Redox Reactions – JEE Main ChemistryPractice Questions & PYQs

The principle of chemical equivalence allows us to equate the reacting species based on their respective n-factors, which simplifies the stoichiometry of redox titrations. To find the equivalents of the permanganate ion acting as the oxidizing agent in a basic medium, we determine its n-factor, which is 3 given the reduction from $\mathrm{MnO}_4^{-}$ to $\mathrm{MnO}_2$. Using 0.2 M and 0.5 L of solution, the number of equivalents is calculated as $0.2 \times 0.5 \times 3 = 0.3$.

For the iodide solution, the n-factor is 1, and the total equivalents available are $1.5 \times 0.5 \times 1 = 0.75$. Comparing these two, the permanganate is the limiting reagent, effectively determining that 0.3 equivalents of molecular iodine are produced. This amount of iodine then reacts with an equivalent quantity of the thiosulfate titrant to reach the end point.

Equating the equivalents of the iodine generated to the equivalents of the thiosulfate consumed, we account for the known volume of 0.3 L and an n-factor of 1 for thiosulfate. The calculation for the unknown molarity $x$ is defined by $x \times 0.3 \times 1 = 0.3$. Solving this yields $x = 1$.

The disproportionation reaction of chlorine in cold, dilute potassium hydroxide is represented by the balanced equation $Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$. Initial amounts are 1 mole of $Cl_2$ and 4 moles of $OH^-$, making $Cl_2$ the limiting reagent since it requires 2 moles of $OH^-$ for complete consumption. Consequently, the reaction consumes 1 mole of $Cl_2$ and 2 moles of $OH^-$, producing 1 mole of $Cl^-$ and 1 mole of $ClO^-$, while leaving 2 moles of $OH^-$ unreacted in the solution. With a constant volume of 2 L, the final concentrations are 0.5 M for $Cl^-$, 0.5 M for $ClO^-$, and 1 M for $OH^-$.

Redox reactions are dictated by the law of equivalence, which requires that the total equivalents of reducing agents equal the total equivalents of the oxidizing agent. In this mixture containing 1 mole each of the four compounds, $\mathrm{FeC}_2\mathrm{O}_4$, $\mathrm{Fe}_2(\mathrm{C}_2\mathrm{O}_4)_3$, and $\mathrm{FeSO}_4$ serve as reducing agents, while $\mathrm{Fe}_2(\mathrm{SO}_4)_3$ contains iron in its stable +3 state and does not participate. By summing the contribution of each species, we find that $\mathrm{FeC}_2\mathrm{O}_4$ provides 3 equivalents, $\mathrm{Fe}_2(\mathrm{C}_2\mathrm{O}_4)_3$ provides 6, and $\mathrm{FeSO}_4$ provides 1, for a cumulative total of 10 equivalents. Since $\mathrm{KMnO}_4$ in an acidic medium has an $n$-factor of 5, we divide the 10 total equivalents by 5 to determine that 2 moles of $\mathrm{KMnO}_4$ are necessary for the reaction.

The redox reaction between permanganate and iodide in a basic medium proceeds according to the following balanced equation:
$2\mathrm{MnO}_4^- + 6\mathrm{I}^- + 4\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{MnO}_2 + 3\mathrm{I}_2 + 8\mathrm{OH}^-$
The initial moles are calculated as $n = M \times V$, yielding 0.1 mol of $\mathrm{MnO}_4^-$ and 0.6 mol of $\mathrm{I}^-$. Given that the stoichiometry requires 3 moles of $\mathrm{I}^-$ per mole of $\mathrm{MnO}_4^-$, the permanganate ion is the limiting reagent, which produces $0.1 \times \frac{3}{2} = 0.15$ mol of $\mathrm{I}_2$. The liberated iodine is titrated with thiosulfate according to the reaction:
$\mathrm{I}_2 + 2\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow 2\mathrm{I}^- + \mathrm{S}_4\mathrm{O}_6^{2-}$
The moles of thiosulfate consumed correspond to $0.15 \times 2 = 0.3$ mol. Using the formula $V = \frac{n}{M}$, the volume of the $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$ solution is $V = \frac{0.3}{0.1} = 3$ L.

