If KMnO_4 is reduced by oxalic acid in the acidic medium, then the ox…

The correct answer is option 3. +7 to +2.

To determine how the oxidation number of manganese (\(Mn\)) changes during the reaction between potassium permanganate (\(KMnO_4\)) and oxalic acid (\(H_2C_2O_4\)) in an acidic medium, let us delve into the details of the reaction.

Identifying the Oxidation States

a. In \(KMnO_4\):

Potassium (\(K\)) has a fixed oxidation state of +1.Oxygen (\(O\)) typically has an oxidation state of -2.

Let us determine the oxidation state of \(Mn\) in \(KMnO_4\).

\(\text{Overall charge of } KMnO_4 = 0 \quad (\text{since it's a neutral compound})\)

\(\text{Let oxidation state of } Mn = x\)

\(\begin{align*}
\text{Charge balance equation:} \\
(+1) + x + 4(-2) &= 0 \\
+1 + x - 8 &= 0 \\
x - 7 &= 0 \\
x &= +7
\end{align*}\)

Conclusion: In \(KMnO_4\), \(Mn\) has an oxidation state of +7.

b. In the Reduced Form (\(Mn^{2+}\)):

When \(KMnO_4\) acts as an oxidizing agent in an acidic medium, it gets reduced to \(Mn^{2+}\).

Oxidation state of \(Mn\) in \(Mn^{2+}\) = +2

The Redox Process

a. Reduction Half-Reaction:

\(MnO_4^- \rightarrow Mn^{2+}\)

b. Oxidation Half-Reaction:

Oxalic acid (\(H_2C_2O_4\)) gets oxidized to carbon dioxide (\(CO_2\)).

\(H_2C_2O_4 \rightarrow CO_2\)

From the above identification:

Initial Oxidation State of \(Mn\) in \(KMnO_4\): +7

Final Oxidation State of \(Mn\) in \(Mn^{2+}\): +2

\(\text{Change in oxidation state} = +7 \rightarrow +2\)

\(\Delta \text{Oxidation Number} = +7 - (+2) = +5\)

This indicates that \(Mn\) is reduced by 5 units during the reaction.

For a complete understanding, let's balance the redox reaction in an acidic medium.

Half-Reactions:

Reduction: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)

Oxidation: \(H_2C_2O_4 \rightarrow 2CO_2 + 2H^+ + 2e^-\)

Multiplying to Equalize Electrons: To balance the electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:

\(\begin{align*}
\text{Reduction:} \quad & 2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O \\
\text{Oxidation:} \quad & 5H_2C_2O_4 \rightarrow 10CO_2 + 10H^+ + 10e^-
\end{align*}\)

Combining the Half-Reactions:

\(2MnO_4^- + 16H^+ + 5H_2C_2O_4 \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2 + 10H^+\)

Simplifying:

\(2MnO_4^- + 6H^+ + 5H_2C_2O_4 \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2\)

Conclusion

Initial Oxidation State of \(Mn\): +7

Final Oxidation State of \(Mn\): +2

Thus, the oxidation number of \(Mn\) changes from +7 to +2 during the reduction of \(KMnO_4\) by oxalic acid in an acidic medium.