`Cr_(2)O_(7)^(-2)+I^(-)+H^(+)rarrCr^(+3)+I_(2)+H_(2)O`
The equivalent weight of the reductant in the above equation is :- (At. wt. of Cr = 52, I = 127)

Options

A. 26
B. 127
C. 63.5
D. 10.4

Explanation

To find the equivalent weight of the reductant in the given redox reaction, we can follow these steps:

### Step 1: Identify the Reductant
In the reaction:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{I}^- + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{I}_2 + \text{H}_2\text{O} \]

We need to identify which species is being oxidized and which is being reduced.

- Chromium (Cr) is reduced from +6 in \(\text{Cr}_2\text{O}_7^{2-}\) to +3 in \(\text{Cr}^{3+}\).
- Iodide ion (\(I^-\)) is oxidized to iodine (\(I_2\)), where the oxidation state changes from -1 to 0.

Thus, the reductant in this reaction is \(I^-\).

### Step 2: Determine the Change in Oxidation State
Next, we calculate the change in oxidation number for the iodide ion:
- Initial oxidation state of \(I^-\) = -1
- Final oxidation state in \(I_2\) = 0

The change in oxidation state (\(\Delta\)) for iodine is:
\[ \Delta = 0 - (-1) = 1 \]

### Step 3: Calculate the Molar Mass of the Reductant
The molar mass of \(I_2\) (since two iodide ions combine to form one molecule of iodine) is calculated as follows:
- Atomic weight of Iodine (I) = 127 g/mol
- Therefore, the molar mass of \(I_2\) = \(2 \times 127 = 254\) g/mol.

### Step 4: Calculate the Equivalent Weight
The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\Delta \text{Oxidation Number}} \]

Substituting the values we have:
\[ \text{Equivalent Weight of } I^- = \frac{254 \text{ g/mol}}{1} = 254 \text{ g/equiv} \]

### Conclusion
Thus, the equivalent weight of the reductant \(I^-\) in the reaction is **254 g/equiv**.