Problem 18 - A Complete Resource Book in Chemistry for JEE Main

Question

For the redox reaction \(\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\), the correct coefficients of the reactants for the balanced reaction are (a) (b) (c) \(\begin{array}{clc}\mathrm{MnO}_{4}^{-} & \mathrm{C}_{2} \mathrm{O}_{4}^{2-} & \mathrm{H}^{+} \\ 2 & 5 & 16 \\ 16 & 5 & 2 \\ 5 & 16 & 2 \\ 2 & 16 & 5\end{array}\) (d)

Step-by-Step Solution

Answer

The correct coefficients are 2, 5, and 16.

Step 1: Determine Oxidation States

Identify the oxidation states of elements in reactants and products. In \( \mathrm{MnO}_4^- \), Mn is +7; in \( \mathrm{Mn}^{2+} \), Mn is +2. In \( \mathrm{C}_2\mathrm{O}_4^{2-} \), C is +3; in \( \mathrm{CO}_2 \), C is +4. Mn is reduced and C is oxidized.

Step 2: Write Half-Reactions

Separate the equation into two half-reactions. Reduction: \( \mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O} \). Oxidation: \( \mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2 + 2e^- \).

Step 3: Balance Electrons

Adjust each half-reaction so the number of electrons lost in oxidation equals the number gained in reduction. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2.

Step 4: Combine Half-Reactions

Add the balanced half-reactions together. Use the electron balance as a guide: \( 2(\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}) \) and \( 5(\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2 + 2e^-) \).

Step 5: Balance Full Reaction

Combine to yield the full balanced reaction: \( 2\mathrm{MnO}_4^- + 5\mathrm{C}_2\mathrm{O}_4^{2-} + 16\mathrm{H}^+ \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O} \).

Step 6: Determine Coefficients

The coefficients for the reactants \( \mathrm{MnO}_4^- \), \( \mathrm{C}_2\mathrm{O}_4^{2-} \), and \( \mathrm{H}^+ \) are 2, 5, and 16, respectively.

Key Concepts

Understanding Oxidation States

Oxidation states, also known as oxidation numbers, provide insight into the electron transfer happening in a redox reaction. They help us identify which elements are oxidized and which are reduced. To determine the oxidation state, we look at the charge distribution in each compound. For example, in the permanganate ion \(\mathrm{MnO}_4^-\), manganese (\(\mathrm{Mn}\)) has an oxidation state of +7, while in \(\mathrm{Mn}^{2+}\), it is +2. This indicates that manganese gains electrons, hence it is reduced. Similarly, in oxalate \(\mathrm{C}_2\mathrm{O}_4^{2-}\), carbon has an oxidation state of +3. When carbon is in carbon dioxide \(\mathrm{CO}_2\), it changes to +4, suggesting that carbon has lost electrons and is oxidized. By determining these changes, we can specify the direction of electron transfer and identify which species act as oxidizing agents and which act as reducing agents.

Breaking Down the Reaction into Half-Reactions

Redox reactions involve two parts which we call half-reactions, one for oxidation and one for reduction. Each half-reaction shows either the loss or gain of electrons separately. The reduction half-reaction involves manganese:

• \(\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}\).

This represents manganese gaining electrons and being reduced. The oxidation half-reaction occurs with oxalate ion:

• \( \mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2 + 2e^- \).

This indicates carbon losing electrons and being oxidized. By writing the separate half-reactions, it is easier to see and balance the flow of electrons between them, ensuring both the number and type of atoms and the net charge are conserved.

Achieving Electron Balance

Electron balance is essential in redox reactions to comply with the law of conservation of charge. In the context of our half-reactions, manganese gains 5 electrons in its reduction half, while the carbon in the oxalate ion loses 2 electrons during oxidation. To balance these electrons, we need equal numbers of electrons lost and gained. Multiply the oxidation half-reaction by 5, giving

• \(5(\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2 + 2e^-) \),

and the reduction half-reaction by 2:

• \(2(\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O})\).

With this multiplication, both parts have 10 electrons transferring, maintaining electron balance. Thus, ensuring that the electrons lost from oxidation are exactly matched by the electrons gained in reduction.

Balancing the Full Chemical Equation

After balancing the half-reactions with respect to electron transfer, the next step is to combine them into a single, balanced chemical equation. When we add the adjusted half-reactions together, we ensure no electrons are left unchecked. The equation becomes: \[ 2\mathrm{MnO}_4^- + 5\mathrm{C}_2\mathrm{O}_4^{2-} + 16\mathrm{H}^+ \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O} \] This equation is balanced as it contains the same number of each atom on both sides, and the charges also match. Guidelines used here include:

• Arranging coefficients to even out atoms
• Ensuring the charges on both sides equalize

This approach assists in comprehending the roles of each reactant and product. Effectively balancing chemical equations is crucial for understanding the stoichiometry and energy relationships of chemical reactions.