What volume of 0.05 M `K_(2)Cr_(2)O_(7)` in acidic medium is needed for completel oxidation of 200 " mL of " 0.6 M `FeC_(2)O_(4)` solution?

Options

A. 1.2 mL
B. 1.2 L
C. 120 mL
D. 800 mL

Explanation

To solve the problem of determining the volume of 0.05 M \( K_2Cr_2O_7 \) needed for the complete oxidation of 200 mL of 0.6 M \( FeC_2O_4 \) solution, we can follow these steps:

### Step 1: Write the balanced chemical equations
The first step is to write the balanced chemical equations for the reactions involved.

1. The reaction of \( K_2Cr_2O_7 \) in acidic medium can be represented as:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
Here, each \( Cr \) goes from an oxidation state of +6 to +3, meaning it gains 6 electrons.

2. The oxidation of \( FeC_2O_4 \) can be represented as:
\[ Fe^{2+} \rightarrow Fe^{3+} + e^- \]
Here, each \( Fe \) goes from an oxidation state of +2 to +3, meaning it loses 1 electron.

### Step 2: Determine the n-factor for each reactant
- For \( K_2Cr_2O_7 \) (Chromate ion):
- The n-factor \( N_1 \) is 6 (as it gains 6 electrons).

- For \( FeC_2O_4 \) (Oxalate ion):
- The n-factor \( N_2 \) is 3 (as it loses 3 electrons).

### Step 3: Use the stoichiometric relationship
Using the stoichiometric relationship:
\[ N_1 \cdot V_1 = N_2 \cdot V_2 \]
Where:
- \( N_1 \) = n-factor of \( K_2Cr_2O_7 \) = 6
- \( V_1 \) = volume of \( K_2Cr_2O_7 \) (what we need to find)
- \( N_2 \) = n-factor of \( FeC_2O_4 \) = 3
- \( V_2 \) = volume of \( FeC_2O_4 \) = 200 mL = 0.2 L

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ 6 \cdot V_1 = 3 \cdot 0.2 \]
\[ 6 \cdot V_1 = 0.6 \]

### Step 5: Solve for \( V_1 \)
Now, solve for \( V_1 \):
\[ V_1 = \frac{0.6}{6} = 0.1 \text{ L} = 100 \text{ mL} \]

### Step 6: Convert to the required volume of \( K_2Cr_2O_7 \)
Since the molarity of \( K_2Cr_2O_7 \) is given as 0.05 M, we can use this to find the volume needed:
\[ \text{Volume of } K_2Cr_2O_7 = \frac{0.1 \text{ L}}{0.05 \text{ M}} = 2 \text{ L} = 2000 \text{ mL} \]

### Final Answer
The volume of 0.05 M \( K_2Cr_2O_7 \) needed for the complete oxidation of 200 mL of 0.6 M \( FeC_2O_4 \) solution is **2000 mL**.

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