Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following statement is/are correct? (a) \(\mathrm{CrI}_{3}+\mathrm{KOH}+\mathrm{Cl}_{2} \rightarrow \mathrm{K}_{2} \mathrm{CrO}_{4}+\mathrm{KCl}+\mathrm{KIO}_{4}+\mathrm{H}_{2} \mathrm{O}\) So, in balanced chemical reaction coefficient of \(\mathrm{KOH}\) is 64 . (b) If \(\mathrm{Zn}+\mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\) \(\mathrm{MnSO}_{4}+\mathrm{H}_{2}\); then equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{KMnO}_{4}+\) equivalents of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O}, \mathrm{n}\)-factor of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is 12 . (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+}+\mathrm{CO}_{2}+\mathrm{NO}_{3}^{-}, \mathrm{n}\)-factor for \([\mathrm{Fe}(\mathrm{CN})]^{3-}\) is 60 .

Step-by-Step Solution

Answer

None of the statements are fully correct based on chemical balance and equivalents.

Step 1: Examine Reaction (a)

First, balance the given chemical equation: \[ \ce{CrI3 + KOH + Cl2 -> K2CrO4 + KCl + KIO4 + H2O} \]Assign oxidation numbers to the elements in the reactions to identify redox changes. Chromium goes from +3 in \( \ce{CrI3} \) to +6 in \( \ce{K2CrO4} \), iodine goes from -1 in \( \ce{CrI3} \) to +7 in \( \ce{KIO4} \), and chlorine goes from 0 in \( \ce{Cl2} \) to -1 in \( \ce{KCl} \). "KOH" is part of the balancing due to hydroxide ions and adjustment for water produced.

Step 2: Balancing Reaction for KOH

Using the inspection method and ensuring charge and mass balance (especially inspecting changes in oxidation states and balancing oxygen and hydrogen atoms): Balance each element one by one.The correct equation has large coefficients, and while historically calculated, this would yield a much more specific coefficient for \( \ce{KOH} \). Balanced coefficient tends to be improper as 64 is immensely large. Re-assume with a proper example, balancing likely results in fewer multiples.

Step 3: Examine Reaction (b)

Analyze and balance the reaction:\[ \ce{Zn + KMnO4 + H2SO4 -> ZnSO4 + K2SO4 + MnSO4 + H2O} \]Oxidation for manganese changes from +7 in \( \ce{KMnO4} \) to +2 in \( \ce{MnSO4} \). Zinc remains zinc sulfate, and sulfuric acid impacts water production. Equivalents are based on redox change. Calculate equivalents for zinc: Electrons lost/gained properly balance the equation through equivalents, however, it could not be simplified to exact identity equivalency.

Step 4: Examine Reaction (c)

Balance and calculate the \( n \)-factor:\[ \ce{(NH4)2Cr2O7 -> N2 + Cr2O3 + H2O} \]Nitrogen balances from +5 to zero; thus, assign flexibly: Chromium +6 to +3, water ensures H,O balance. \( n \)-factor determination could be erroneous by statement alone, needing detailed oxidation number tally: corrected by context is likely around 6.

Step 5: Examine Reaction (d)

Assess the reaction \[ \ce{[Fe(CN)6]^{3-} -> Fe^{3+} + CO2 + NO3-} \]. Iron is not oxidized; however, potential cyanide transformation makes large electron transfers or fictional response. Given n-factor is escalated senselessly as 60, making errors visible, large due to no probable dissolution so high.

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. This transfer brings changes in the oxidation states of those substances, indicating which has been oxidized and which has been reduced. Oxidation involves the loss of electrons, whereas reduction involves the gain of electrons.

In a chemical equation, one can identify a redox reaction by observing the changes in the oxidation states of the elements involved. For example, in the reaction involving chromium and iodine, chromium changes its oxidation state from +3 to +6, indicating oxidation. Meanwhile, iodine changes from -1 to +7, showing that it is also being oxidized, which is not a typical scenario as usually one element gets reduced and another gets oxidized. This dual oxidation requires careful balancing, ensuring that electron transfer is accurately represented where the oxidized states should be paired with complementary reduced species.

Oxidation States

An oxidation state, often known as an oxidation number, is a figure that illustrates the degree of oxidation of an atom in a compound. Knowing how to assign oxidation states correctly is crucial as it helps recognize how redox reactions proceed.

Here are some guidelines for assigning oxidation states:

• The oxidation state of an element in its natural form is always 0 (for example, in O₂ or Cl₂).
• For monoatomic ions, the oxidation state equals the ion charge (e.g., Na⁺ has an oxidation state of +1).
• Oxygen generally has an oxidation state of -2 in most compounds, but in peroxides, it's -1.
• Hydrogen is typically assigned an oxidation state of +1 unless it forms hydrides with metals, in which case it is -1.
• The sum of the oxidation states in a chemical compound must add up to zero or match the ion charge.

In exercises like balancing chemical reactions, observing the changes in these oxidation numbers helps determine what element is oxidized or reduced. This knowledge is fundamental in identifying which elements are losing or gaining electrons during the reaction.

Balancing Chemical Equations

Balancing chemical equations is fundamental in chemistry as it ensures the law of conservation of mass is observed: matter is neither created nor destroyed in a chemical reaction.

When balancing, each atom involved must have the same quantity on the reactant side as on the product side. Here’s a simple approach to balance chemical equations:

• Write down the unbalanced equation.
• List how many of each type of atom is on either side.
• Adjust coefficients to minimize changes until each element has equal quantities on both sides.
• Balance more complex molecules last, and leave single atoms, such as O₂ or H₂, for the final steps, making balancing simpler.
• Ensure the charge balances in redox reactions, if applicable.

For example, the given reaction a) CrI3+KOH+Cl2 -> K2CrO4+KCl+KIO4+H2O, initially appears complex due to the numerous elements and varied states. Careful monitoring of each atom's counts and their cumulative charges leads to a balanced equation, reflecting similar chemical behavior upon close examination of reactions invariably ensues to appropriate stoichiometric balance.