How many ml of 0.3M `K_2 Cr_2 O_7`(acidic) is required for complete oxidation of 5 ml of 0.2 M `SnC_2 O_4` solution.

Options

A. 3.33 ml
B. 1.11 ml
C. 11 ml
D. 4.44 ml

Answer

To solve the problem of how many mL of 0.3 M K₂Cr₂O₇ (in acidic medium) is required for the complete oxidation of 5 mL of 0.2 M SnC₂O₄ solution, we can follow these steps:

### Step 1: Write the half-reactions
In acidic medium, the half-reaction for K₂Cr₂O₇ is:
\[ \text{Cr}_2\text{O}_7^{2-} + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} \]

For SnC₂O₄, the half-reaction is:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2 \text{e}^- \]

### Step 2: Determine the number of electrons transferred (n-factor)
From the half-reactions:
- The n-factor for K₂Cr₂O₇ (Cr₂O₇²⁻) is 6 (it accepts 6 electrons).
- The n-factor for SnC₂O₄ (C₂O₄²⁻) is 2 (it donates 2 electrons).

### Step 3: Use the equivalence concept
According to the equivalence concept:
\[ \text{milliequivalents of Cr}_2\text{O}_7^{2-} = \text{milliequivalents of C}_2\text{O}_4^{2-} \]

The formula for milliequivalents is:
\[ n_1 V_1 = n_2 V_2 \]
Where:
- \( n_1 \) = n-factor of K₂Cr₂O₇ = 6
- \( V_1 \) = volume of K₂Cr₂O₇ (mL) (unknown)
- \( n_2 \) = n-factor of SnC₂O₄ = 2
- \( V_2 \) = volume of SnC₂O₄ = 5 mL

### Step 4: Calculate the normality
Normality (N) can be calculated as:
\[ N = n \times M \]
Where:
- For K₂Cr₂O₇: \( N_1 = 6 \times 0.3 \, \text{M} = 1.8 \, \text{N} \)
- For SnC₂O₄: \( N_2 = 2 \times 0.2 \, \text{M} = 0.4 \, \text{N} \)

### Step 5: Set up the equation
Now substituting the values into the equivalence equation:
\[ 6 \times 0.3 \times V_1 = 2 \times 0.2 \times 5 \]

### Step 6: Solve for \( V_1 \)
Calculating the right side:
\[ 2 \times 0.2 \times 5 = 2 \]

Now substituting back:
\[ 1.8 \times V_1 = 2 \]
\[ V_1 = \frac{2}{1.8} \]
\[ V_1 \approx 1.11 \, \text{mL} \]

### Conclusion
Thus, the volume of 0.3 M K₂Cr₂O₇ required for the complete oxidation of 5 mL of 0.2 M SnC₂O₄ solution is approximately **1.11 mL**.

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