In the following redox reactionn,
`Cr_(2)O_(7)^(2-) + Fe^(2+) to Fe^(3+) + Cr^(3+)`
1 mole of `Cr_(2)O_(7)^(2-)` oxidises.

Explanation

To solve the problem, we need to analyze the given redox reaction:

**Reaction:**
\[ \text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{Cr}^{3+} \]

We need to determine how many moles of \(\text{Fe}^{2+}\) are oxidized by 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\).

### Step 1: Identify the oxidation states
First, we need to find the oxidation states of chromium in \(\text{Cr}_2\text{O}_7^{2-}\) and the change it undergoes to \(\text{Cr}^{3+}\).

- In \(\text{Cr}_2\text{O}_7^{2-}\):
- Let the oxidation state of Cr be \(x\).
- The equation for the oxidation state is:
\[ 2x + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = 12 \implies x = +6 \]
- Thus, the oxidation state of Cr in \(\text{Cr}_2\text{O}_7^{2-}\) is +6.

- In \(\text{Cr}^{3+}\):
- The oxidation state is +3.

### Step 2: Calculate the change in oxidation state for chromium
The change in oxidation state for each chromium atom is:
\[ +6 \text{ (in Cr}_2\text{O}_7^{2-}) \rightarrow +3 \text{ (in Cr}^{3+}) \]
The change is:
\[ 6 - 3 = 3 \]
Since there are 2 chromium atoms in \(\text{Cr}_2\text{O}_7^{2-}\), the total change for both chromium atoms is:
\[ 2 \times 3 = 6 \]

### Step 3: Determine the n-factor for \(\text{Cr}_2\text{O}_7^{2-}\)
The n-factor (valency factor) for \(\text{Cr}_2\text{O}_7^{2-}\) is 6, as calculated from the total change in oxidation state.

### Step 4: Analyze the oxidation of iron
Next, we look at the oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\):
- The change in oxidation state for iron is:
\[ +2 \text{ (in Fe}^{2+}) \rightarrow +3 \text{ (in Fe}^{3+}) \]
The change is:
\[ 3 - 2 = 1 \]
Thus, the n-factor for \(\text{Fe}^{2+}\) is 1.

### Step 5: Set up the equivalence equation
Now we can set up the equivalence equation based on the number of equivalents:
\[ \text{Equivalence of } \text{Cr}_2\text{O}_7^{2-} = \text{Equivalence of } \text{Fe}^{2+} \]
Using the formula for equivalence:
\[ \text{Equivalence} = n \times \text{moles} \]
We have:
\[ 6 \times 1 = 1 \times x \]
Where \(x\) is the moles of \(\text{Fe}^{2+}\) oxidized.

### Step 6: Solve for \(x\)
From the equation:
\[ 6 = 1 \times x \implies x = 6 \]

### Conclusion
Thus, 6 moles of \(\text{Fe}^{2+}\) are oxidized by 1 mole of \(\text{Cr}_2\text{O}_7^{2-}\).

**Final Answer:** 6 moles of \(\text{Fe}^{2+}\) are oxidized.

---