Question

Balance the following half-reaction for Mo^3+ -> MoO2^2+ in acidic solution.

Answer 1

To balance the half-reaction of \( \text{Mo}^{3+} \) to \( \text{MoO}_2^{2+} \) in acidic solution, follow these steps:

1. **Write the unbalanced half-reaction:**
\[
\text{Mo}^{3+} \rightarrow \text{MoO}_2^{2+}
\]

2. **Balance the molybdenum atoms:**
There is one molybdenum atom on both sides, so it is already balanced.

3. **Balance the oxygen atoms:**
There are two oxygen atoms in \( \text{MoO}_2^{2+} \). To balance the oxygen, we can add water (\( \text{H}_2\text{O} \)) to the left side:
\[
\text{Mo}^{3+} + 2 \text{H}_2\text{O} \rightarrow \text{MoO}_2^{2+}
\]

4. **Balance the hydrogen atoms:**
We added 4 hydrogens on the left side (from 2 \( \text{H}_2\text{O} \)). To balance the hydrogens, we need to add \( 4 \text{H}^+ \) to the right side:
\[
\text{Mo}^{3+} + 2 \text{H}_2\text{O} \rightarrow \text{MoO}_2^{2+} + 4 \text{H}^+
\]

5. **Balance the charges:**
The left side has a charge of +3 (from \( \text{Mo}^{3+} \)), and the right side has a charge of +2 (from \( \text{MoO}_2^{2+} \)) plus +4 (from \( 4 \text{H}^+ \)), totaling +6. To balance the charges, we add 3 electrons (\( 3e^- \)) to the left side:
\[
\text{Mo}^{3+} + 2 \text{H}_2\text{O} + 3 e^- \rightarrow \text{MoO}_2^{2+} + 4 \text{H}^+
\]

Now the half-reaction is balanced in terms of both mass and charge:
\[
\text{Mo}^{3+} + 2 \text{H}_2\text{O} + 3 e^- \rightarrow \text{MoO}_2^{2+} + 4 \text{H}^+
\]

This is the balanced half-reaction in acidic solution.