To write the balanced equation for the oxidation of Fe²⁺ ions to Fe³⁺ ions by dichromate ions (Cr₂O₇²⁻), we first need to identify the half-reactions and then balance them. The dichromate ion is reduced in this reaction.
Step 1: Write the half-reactions
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Oxidation half-reaction (Fe²⁺ to Fe³⁺): \[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]
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Reduction half-reaction (Cr₂O₇²⁻ to Cr³⁺): \[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 e^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]
Step 2: Balance the electrons
We need to balance the number of electrons transferred in both half-reactions. Since the oxidation half-reaction produces 1 electron and the reduction half-reaction consumes 6 electrons, we can multiply the oxidation half-reaction by 6:
\[ 6 \text{Fe}^{2+} \rightarrow 6 \text{Fe}^{3+} + 6 e^- \]
Step 3: Combine the half-reactions
Now we can combine the balanced half-reactions:
\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]
Final Balanced Equation
The final balanced equation for the oxidation of Fe²⁺ ions to Fe³⁺ ions by dichromate ions is:
\[ 6 \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ \rightarrow 6 \text{Fe}^{3+} + 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \]
This equation is now balanced in terms of mass and charge.