Complete and Balance the Following Equation:
\[
5 \text{C}_2 \text{O}_4^{2-} + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow \; ?
\]
Step 1: Identify the Reaction Type
This is a redox reaction involving the reduction of MnO₄⁻ (permanganate ion) and the oxidation of C₂O₄²⁻ (oxalate ion). Manganese is reduced, and oxalate is oxidized, with H⁺ ions balancing the equation in acidic medium.
Step 2: Write the Half-Reactions
- The reduction half-reaction for permanganate ion (MnO₄⁻) in acidic medium is:
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2O
\]
- The oxidation half-reaction for oxalate ion (C₂O₄²⁻) is:
\[
\text{C}_2\text{O}_4^{2-} \rightarrow 2 \text{CO}_2 + 2e^-
\]
Each oxalate ion (C₂O₄²⁻) loses 2 electrons, and each permanganate ion (MnO₄⁻) gains 5 electrons.
Step 3: Balance the Half-Reactions
- The reduction half-reaction involves the transfer of 5 electrons, so we multiply it by 2.
- The oxidation half-reaction involves the transfer of 2 electrons, so we multiply it by 5.
The half-reactions become:
- Reduction:
\[
2 \text{MnO}_4^- + 16 \text{H}^+ + 10e^- \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2O
\]
- Oxidation:
\[
5 \text{C}_2\text{O}_4^{2-} \rightarrow 10 \text{CO}_2 + 10e^-
\]
Step 4: Combine the Half-Reactions
Now, add the two half-reactions together, canceling the electrons on both sides:
\[
5 \text{C}_2\text{O}_4^{2-} + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2O
\]
Conclusion:
The balanced equation is:
\[
5 \text{C}_2\text{O}_4^{2-} + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2O
\]