Problem 156 - A Complete Resource Book in Chemistry for JEE Main

Question

The value of \(\mathrm{n}\) (the number of electrons) in the half equation in which \(\mathrm{Cr}(\mathrm{OH})_{4}^{-}\)is oxidized to chromate ion in alkali medium is

Step-by-Step Solution

Answer

The value of \( n \) for the oxidation of \( \mathrm{Cr(OH)}_{4}^{-} \) to chromate ion is 1.

Step 1: Write the Half Reaction Equation

First, let's write down the half reaction of the oxidation of \ \mathrm{Cr(OH)}_{4}^{-} \ to \ \mathrm{CrO}^{2-}_{4} \ in an alkaline medium initially without balancing hydrogen (H) and oxygen (O) atoms: \[ \mathrm{Cr(OH)}_{4}^{-} \rightarrow \mathrm{CrO}_{4}^{2-} \]

Step 2: Balance Chromium Atoms

In the equation, there is one chromium atom on both the reactant and product sides, so chromium is already balanced.\[ \mathrm{Cr(OH)}_{4}^{-} \rightarrow \mathrm{CrO}_{4}^{2-} \]

Step 3: Balance Oxygen Atoms

On the left side, we have 4 OH groups, which means 4 oxygen atoms, and on the right side, we have 4 oxygen atoms in the chromate ion. Therefore, the oxygen atoms are also balanced.\[ \mathrm{Cr(OH)}_{4}^{-} \rightarrow \mathrm{CrO}_{4}^{2-} \]

Step 4: Balance Hydrogen Atoms

The left side has 4 OH groups, which contribute 4 hydrogen atoms. Since there are no hydrogens in the products, add water molecules to the right side to balance hydrogens:\[ \mathrm{Cr(OH)}_{4}^{-} \rightarrow \mathrm{CrO}_{4}^{2-} + 2H_2O \]

Step 5: Balance Charges by Adding Electrons

Now, consider the charges. On the left, \((\mathrm{Cr(OH)}_{4}^{-})\) has a charge of \(-1\). On the right, \((\mathrm{CrO}_{4}^{2-})\) has a charge of \(-2\). Add electrons to make both sides equal: \[ \mathrm{Cr(OH)}_{4}^{-} \rightarrow \mathrm{CrO}_{4}^{2-} + 2H_{2}O + e^{-} \]This confirms that we need to add 1 electron to balance the charges.

Key Concepts

Half-Reaction Method

The half-reaction method is a key technique in redox chemistry, especially when dealing with complex redox reactions in alkaline or acidic solutions. It involves splitting the overall redox reaction into two half-reactions: one for oxidation and one for reduction. This makes it easier to balance the equation, as each half-reaction can be addressed separately. In the given exercise, we focused on the half-reaction where chromium in the form of \(\mathrm{Cr(OH)}_4^- \) is oxidized to \(\mathrm{CrO}_4^{2-} \). To effectively use the half-reaction method, follow these steps:

• Identification of Half-Reactions: Separate the full reaction into two halves based on electron transfer direction - an oxidation half and a reduction half.
• Balancing Atoms: Balance all elements except oxygen and hydrogen first.
• Balancing Oxygen: Balance oxygen atoms by adding water \(H_2O \) to the needed side.
• Balancing Hydrogen: Add \(OH^- \) in alkaline media, or \(H^+ \) in acidic media, to balance hydrogen atoms.
• Charge Balancing: Add electrons to one side to balance the charges.

This structured approach simplifies complex redox reactions, allowing you to focus on the details step-by-step while ensuring the final equation is correct and balanced.

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry, fundamental for ensuring that the matter is conserved throughout chemical reactions. In a balanced equation, the number of atoms for each element, as well as the net charge, remains the same on both the reactant and product sides. This ensures the law of conservation of mass is upheld. To balance the given half-reaction of \(\mathrm{Cr(OH)}_4^- \rightarrow \mathrm{CrO}_4^{2-} \), we follow a systematic approach:

• Start with Major Atoms: Begin by balancing atoms of elements that appear less frequently in the reaction - often, these are the central atoms like chromium.
• Balance Oxygen with Water: Compare the number of oxygen atoms on both sides. If imbalanced, adjust by adding \(H_2O \) molecules.
• Balance Hydrogen with \(OH^- \): In alkaline conditions, add \(OH^- \) to balance out any disparities in hydrogen atoms.
• Charge Balance with Electrons: As seen in the step-by-step solution, add electrons to ensure the same total charge on both sides of the equation.

This logical process ensures everything adds up, both in terms of mass and charge, granting a clear pathway to correctly balanced chemical equations.

Oxidation-Reduction Chemistry

Oxidation-reduction, or redox, chemistry revolves around the transfer of electrons between species. This kind of reaction is critical as it underpins a wide range of processes, from the metabolic reactions in living organisms to industrial chemical production.In any redox reaction, there are two concurrent processes: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons. In the given scenario of \(\mathrm{Cr(OH)}_4^- \) oxidizing to \(\mathrm{CrO}_4^{2-} \), oxidation occurs as chromium changes its oxidation state by losing electrons.

• Oxidation: Increases oxidation state by releasing electrons. Here, \(\mathrm{Cr(OH)}_4^- \) diminishes its electron count in favor of becoming \(\mathrm{CrO}_4^{2-} \).
• Reduction: Decreases oxidation state by gaining electrons. While the companion half-reaction is not detailed here, it would involve electron gain.

In specifying the movement of electrons, determining oxidation states can help track which substance is oxidized and which is reduced. The act of adding electrons during equation balancing is directly tied to these oxidation and reduction processes, ultimately ensuring the chemical integrity of the reaction.