Question

Complete the following ionic equation:
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ →

Hint:

When balancing redox reactions, start by balancing the atoms, then the charges, and use electrons to balance the charge.

Answer 1

This equation represents the reduction of dichromate ions (\( \text{Cr}_2\text{O}_7^{2-} \)) to chromium(III) ions. The chromium oxidation state transitions from \( +6 \) in dichromate to \( +3 \) in chromium(III) ions. The balanced ionic equation is:\[\text{Cr}_2\text{O}_7^{2-} + 14 \, \text{H}^+ + 6 \, \text{e}^- \rightarrow 2 \, \text{Cr}^{3+} + 7 \, \text{H}_2\text{O}\]

Step 1: Ensure chromium atoms are balanced. The presence of two chromium atoms in \( \text{Cr}_2\text{O}_7^{2-} \) necessitates the formation of two \( \text{Cr}^{3+} \) ions on the product side.

Step 2: Balance oxygen atoms. Seven oxygen atoms in \( \text{Cr}_2\text{O}_7^{2-} \) are balanced by adding seven water molecules (H$_2$O) to the product side.

Step 3: Balance hydrogen atoms. The fourteen hydrogen atoms present in the seven water molecules on the product side are balanced by introducing fourteen \( \text{H}^+ \) ions to the reactant side. Step 4: Balance the charges. The net positive charge on the reactant side (contributed by \( \text{H}^+ \) ions) is neutralized by adding six electrons.