x moles of potassium dichromate oxidise 1 mole of ferrous oxalate, in acidic medium. Hence x is

Options

A. 3
B. 1.5
C. 0.5
D. `1.0`

Explanation

To solve the problem of how many moles of potassium dichromate (K₂Cr₂O₇) are required to oxidize 1 mole of ferrous oxalate (FeC₂O₄) in acidic medium, we need to follow these steps:

### Step 1: Write the half-reactions
In acidic medium, potassium dichromate (K₂Cr₂O₇) acts as an oxidizing agent and ferrous oxalate (FeC₂O₄) is oxidized to ferric oxalate (Fe₂(C₂O₄)₃).

The half-reaction for the reduction of dichromate ions (Cr₂O₇²⁻) can be written as:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

The half-reaction for the oxidation of ferrous ions (Fe²⁺) can be written as:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \]

### Step 2: Balance the electrons
From the half-reaction of Cr₂O₇²⁻, we see that 6 electrons are required to reduce 1 mole of Cr₂O₇²⁻.

For the oxidation of ferrous ions, 1 mole of Fe²⁺ produces 1 mole of Fe³⁺ and releases 1 electron.

### Step 3: Determine the stoichiometry
To find out how many moles of Cr₂O₇²⁻ are needed to oxidize 1 mole of Fe²⁺, we can set up the following relationship based on the number of electrons:

- 6 moles of electrons are produced from the oxidation of 6 moles of Fe²⁺.
- Therefore, to oxidize 1 mole of Fe²⁺, we need:
\[ \frac{1}{6} \text{ moles of } \text{Cr}_2\text{O}_7^{2-} \]

### Step 4: Relate to the original question
Since we are asked how many moles of K₂Cr₂O₇ (which provides Cr₂O₇²⁻) are needed to oxidize 1 mole of ferrous oxalate, we can conclude that:
\[ x = \frac{1}{6} \]

### Final Answer
Thus, the value of \( x \) is \( \frac{1}{6} \).

---