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 will follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation states**:
- In `FeC2O4`, iron (`Fe`) is in the +2 oxidation state as `FeC2O4` can be broken down into `Fe^(2+)` and `C2O4^(2-)`.
- The `C2O4^(2-)` (oxalate ion) will be oxidized to carbon dioxide (`CO2`), where carbon is in the +4 oxidation state.

2. **Determine the oxidation half-reaction**:
- The oxidation of `Fe^(2+)` to `Fe^(3+)` involves the loss of 1 electron:
\[ Fe^{2+} \rightarrow Fe^{3+} + e^- \]
- The oxidation of `C2O4^(2-)` to `CO2` involves the loss of 2 electrons for each `C2O4^(2-)`:
\[ C2O4^{2-} \rightarrow 2CO2 + 2e^- \]

3. **Balance the oxidation half-reactions**:
- For 1 mole of `FeC2O4`, we have:
- 1 mole of `Fe^(2+)` will produce 1 mole of `Fe^(3+)` and release 1 electron.
- 1 mole of `C2O4^(2-)` will produce 2 moles of `CO2` and release 2 electrons.

4. **Combine the half-reactions**:
- The total electrons released from 1 mole of `FeC2O4` is 3 electrons (1 from `Fe` and 2 from `C2O4`).
- The reduction half-reaction for `Cr2O7^(2-)` in acidic medium is:
\[ Cr2O7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H2O \]
- This indicates that 1 mole of `Cr2O7^(2-)` can accept 6 electrons.

5. **Calculate the moles of `Cr2O7^(2-)` needed**:
- Since 3 moles of electrons are produced from the oxidation of 1 mole of `FeC2O4`, we need to find how many moles of `Cr2O7^(2-)` are required to accept these 3 electrons:
\[ \text{Moles of } Cr2O7^{2-} = \frac{3 \text{ electrons}}{6 \text{ electrons per } Cr2O7^{2-}} = \frac{1}{2} \text{ moles} \]

### Conclusion:
Thus, the value of `x` (the moles of `Cr2O7^(2-)` required) is **0.5 moles**.