In the reaction, VO + Fe2O3 → FeO + V2O5, what is the n – factor for V2O5?

To determine the \( n \)-factor for \(\text{V}_2\text{O}_5\) in the reaction:

\[
VO + Fe_2O_3 \to FeO + V_2O_5,
\]

we need to follow these steps:

### 1. Assign Oxidation States:
- **Vanadium (V):**
- In \(VO\), vanadium (V) is in the \(+2\) oxidation state because oxygen is \( -2 \).
\[
x + (-2) = 0 \quad \Rightarrow \quad x = +2.
\]
- In \(\text{V}_2\text{O}_5\), vanadium is in the \(+5\) oxidation state.
\[
2x + 5(-2) = 0 \quad \Rightarrow \quad 2x - 10 = 0 \quad \Rightarrow \quad 2x = 10 \quad \Rightarrow \quad x = +5.
\]

### 2. Determine the Change in Oxidation State:
- Vanadium changes from \(+2\) in \(VO\) to \(+5\) in \(\text{V}_2\text{O}_5\).
- The change in oxidation state for one vanadium atom is:
\[
5 - 2 = 3.
\]
- Since there are two vanadium atoms in \(\text{V}_2\text{O}_5\), the total change is:
\[
2 \times 3 = 6.
\]

### 3. Calculate the \( n \)-factor:
The \( n \)-factor for a compound is the total change in the oxidation state of the atoms involved in the reaction.

Therefore, the \( n \)-factor for \(\text{V}_2\text{O}_5\) is **6**.