In acidic medium, equivalent weight of `K_(2)Cr_(2)O_(7)` (molecular weight `= M`) is

Options

A. M/3
B. M/4
C. M/6
D. M/2

Answer

To find the equivalent weight of potassium dichromate (K₂Cr₂O₇) in acidic medium, we can follow these steps:

### Step 1: Understand Equivalent Weight
The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{N Factor}} \]
where the N factor is the number of moles of electrons exchanged in the reaction.

### Step 2: Determine the Molecular Weight
Given that the molecular weight of potassium dichromate is \( M \).

### Step 3: Write the Reaction in Acidic Medium
In acidic medium, potassium dichromate reacts with hydrogen ions (H⁺) and is reduced. The reaction can be simplified as follows:
\[ K_2Cr_2O_7 + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
This shows that each mole of potassium dichromate accepts 6 moles of electrons.

### Step 4: Calculate the N Factor
From the reaction, we can see that the change in oxidation state of chromium is from +6 in K₂Cr₂O₇ to +3 in Cr³⁺. Since there are 2 chromium atoms in potassium dichromate, the total change in oxidation state is:
\[ 2 \times (6 - 3) = 6 \]
Thus, the N factor for potassium dichromate in acidic medium is 6.

### Step 5: Calculate the Equivalent Weight
Now, we can substitute the values into the equivalent weight formula:
\[ \text{Equivalent Weight} = \frac{M}{6} \]

### Final Answer
The equivalent weight of potassium dichromate (K₂Cr₂O₇) in acidic medium is:
\[ \frac{M}{6} \]

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