Consider the following reaction,
xMnO4- + yC2O4(2-) + zH+ - xMn2+ + 2yCO2 + (z/2)H2O
The values of

Q: Why do we add $\text{H}^+$ and $\text{H}_2\text{O}$ to the reaction?

In an acidic medium, oxygen atoms are balanced by adding water ($\text{H}_2\text{O}$) molecules to the side deficient in oxygen. Then, to balance the hydrogen atoms introduced by the water, we add protons ($\text{H}^+$) to the opposite side.

Q: How did we calculate the oxidation state of Carbon as $+3$ in the oxalate ion?

Let the oxidation state of Carbon be $x$. The oxalate ion is $\text{C}_2\text{O}_4^{2-}$. Since oxygen is typically $-2$, we set up the equation: $2x + 4(-2) = -2$. Solving this gives $2x - 8 = -2$, which means $2x = 6$, so $x = +3$.

Q: Why did we multiply the oxidation half by 5 and the reduction half by 2?

In any redox reaction, the total number of electrons lost must exactly equal the total number of electrons gained. The oxidation half loses 2 electrons, and the reduction half gains 5. The lowest common multiple of 2 and 5 is 10. So, we multiply by 5 and 2 respectively to make the electron transfer equal to 10.

Q: Can I just use the n-factor method instead of writing out the full half-reactions?

Absolutely! For objective exams like JEE, the n-factor method is a massive time-saver. The n-factor for $\text{MnO}_4^-$ is 5, and for $\text{C}_2\text{O}_4^{2-}$ it is 2. Cross-multiplying these n-factors directly gives you the stoichiometric coefficients $x=2$ and $y=5$.

Q: What if this reaction was taking place in a basic medium instead of an acidic one?

If it were a basic medium, we would balance the oxygen and hydrogen atoms using $\text{OH}^-$ and $\text{H}_2\text{O}$. However, the core electron transfer (the n-factors) would remain exactly the same, so the coefficients $x$ and $y$ would not change!

Question

Consider the following reaction,
$x	ext{MnO}_4^- + y	ext{C}_2	ext{O}_4^{2-} + z	ext{H}^+ \longrightarrow x	ext{Mn}^{2+} + 2y	ext{CO}_2 + \frac{z}{2}	ext{H}_2	ext{O}$
The values of $x$, $y$ and $z$ in the reaction are, respectively

Options:

(A) 5, 2 and 16

(B) 2, 5 and 8

(C) 2, 5 and 16

(D) 5, 2 and 8

Sigma Percentile · Accepted Answer

Balancing redox reactions can sometimes feel like an intricate dance of electrons, but once you understand the rhythm, it becomes incredibly satisfying. Let's break down this classic reaction between permanganate and oxalate ions in an acidic medium.

### Analyzing the Setup

The first step in any redox problem is to identify the key players: who is losing electrons and who is gaining them. We do this by calculating the oxidation states of the central atoms.

In the permanganate ion, $$	ext{MnO}_4^-$$, oxygen is typically at a $-2$ oxidation state. Setting up the equation $x + 4(-2) = -1$, we find that Manganese is sitting at a highly oxidized state of $+7$.

In the oxalate ion, $$	ext{C}_2	ext{O}_4^{2-}$$, we set up a similar equation: $2x + 4(-2) = -2$. Solving this gives us an oxidation state of $+3$ for Carbon.

Looking at the products, Manganese is reduced to $$	ext{Mn}^{2+}$$ (oxidation state $+2$), and Carbon is oxidized to $$	ext{CO}_2$$ (oxidation state $+4$).

### The Master Equation: Half-Reactions

To keep things organized, we split the overall reaction into two half-reactions: oxidation and reduction.

