In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in p

Q: Why don't we need to consider the permanganate ion ($\text{MnO}_4^-$) at all?

The question specifically asks for the number of electrons involved in producing *one molecule of $\text{CO}_2$*. Since $\text{CO}_2$ is solely produced from the oxidation of the oxalate ion, the reduction half-reaction of permanganate is irrelevant to this specific ratio.

Q: How do we know the oxidation state of carbon in oxalate is +3?

Let the oxidation state of carbon be $x$. In $\text{C}_2\text{O}_4^{2-}$, the sum of oxidation states is $2x + 4(-2) = -2$. Solving this gives $2x - 8 = -2 \implies 2x = 6 \implies x = +3$.

Q: Why does one oxalate ion produce two $\text{CO}_2$ molecules?

Conservation of mass! One oxalate ion ($\text{C}_2\text{O}_4^{2-}$) contains two carbon atoms. To balance the carbon atoms on both sides of the half-reaction, it must produce two molecules of $\text{CO}_2$.

Q: What if the question asked for the n-factor of oxalate?

The n-factor is the total number of electrons lost or gained per mole of the reactant. For oxalate ($\text{C}_2\text{O}_4^{2-}$), it loses 2 electrons per mole. So, its n-factor is 2.

Q: Is the medium (acidic/basic) important for this specific calculation?

For this specific question, no. The oxidation of oxalate to $\text{CO}_2$ involves the same electron transfer regardless of the medium. The medium only affects how we balance oxygen and hydrogen atoms in the overall reaction.

Question

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of $	ext{CO}_2$ is

Options:

(A) 2

(B) 5

(C) 1

(D) 10

Sigma Percentile · Accepted Answer

### The Trap of Overcomplication

Many students see "oxalate with permanganate in acidic medium" and immediately start writing out the massive, fully balanced redox equation. They calculate the n-factors, cross-multiply, and balance the water and hydrogen ions.

But wait! Is all that really necessary?

**Always read the question carefully.** It asks for the number of electrons involved in producing *one molecule of $	ext{CO}_2$*. Since $	ext{CO}_2$ comes exclusively from the oxalate ion, we can completely ignore the permanganate!

### Analyzing the Half-Reaction

Let's focus solely on the oxidation of the oxalate ion ($	ext{C}_2	ext{O}_4^{2-}$).

First, we need to determine the oxidation states of carbon. In the oxalate ion, the sum of oxidation states is $-2$. With four oxygen atoms at $-2$ each, the two carbon atoms must contribute $+6$ in total. Therefore, each carbon atom is in a **$+3$ oxidation state**.

When oxalate is oxidized, it forms carbon dioxide ($	ext{CO}_2$). In $	ext{CO}_2$, the two oxygen atoms contribute $-4$, meaning the single carbon atom is in a **$+4$ oxidation state**.

### The Master Equation

Since the oxidation state of carbon goes from $+3$ to $+4$, each carbon atom loses exactly $1$ electron.

Now, let's look at the stoichiometry. A single oxalate ion ($	ext{C}_2	ext{O}_4^{2-}$) contains two carbon atoms. To balance the carbon atoms, it must produce two molecules of $	ext{CO}_2$.

Because there are two carbon atoms, and each loses one electron, the total number of electrons lost in this half-reaction is $2$.

The balanced half-reaction is:

$	ext{C}_2	ext{O}_4^{2-} ightarrow 2	ext{CO}_2 + 2e^-$

### Final Calculation

We are almost there! The balanced half-reaction tells us a very simple story:

**Producing $2$ molecules of $	ext{CO}_2$ involves the transfer of $2$ electrons.**

Using a simple unitary method, we can find the answer for a single molecule. If $2$ molecules require $2$ electrons, then producing **$1$ molecule of $	ext{CO}_2$ involves exactly $1$ electron**.

And that's it! By focusing only on the relevant part of the chemical process, we bypassed a mountain of unnecessary algebra. Smart work always wins!

Animated Solution for Chemistry - Redox Reactions: In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of $CO_{2}$ is

Select Answer:

Visualized Solution

The Oxidation Half-Reaction

Oxidation State of Carbon

Balancing the Half-Reaction

Electrons per $CO_{2}$ Molecule

Conclusion

Pro-Tip for Redox Reactions

The Sigma Insight: Chemical Equation Balancing

The Trap of Overcomplication

Analyzing the Half-Reaction

The Master Equation

Final Calculation

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