Step 1: Identify the Oxidation and Reduction Half-Reactions
The given reaction consists of multiple species undergoing oxidation and reduction:
1. Reduction Half-Reaction (Chromium):
\[
\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}
\]
Chromium changes from \( +6 \) (in \( \text{Cr}_2\text{O}_7^{2-} \)) to \( +3 \) (in \( \text{Cr}^{3+} \)).
Since each chromium atom gains 3 electrons, and there are 2 Cr atoms, the total electrons gained:
\[
2 \times 3 = 6 \text{ electrons}.
\]
2. Oxidation Half-Reaction (Iron):
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}
\]
Iron changes from +2 to +3, meaning it loses 1 electron per Fe atom.
3. Oxidation Half-Reaction (Oxalate Ion):
\[
\text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2
\]
Each carbon in \( \text{C}_2\text{O}_4^{2-} \) changes from +3 to +4, losing 1 electron per carbon atom.
Since there are 2 carbon atoms, the total electrons lost:
\[
2 \times 1 = 2 \text{ electrons}.
\]
Step 2: Balancing the Electrons
The total electrons gained in the reduction step = 6.
The total electrons lost = 1 (Fe) + 2 (C) = 3.
To balance the loss and gain, we need 3 electrons.
\[
\text{Total electrons involved} = 3.
\]
Final Answer: The total number of electrons involved in the redox reaction is 3.