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: Identify and balance the half-reactions.
This is a redox reaction in an acidic medium. We will use the ion-electron method to balance it.
Oxidation Half-Reaction:
The oxalate ion is oxidized to carbon dioxide.
\[ \text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2 \]

• Balance atoms other than O and H (Carbon):
  \[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 \]
• Oxygen is already balanced.
• Balance the charge by adding electrons to the more positive side:
  \[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^- \]

Reduction Half-Reaction:
The permanganate ion is reduced to manganese(II) ion.
\[ \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} \]

• Manganese is balanced.
• Balance Oxygen by adding H\(_2\)O:
  \[ \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
• Balance Hydrogen by adding H\(^+\):
  \[ \text{MnO}_4^{-} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
• Balance the charge by adding electrons:
  (LHS charge = -1 + 8 = +7; RHS charge = +2). Add 5e\(^-\) to LHS.
  \[ \text{MnO}_4^{-} + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

Step 2: Combine the half-reactions.
To combine the reactions, the number of electrons lost must equal the number of electrons gained. The LCM of 2 and 5 is 10.
Multiply the oxidation half-reaction by 5:
\[ 5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10e^- \] Multiply the reduction half-reaction by 2:
\[ 2\text{MnO}_4^{-} + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \] Step 3: Add the balanced half-reactions and cancel the electrons.
\[ 5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^{-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2 \] Step 4: Determine the values of 'a' and 'x'.
Comparing the balanced equation with the given template:
a C\(_2\)O\(_4\)\(^{2-}\) + b MnO\(_4\)\(^{-}\) + c H\(^+\) \(\rightarrow\) x Mn\(^{2+}\) + y H\(_2\)O + z CO\(_2\)
We find that the coefficient for C\(_2\)O\(_4\)\(^{2-}\) is a = 5.
The coefficient for Mn\(^{2+}\) is x = 2.
Step 5: Final Answer.
The values of a and x are 5 and 2, respectively.