A given amount of `Fe^(2+)` is oxidised by x mol of `MnO_4^–` in acidic medium. Calculate mol of `Cr_2O_7^(2–)` required same amount of `Fe^(2+)` in acidic medium

Answer

To solve the problem, we need to determine how many moles of `Cr_2O_7^(2-)` are required to oxidize the same amount of `Fe^(2+)` that is oxidized by `x` moles of `MnO_4^-` in acidic medium.

### Step-by-Step Solution:

1. **Understanding the Reaction of `MnO_4^-` with `Fe^(2+)`:**
In acidic medium, the half-reaction for the oxidation of `Fe^(2+)` to `Fe^(3+)` is:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
This means that 1 mole of `Fe^(2+)` loses 1 electron.

2. **Determining the Electrons Transferred by `MnO_4^-`:**
The half-reaction for the reduction of `MnO_4^-` in acidic medium is:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
This indicates that 1 mole of `MnO_4^-` can accept 5 electrons.

3. **Calculating the Moles of `Fe^(2+)` Oxidized by `x` Moles of `MnO_4^-`:**
If `x` moles of `MnO_4^-` are used, the total number of moles of electrons transferred is:
\[ \text{Total electrons} = x \times 5 \]
Since each mole of `Fe^(2+)` loses 1 electron, the number of moles of `Fe^(2+)` oxidized is equal to the total number of electrons:
\[ \text{Moles of } Fe^{2+} = x \times 5 \]

4. **Understanding the Reaction of `Cr_2O_7^{2-}` with `Fe^(2+)`:**
The half-reaction for the reduction of `Cr_2O_7^{2-}` in acidic medium is:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
This means that 1 mole of `Cr_2O_7^{2-}` accepts 6 electrons.

5. **Calculating the Moles of `Cr_2O_7^{2-}` Required:**
To find out how many moles of `Cr_2O_7^{2-}` are needed to oxidize the same amount of `Fe^(2+)`, we can set up the following relationship:
\[ \text{Moles of } Cr_2O_7^{2-} = \frac{\text{Total electrons}}{6} = \frac{x \times 5}{6} \]

### Final Answer:
The number of moles of `Cr_2O_7^{2-}` required to oxidize the same amount of `Fe^(2+)` is:
\[ \text{Moles of } Cr_2O_7^{2-} = \frac{5x}{6} \]