41 \, \text{V} \, (\text{reverse of the reduction}) \]
- The reduction reaction for iodine remains the same.

3. **Calculate the standard cell potential (E°cell):**
- The cell potential is calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Here, the cathode reaction (reduction) is I₂ to I⁻, and the anode reaction (oxidation) is Cr²⁺ to Cr³⁺.
- Plugging in the values:
\[ E^\circ_{\text{cell}} = 0.536 \, \text{V} - (-0.41 \, \text{V}) = 0.536 \, \text{V} + 0.41 \, \text{V} = 0.946 \, \text{V} \]

4. **Interpret the result:**
- Since \(E^\circ_{\text{cell}} > 0\), the reaction is spontaneous. This indicates that the reaction will proceed in the forward direction.

5. **Identify the favorable redox reaction:**
- The favorable redox reaction is:
\[ \text{Cr}^{2+} \rightarrow \text{Cr}^{3+} + e^- \quad \text{(oxidation)} \]
\[ \text{I}_2 + 2e^- \rightarrow 2\text{I}^- \quad \text{(reduction)} \]

### Conclusion:
The favorable redox reaction is that Cr²⁺ is oxidized to Cr³⁺ while I₂ is reduced to I⁻. Therefore, the correct option is **option number 1**.