1 mol of `FeC_(2)O_(4)` is oxidized by x mol of `Cr_(2)O_(7)^(2-)` in acidic medium, x is :

Options

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

Answer

To solve the problem of how many moles of `Cr2O7^(2-)` are required to oxidize 1 mole of `FeC2O4` in acidic medium, we can follow these steps:

### Step 1: Write the half-reaction for the reduction of dichromate ions.
In acidic medium, the reduction half-reaction for `Cr2O7^(2-)` is:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

### Step 2: Write the half-reaction for the oxidation of `FeC2O4`.
The oxidation half-reaction for `FeC2O4` (ferrous oxalate) is:
\[ FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^- \]

### Step 3: Balance the number of electrons transferred.
To balance the electrons transferred in both half-reactions, we need to multiply the oxidation half-reaction by 2 so that both half-reactions involve 6 electrons:
\[ 2(FeC_2O_4 \rightarrow Fe^{3+} + 2CO_2 + 3e^-) \]
This gives us:
\[ 2FeC_2O_4 \rightarrow 2Fe^{3+} + 4CO_2 + 6e^- \]

### Step 4: Combine the balanced half-reactions.
Now we can add the two half-reactions:
1. Reduction half-reaction:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
2. Oxidation half-reaction (multiplied by 2):
\[ 2FeC_2O_4 \rightarrow 2Fe^{3+} + 4CO_2 + 6e^- \]

Adding these gives:
\[ Cr_2O_7^{2-} + 2FeC_2O_4 + 14H^+ \rightarrow 2Cr^{3+} + 2Fe^{3+} + 4CO_2 + 7H_2O \]

### Step 5: Determine the stoichiometry.
From the balanced equation, we can see that:
- 1 mole of `Cr2O7^(2-)` reacts with 2 moles of `FeC2O4`.
- Therefore, 1 mole of `FeC2O4` will react with \( \frac{1}{2} \) mole of `Cr2O7^(2-)`.

### Conclusion:
Thus, the value of \( x \) (the moles of `Cr2O7^(2-)` needed to oxidize 1 mole of `FeC2O4`) is:
\[ x = 0.5 \text{ moles} \]