Question

\[ 5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \rightarrow \; ? \]

Hint:

Permanganate reductions: Acidic → Mn\(^{2+}\) Neutral → MnO\(_2\) Alkaline → MnO\(_4^{2-}\)

Answer 1

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: \[ 5 \text{C}_2\text{O}_4^{2-} + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2O \]