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 is a half-reaction for the reduction of dichromate ions (\( \text{Cr}_2\text{O}_7^{2-} \)) to chromium(III) ions. The oxidation states of chromium change from \( +6 \) in \( \text{Cr}_2\text{O}_7^{2-} \) to \( +3 \) in \( \text{Cr}^{3+} \). The ionic equation is as follows: \[ \text{Cr}_2\text{O}_7^{2-} + 14 \, \text{H}^+ + 6 \, \text{e}^- \rightarrow 2 \, \text{Cr}^{3+} + 7 \, \text{H}_2\text{O} \]

Step 1: Balance the chromium atoms. In \( \text{Cr}_2\text{O}_7^{2-} \), there are two chromium atoms, so we need to produce two \( \text{Cr}^{3+} \) ions on the right side of the equation.


Step 2: Balance the oxygen atoms. In \( \text{Cr}_2\text{O}_7^{2-} \), there are 7 oxygen atoms, so we add 7 molecules of water (H$_2$O) on the right side to balance the oxygens.


Step 3: Balance the hydrogen atoms. Since there are 7 water molecules, there are 14 hydrogen atoms on the right side, so we add 14 \( \text{H}^+ \) ions on the left side.
Step 4: Finally, balance the charges by adding 6 electrons on the left side, as the total positive charge on the left side (from the \( \text{H}^+ \) ions) needs to be balanced by the electrons.