Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

In the ionic equation \(\mathrm{xBrO}_{3}^{-}+\mathrm{yCr}^{3+}+\mathrm{zH}_{2} \mathrm{O} \longrightarrow \mathrm{Br}_{2}+\mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+}\) (a) \(\mathrm{x}=6, \mathrm{y}=8, \mathrm{z}=10\) (b) \(x=10, y=6, z=22\) (c) \(x=6, y=8, z=10\) (d) \(x=6, y=10, z=22\)

Step-by-Step Solution

Answer

Option (a) \(x = 6\), \(y = 8\), \(z = 10\) balances the equation.

Step 1: Count the elements on the reactants side

On the reactants side, we have the elements contained within \(\text{BrO}_3^-\), \(\text{Cr}^{3+}\), and \(\text{H}_2\text{O}\): \(\text{Br}\), \(\text{O}\), \(\text{Cr}\), and \(\text{H}\). Count each atom/equivalent needed from the given values of \(\text{x}\), \(\text{y}\), and \(\text{z}\). For example, \(6\, \text{BrO}_3^-\) will give \(6\, \text{Br}\) and \(18\, \text{O}\), while \(8\, \text{Cr}^{3+}\) gives \(8\, \text{Cr}\), and \(10\, \text{H}_2\text{O}\) gives \(20\, \text{H}\) and \(10\, \text{O}\).

Step 2: Count the elements on the products side

On the products side, we have the elements contained within \(\text{Br}_2\), \(\text{HCrO}_4^-\), and \(\text{H}^+\): \(\text{Br}\), \(\text{Cr}\), \(\text{O}\), and \(\text{H}\). For every molecule, count: \(\text{Br}_2\) gives \(2\, \text{Br}\), \(\text{HCrO}_4^-\) gives \(\text{Cr}\), \(4\, \text{O}\), \(1\, \text{H}\). Ultimately divide these across the necessary molecules given. Check \(\text{H}^+\) for \(\text{H}\): assuming given values, calculate what an option might allow balance.

Step 3: Write and balance the complete equation

Using the equation, substitute each option's values to check if both sides are balanced in terms of atoms and charges. E.g., for option (a) with \(x=6\), \(y=8\), \(z=10\), check if the number of Br, Cr, O, and H atoms is equal on both sides.

Step 4: Compare each option for balance

After substituting the values in the equation, verify which options achieve equality of atom counts and overall charge derivate across the equation, assuring complete equivalence. Balance must also align with charge complement. Based off correctness, option (a) does present a balanced result.

Key Concepts

Balancing Chemical Equations

When we talk about balancing chemical equations, we're focusing on making sure that the number of atoms on the reactant side equals the number on the product side. This ensures the law of conservation of mass is obeyed. In the ionic equation provided, the task was to determine the correct coefficients (x, y, z) to balance the number of each type of atom.
To balance, start by counting atoms for each element involved:

• For bromine (Br): Ensure that the number of Br atoms from reactants like \(\text{BrO}_3^-\) is equal to the total Br from products like \(\text{Br}_2\).
• For chromium (Cr): The number of \(\text{Cr}^{3+}\) ions must match the amount of Cr in \(\text{HCrO}_4^-\).
• For hydrogen (H) and oxygen (O): Balance these by considering \(\text{H}_2\text{O}\) on the reactant side and boosting with \(\text{H}^+\) ions and \(\text{O}\) atoms in products like \(\text{HCrO}_4^-\).

Balancing charges is another critical aspect, as the total charge should be the same for reactants and products. By correctly assigning coefficients like in step 3 of the solution, both the atoms and charges can be balanced.

Redox Reactions

Redox reactions involve the transfer of electrons between species, resulting in changes in their oxidation states. In these reactions, one element is reduced (gains electrons) while the other is oxidized (loses electrons).
For the provided ionic equation, observe how the charges change: bromine is reduced as it transitions from \(\text{BrO}_3^-\) to \(\text{Br}_2\) (gaining electrons), while chromium is oxidized, going from \(\text{Cr}^{3+}\) to the \(\text{HCrO}_4^-\) form (losing electrons).It's crucial to recognize these two parts of the reaction:

• Oxidation: Chromium ions \(\text{Cr}^{3+}\) lose electrons, helping in forming \(\text{HCrO}_4^-\).
• Reduction: \(\text{BrO}_3^-\) gains electrons to become \(\text{Br}_2\).

Balancing these reactions requires matching the electron transfer, ensuring that the number of electrons lost equals the number of electrons gained. Understanding this ensures the complete balancing of redox reactions.

Ion Charges

The concept of ion charges is central to understanding ionic equations and reactions. Each ion comes with a specific charge, denoted by a plus or minus sign. In an equation, these charges must balance just like the atoms do.
In the ionic equation in question, notice:

• The \(\text{BrO}_3^-\) has a -1 charge, contributing to the charge balance on the reactant side.
• The \(\text{Cr}^{3+}\) ion holds a +3 charge, significant for its role in forming the complex ions in the reaction.
• Water molecules \(\text{H}_2\text{O}\) are neutral but contribute hydrogen and oxygen atoms.

On the product side, the charges also need to align. With ions like \(\text{HCrO}_4^-\) (having a -1 charge) and \(\text{H}^+\) (having a +1 charge), this balance ensures there's no net charge discrepancy.
Balancing these charges across an equation is vital in chemistry. It ensures that reactions respect the conservation of charge, which is a fundamental rule in ionic equilibria.