Direct titration of `I_(2)` with a reducing agent is called iodimetry. If `I_(2)` is leberated by the oxidation of `I_(ɵ)` ion by a strong oxidising agent in neutral or acidic medium, the liberated `I_(2)` is then titrated with reducing agent. Iodometry is used to estimate the strngth of the oxidising agent. For example, in the estimation of `Cu^(2+)` with `S_(2)O_(3)^(2-)`
`Cu^(2+)+I^(ɵ)toCuI_(2)+I_(2)` (iodometry)
`I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(ɵ)` (iodimetry)
Strach is used as an indicator at the end point, which forms bluecoloured complex with `I_(3)^(ɵ)` Disappearance of blue colourindicates the end point whe free `I_(2)` in not present.
Q. In the reaction
`2CuSO_(4)+4KItoCu_(2)I_(2)+2K_(2)SO_(4)+I_(2)`
The equivalent weight of `CuSO_(4)` is
`(Mw=159.5g mol^(-1))`

Options

A. `Mw`
B. `(Mw)/(2)`
C. `(Mw)/(4)`
D. `(Mw)/(8)`

Explanation

To find the equivalent weight of \( \text{CuSO}_4 \) in the given reaction, we can follow these steps:

### Step 1: Write the balanced reaction
The reaction given is:
\[ 2 \text{CuSO}_4 + 4 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + 2 \text{K}_2\text{SO}_4 + \text{I}_2 \]

### Step 2: Identify the oxidation states
In this reaction, copper (\( \text{Cu} \)) is in the +2 oxidation state in \( \text{CuSO}_4 \) and remains in the +2 oxidation state in \( \text{Cu}_2\text{I}_2 \). Therefore, there is no change in the oxidation state of copper.

### Step 3: Determine the number of electrons transferred
Since \( \text{Cu}^{2+} \) does not change its oxidation state, we need to consider the iodine (\( \text{I} \)) in the reaction. The iodide ion (\( \text{I}^- \)) is oxidized to \( \text{I}_2 \). Each \( \text{I}^- \) loses one electron to form \( \text{I}_2 \).

In the reaction:
- 4 \( \text{I}^- \) ions produce 2 \( \text{I}_2 \), which means a total of 4 electrons are transferred (2 electrons for each \( \text{I}_2 \)).

### Step 4: Calculate the n-factor
The n-factor is defined as the number of moles of electrons transferred per mole of the substance. In this case, since 2 moles of \( \text{CuSO}_4 \) are involved and 4 electrons are transferred:
\[ \text{n-factor} = \frac{\text{Total electrons transferred}}{\text{Moles of } \text{CuSO}_4} = \frac{4}{2} = 2 \]

### Step 5: Calculate the equivalent weight
The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
The molecular weight of \( \text{CuSO}_4 \) is given as \( 159.5 \, \text{g/mol} \). Thus, we can substitute the values:
\[ \text{Equivalent weight of } \text{CuSO}_4 = \frac{159.5 \, \text{g/mol}}{2} = 79.75 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of \( \text{CuSO}_4 \) is \( 79.75 \, \text{g/equiv} \).

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