Step 1: Identify Oxidation and Reduction Half-Reactions
The reaction involves the following species undergoing oxidation and reduction:1. Reduction Half-Reaction (Chromium): \[ \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} \] Chromium's oxidation state changes from \( +6 \) in \( \text{Cr}_2\text{O}_7^{2-} \) to \( +3 \) in \( \text{Cr}^{3+} \). Each chromium atom gains 3 electrons. With 2 chromium atoms, the total electrons gained are: \[ 2 \times 3 = 6 \text{ electrons}. \]2. Oxidation Half-Reaction (Iron): \[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} \] Iron's oxidation state increases from +2 to +3, indicating a loss of 1 electron per Fe atom.3. Oxidation Half-Reaction (Oxalate Ion): \[ \text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 \] In \( \text{C}_2\text{O}_4^{2-} \), each carbon atom's oxidation state changes from +3 to +4, resulting in a loss of 1 electron per carbon atom. With 2 carbon atoms, the total electrons lost are: \[ 2 \times 1 = 2 \text{ electrons}. \]
Step 2: Balance the Electrons
The total electrons gained in the reduction step is 6.
The total electrons lost from oxidation is the sum of electrons lost by iron and oxalate: 1 (Fe) + 2 (C) = 3.To balance the electron transfer, the least common multiple of electrons gained (6) and electrons lost (3) is 6. This requires a multiplier of 1 for the reduction half-reaction and a multiplier of 2 for the oxidation half-reactions. The total electrons involved in the balanced redox reaction are determined by scaling:\[\text{Total electrons involved} = 6.\]Final Answer: The total number of electrons involved in the redox reaction is 6.