**The Oxidation Half:**
The oxalate ion converts to carbon dioxide:
$$ 	ext{C}_2	ext{O}_4^{2-} \longrightarrow 2	ext{CO}_2 $$
Each Carbon atom goes from $+3$ to $+4$, losing $1$ electron. Since there are two Carbon atoms, the total loss is $2$ electrons:
$$ 	ext{C}_2	ext{O}_4^{2-} \longrightarrow 2	ext{CO}_2 + 2e^- $$

**The Reduction Half:**
The permanganate ion converts to Manganese(II):
$$ 	ext{MnO}_4^- \longrightarrow 	ext{Mn}^{2+} $$
Manganese goes from $+7$ to $+2$, which requires a gain of $5$ electrons. To balance the oxygen atoms, we add $4$ water molecules to the right side. Consequently, to balance the hydrogen atoms, we add $8$ protons ($$	ext{H}^+$$) to the left side:
$$ 	ext{MnO}_4^- + 8	ext{H}^+ + 5e^- \longrightarrow 	ext{Mn}^{2+} + 4	ext{H}_2	ext{O} $$

### The Grand Finale: Balancing the Charges

Here is the golden rule of redox reactions: **The total number of electrons lost must exactly equal the total number of electrons gained.**

Currently, our oxidation half loses $2$ electrons, while our reduction half gains $5$ electrons. To equalize this, we find the lowest common multiple, which is $10$. We achieve this by cross-multiplying: we multiply the entire oxidation half-reaction by $5$, and the entire reduction half-reaction by $2$.

$$ 5	ext{C}_2	ext{O}_4^{2-} \longrightarrow 10	ext{CO}_2 + 10e^- $$
$$ 2	ext{MnO}_4^- + 16	ext{H}^+ + 10e^- \longrightarrow 2	ext{Mn}^{2+} + 8	ext{H}_2	ext{O} $$

Adding these two balanced half-reactions together, the electrons cancel out perfectly, leaving us with the final balanced equation:
$$ 2	ext{MnO}_4^- + 5	ext{C}_2	ext{O}_4^{2-} + 16	ext{H}^+ \longrightarrow 2	ext{Mn}^{2+} + 10	ext{CO}_2 + 8	ext{H}_2	ext{O} $$

Comparing this to the given equation $x	ext{MnO}_4^- + y	ext{C}_2	ext{O}_4^{2-} + z	ext{H}^+ \longrightarrow \dots$, we can directly read off the coefficients: **$x = 2$**, **$y = 5$**, and **$z = 16$**.

### The Ninja Technique: n-factor Shortcut

While the half-reaction method is rigorous and foolproof, competitive exams like JEE demand speed. This is where the **n-factor** method shines.

The n-factor is simply the total change in oxidation state per molecule. 
For $$	ext{MnO}_4^-$$, the change is from $+7$ to $+2$, so its n-factor is $5$.
For $$	ext{C}_2	ext{O}_4^{2-}$$, the change is from $+3$ to $+4$ per Carbon atom. Since there are two Carbon atoms, the n-factor is $1 	imes 2 = 2$.

To balance the main reactants, you simply cross-multiply their n-factors! The coefficient for $$	ext{MnO}_4^-$$ becomes the n-factor of oxalate ($2$), and the coefficient for oxalate becomes the n-factor of permanganate ($5$). Instantly, you know that $x=2$ and $y=5$. This shortcut is a powerful tool to have in your arsenal!

Animated Solution for Chemistry - Redox Reactions: Consider the following reaction, $x MnO_{4}^{-} + y C_{2} O_{4}^{2 -} + z H^{+} ⟶ x Mn^{2 +} + 2 y CO_{2} + \frac{z}{2} H_{2} O$ The values of $x$ , $y$ and $z$ in the reaction are, respectively

Select Answer:

Visualized Solution

\text{Oxidation States}

\text{Electron Transfer}

\text{Oxidation Half-Reaction}

\text{Reduction Half-Reaction}

\text{Equating Electrons}

\text{Final Balanced Equation}

\text{The n-factor Shortcut}

The Sigma Insight: Chemical Equation Balancing

Analyzing the Setup

The Master Equation

The Grand Finale

The Ninja Technique: n-factor Shortcut

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