Volume of 0.1 M `K_2Cr_2O_7` required to oxidize 35 ml of 0.5 M `FeSO_4` solution is

Options

A. 29.2 ml
B. 145 ml
C. 175 ml
D. 58.9 ml

Answer

To solve the problem of finding the volume of 0.1 M \( K_2Cr_2O_7 \) required to oxidize 35 mL of 0.5 M \( FeSO_4 \), we can follow these steps:

### Step 1: Write the half-reactions
1. **Oxidation half-reaction for \( FeSO_4 \)**:
\[ Fe^{2+} \rightarrow Fe^{3+} + e^- \]
Here, 1 mole of \( Fe^{2+} \) loses 1 electron.

2. **Reduction half-reaction for \( K_2Cr_2O_7 \)**:
\[ Cr_2O_7^{2-} + 14 H^+ + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2O \]
Here, 1 mole of \( Cr_2O_7^{2-} \) gains 6 electrons.

### Step 2: Determine the moles of \( FeSO_4 \)
To find the moles of \( FeSO_4 \) in 35 mL of 0.5 M solution:
\[ \text{Moles of } FeSO_4 = \text{Molarity} \times \text{Volume (in L)} = 0.5 \, \text{mol/L} \times 0.035 \, \text{L} = 0.0175 \, \text{mol} \]

### Step 3: Calculate the moles of electrons transferred
Since each mole of \( Fe^{2+} \) loses 1 electron, the total moles of electrons transferred from \( FeSO_4 \) is equal to the moles of \( FeSO_4 \):
\[ \text{Moles of electrons} = 0.0175 \, \text{mol} \]

### Step 4: Relate moles of \( K_2Cr_2O_7 \) to moles of electrons
From the reduction half-reaction, 1 mole of \( K_2Cr_2O_7 \) accepts 6 moles of electrons. Therefore, the moles of \( K_2Cr_2O_7 \) required can be calculated as:
\[ \text{Moles of } K_2Cr_2O_7 = \frac{\text{Moles of electrons}}{6} = \frac{0.0175}{6} \approx 0.0029167 \, \text{mol} \]

### Step 5: Calculate the volume of \( K_2Cr_2O_7 \) solution required
Using the molarity of \( K_2Cr_2O_7 \) (0.1 M), we can find the volume:
\[ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.0029167 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.029167 \, \text{L} = 29.167 \, \text{mL} \]

### Step 6: Round the answer
Rounding to the appropriate significant figures, we find:
\[ \text{Volume of } K_2Cr_2O_7 \approx 29.2 \, \text{mL} \]

### Final Answer
The volume of 0.1 M \( K_2Cr_2O_7 \) required to oxidize 35 mL of 0.5 M \( FeSO_4 \) is approximately **29.2 mL**.

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