Understanding Oxidation Number Change in Redox Reactions

This question asks about the change in the oxidation number of Manganese (Mn) when Potassium Permanganate (KMnO₄) reacts with oxalic acid in an acidic medium. This is a classic example of a redox reaction where one species is oxidized (loses electrons, oxidation number increases) and another is reduced (gains electrons, oxidation number decreases).

Determining the Initial Oxidation State of Manganese in KMnO₄

To find the initial oxidation state of Mn in KMnO₄, we use the known oxidation states of Potassium (K) and Oxygen (O).

• Potassium (K) is an alkali metal, so its oxidation state is +1.
• Oxygen (O) usually has an oxidation state of -2, except in peroxides or with fluorine.
• KMnO₄ is a neutral compound, so the sum of the oxidation states of all atoms is zero.

Let the oxidation state of Mn be $x$. The equation is:

$\text{Oxidation state of K} + \text{Oxidation state of Mn} + 4 \times (\text{Oxidation state of O}) = 0$

$+1 + x + 4 \times (-2) = 0$

$+1 + x - 8 = 0$

$x - 7 = 0$

$x = +7$

So, the oxidation number of Mn in KMnO₄ is +7.

Manganese Reduction in Acidic Medium

Potassium Permanganate (KMnO₄) contains the permanganate ion ($\text{MnO}_4^{-}$). The permanganate ion is a strong oxidizing agent, especially in acidic solution. When it acts as an oxidizing agent, it gets reduced. The product of the reduction of the permanganate ion in a strongly acidic medium is typically the Manganese(II) ion ($\text{Mn}^{2+}$).

The Manganese(II) ion ($\text{Mn}^{2+}$) has an oxidation state equal to its charge, which is +2.

Determining the Final Oxidation State of Manganese

In the reaction between KMnO₄ and oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) in acidic medium, the $\text{MnO}_4^{-}$ ion is reduced to $\text{Mn}^{2+}$ ions. These $\text{Mn}^{2+}$ ions often appear as salts like Manganese(II) sulfate ($\text{MnSO}_4$) if sulfuric acid is used as the acidic medium.

Let's confirm the oxidation state of Mn in $\text{MnSO}_4$. The sulfate ion ($\text{SO}_4^{2-}$) has a charge of -2. For $\text{MnSO}_4$ to be neutral, the charge on the Mn ion must be +2.

$\text{Oxidation state of Mn} + \text{Charge of } \text{SO}_4^{2-} = 0$

$x + (-2) = 0$

$x = +2$

The oxidation number of Mn in the product is +2.

Meanwhile, the oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) is oxidized to carbon dioxide ($\text{CO}_2$). This is why KMnO₄ is reduced.

Summarizing the Change in Oxidation Number

The initial oxidation number of Mn in KMnO₄ is +7. The final oxidation number of Mn in the reduced product (like $\text{MnSO}_4$) is +2.

Therefore, the oxidation number of Mn changes from +7 to +2.

Balanced Chemical Equation Example

Here is a balanced chemical equation for the reaction in acidic medium (using $\text{H}_2\text{SO}_4$ as acid):

$2\text{KMnO}_4(aq) + 5\text{H}_2\text{C}_2\text{O}_4(aq) + 3\text{H}_2\text{SO}_4(aq) \rightarrow \text{K}_2\text{SO}_4(aq) + 2\text{MnSO}_4(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l)$

In this reaction:

• Mn changes from +7 (in KMnO₄) to +2 (in MnSO₄). It gains 5 electrons per Mn atom, so it is reduced.
• Carbon in oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) changes oxidation state. In $\text{H}_2\text{C}_2\text{O}_4$, H is +1, O is -2. $2(1) + 2x + 4(-2) = 0 \Rightarrow 2 + 2x - 8 = 0 \Rightarrow 2x = 6 \Rightarrow x = +3$. In $\text{CO}_2$, O is -2. $y + 2(-2) = 0 \Rightarrow y - 4 = 0 \Rightarrow y = +4$. Carbon changes from +3 to +4. It loses 1 electron per C atom (or 2 electrons per $\text{C}_2\text{O}_4^{2-}$ ion), so it is oxidized.

This confirms the reduction of Mn and the oxidation of oxalic acid.

Comparing with Options

The change in oxidation number of Mn is from +7 to +2. Let's look at the options:

Our calculated change (+7 to +2) matches Option 3.

Revision Table: Oxidation States

Additional Information: Permanganate Reactions

The reduction product of the permanganate ion ($\text{MnO}_4^{-}$) depends on the pH of the solution:

• Strongly acidic medium (like with H₂SO₄): $\text{MnO}_4^{-}$ (Mn is +7) is reduced to $\text{Mn}^{2+}$ (Mn is +2). This is the most common scenario and results in a significant color change from purple (MnO₄^-) to colorless (Mn²⁺).
• Weakly acidic or neutral medium: $\text{MnO}_4^{-}$ (Mn is +7) is typically reduced to $\text{MnO}_2$ (Manganese dioxide) (Mn is +4). $\text{MnO}_2$ is a brown precipitate.
• Strongly alkaline medium: $\text{MnO}_4^{-}$ (Mn is +7) is often reduced to $\text{MnO}_4^{2-}$ (Manganate ion) (Mn is +6). The manganate ion is green. Further reduction to $\text{MnO}_2$ can occur with strong reducing agents.

In this specific question, the reaction is in an acidic medium, which confirms the reduction of Mn from +7 to +2.