For a given reductant , ratio of volumes of 0.2 M `KMnO_(4)` and `1M K_(2)Cr_(2)O_(7)` in acidic medium will be

Answer

To solve the problem of finding the ratio of volumes of 0.2 M KMnO4 and 1 M K2Cr2O7 in acidic medium for a given reductant, we will follow these steps:

### Step 1: Understand the concept of milliequivalents
Milliequivalents (mEq) are used to express the reactive capacity of a substance in a chemical reaction. The milliequivalent of a substance can be calculated using the formula:
\[ \text{mEq} = \text{moles} \times \text{n factor} \]

### Step 2: Determine the n-factor for KMnO4
In acidic medium, KMnO4 is reduced from Mn(VII) to Mn(II). The change in oxidation state is:
- Initial oxidation state of Mn in KMnO4 = +7
- Final oxidation state of Mn in Mn2+ = +2

Thus, the change in oxidation state (n-factor) for KMnO4 is:
\[ n_{\text{KMnO4}} = 7 - 2 = 5 \]

### Step 3: Determine the n-factor for K2Cr2O7
In acidic medium, K2Cr2O7 is reduced from Cr(VI) to Cr(III). The change in oxidation state is:
- Initial oxidation state of Cr in K2Cr2O7 = +6
- Final oxidation state of Cr in Cr3+ = +3

Thus, the change in oxidation state (n-factor) for K2Cr2O7 is:
\[ n_{\text{K2Cr2O7}} = 6 - 3 = 3 \]

Since there are 2 moles of Cr in K2Cr2O7, we multiply the n-factor by 2:
\[ n_{\text{K2Cr2O7}} = 2 \times 3 = 6 \]

### Step 4: Set up the equation for milliequivalents
According to the law of equivalence:
\[ \text{mEq of KMnO4} = \text{mEq of K2Cr2O7} \]

Using the formula for milliequivalents:
\[ M_1 \cdot V_1 \cdot n_{\text{KMnO4}} = M_2 \cdot V_2 \cdot n_{\text{K2Cr2O7}} \]

Substituting the known values:
- \( M_1 = 0.2 \, \text{M} \) (for KMnO4)
- \( n_{\text{KMnO4}} = 5 \)
- \( M_2 = 1 \, \text{M} \) (for K2Cr2O7)
- \( n_{\text{K2Cr2O7}} = 6 \)

The equation becomes:
\[ 0.2 \cdot V_1 \cdot 5 = 1 \cdot V_2 \cdot 6 \]

### Step 5: Rearrange the equation to find the volume ratio
Rearranging the equation gives:
\[ V_1 \cdot 5 = \frac{6}{0.2} \cdot V_2 \]
\[ V_1 \cdot 5 = 30 \cdot V_2 \]
\[ \frac{V_1}{V_2} = \frac{30}{5} = 6 \]

### Step 6: Conclusion
Thus, the ratio of volumes \( V_1 : V_2 \) is:
\[ V_1 : V_2 = 6 : 1 \]