Question

a C\(_2\)O\(_4\)\(^{2-}\) + b MnO\(_4\)\(^{-}\) + c H\(^+\) \(\rightarrow\) x Mn\(^{2+}\) + y H\(_2\)O + z CO\(_2\)
a and x respectively are

• A. 5, 2
• B. 4, 1
• C. 3, 2
• D. 4, 2

Hint:

For balancing redox reactions, the ion-electron method is systematic.

• Split into half-reactions.
• Balance atoms other than O and H.
• Balance O with H\(_2\)O.
• Balance H with H\(^+\) (in acidic medium).
• Balance charge with e\(^-\).
• Equalize e\(^-\) in both half-reactions and add them up.

Always double-check that both atoms and charge are balanced in the final equation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a redox reaction happening in an acidic medium involving the oxidation of oxalate ions by permanganate ions.
Step 2: Key Formula or Approach:
Balance the redox equation using the ion-electron (half-reaction) method by separately balancing the oxidation and reduction halves, then equalizing the electrons transferred.
Step 3: Detailed Explanation:
1. Oxidation half-reaction: Oxalate ion oxidizes to carbon dioxide.
$C_{2}O_{4}^{2-} \rightarrow 2CO_{2} + 2e^{-}$ (Loss of 2 electrons per oxalate ion)

2. Reduction half-reaction: Permanganate ion reduces to Manganese(II) in acidic medium.
$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$ (Gain of 5 electrons per permanganate ion)

3. Balancing electrons:
Multiply the oxidation half-reaction by 5:
$5C_{2}O_{4}^{2-} \rightarrow 10CO_{2} + 10e^{-}$
Multiply the reduction half-reaction by 2:
$2MnO_{4}^{-} + 16H^{+} + 10e^{-} \rightarrow 2Mn^{2+} + 8H_{2}O$

4. Summing the reactions:
$5C_{2}O_{4}^{2-} + 2MnO_{4}^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$
Comparing with the given equation: $a = 5$, $b = 2$, $c = 16$, $x = 2$, $y = 8$, $z = 10$.
Step 4: Final Answer:
The values of a and x are 5 and 2 respectively.