In acidic medium, the dichromate ion ($Cr_2O_7^{2-}$) functions as a potent oxidizing agent that is reduced upon reacting with iodide ions ($I^-$). The chemical change is described by the redox equation $Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$, where the chromium atom \in the dichromate ion is reduced from an oxidation state of +6 to the +3 state found \in the chromic ion. Since the chromium product exists as the $Cr^{3+}$ species, the oxidation state of chromium in the final product is +3.

The oxidizing power of halogens decreases down the group following the sequence $\mathrm{F}_2 > \mathrm{Cl}_2 > \mathrm{Br}_2 > \mathrm{I}_2$, which means a stronger halogen can displace a weaker one from its aqueous salt solution. This hierarchy forms the basis of the "Layer test," in which chlorine water acts as an oxidizing agent to displace bromine or iodine from bromide or iodide solutions through a standard displacement redox process. Because this analytical technique relies on displacement reactions such as $\mathrm{NaBr} + \mathrm{Cl}_2 \rightarrow \mathrm{Br}_2 + \mathrm{NaCl}$ and $\mathrm{NaI} + \mathrm{Cl}_2 \rightarrow \mathrm{I}_2 + \mathrm{NaCl}$, both statements regarding the oxidizing capacity and the nature of the test are scientifically accurate.

The equivalence of a reactant in a redox titration is determined by the product of its molarity, volume, and $n$-factor, and at the equivalence point, the number of equivalents of the oxidizing agent, potassium dichromate, equals the number of equivalents of the reducing agent, Mohr's salt. The $n$-factor for potassium dichromate is 6 due to the reduction of $Cr^{6+}$ to $Cr^{3+}$, while the $n$-factor for Mohr's salt is 1 due to the oxidation of $Fe^{2+}$ to $Fe^{3+}$. Applying the relationship $M_1 V_1 n_1 = M_2 V_2 n_2$ with $M_1 = x \times 10^{-3}$, $V_1 = 200$, $n_1 = 6$, $M_2 = 0.6$, $V_2 = 750$, and $n_2 = 1$ results in:

$(x \times 10^{-3}) \times 200 \times 6 = 0.6 \times 750 \times 1$

Simplification of the equation leads to $1.2x = 450$, which gives $x = 375$.

Acidified potassium dichromate functions as an oxidizing agent, undergoing a reduction characterized by the half-reaction $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$, which necessitates the transfer of 6 electrons per mole of dichromate. For the oxidation of iodide ions, the balanced redox reaction is $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{I}^- \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{I}_2 + 7\mathrm{H}_2\mathrm{O}$, where 6 electrons are released from 6 moles of $\mathrm{I}^-$, identifying $X = 6$. Similarly, for the oxidation of sulfide ions, the balanced reaction is $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 3\mathrm{S}^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{S} + 7\mathrm{H}_2\mathrm{O}$, where 6 electrons are released from 3 moles of $\mathrm{S}^{2-}$, identifying $Y = 6$. The summation of these values yields $X + Y = 6 + 6 = 12$.

Potassium permanganate ($\mathrm{KMnO}_4$) and potassium dichromate ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$) act as powerful oxidizing agents in acidic solutions, typically driving the conversion of common reducing agents to higher oxidation states. Both reagents effectively oxidize iodide ions ($\mathrm{I}^{-}$) to iodine ($\mathrm{I}_2$), sulfide ions ($\mathrm{S}^{2-}$) to elemental sulfur ($\mathrm{S}$), and ferrous ions ($\mathrm{Fe}^{2+}$) to ferric ions ($\mathrm{Fe}^{3+}$). While some oxidants can achieve more extensive transformations, such as converting iodide to iodate or thiosulfate to sulfate, these specific reactions are not standard shared characteristics for both of these agents under typical acidic laboratory conditions.

A disproportionation reaction requires the central atom to simultaneously increase and decrease its oxidation state, which is only possible if that atom is currently in an intermediate oxidation state. If an element has already reached its maximum possible oxidation state, such as $+7$ for chlorine, it cannot be further oxidized, thereby preventing the necessary electron loss for disproportionation. Since the chlorine atom in $\mathrm{ClO}_4^-$ exists in this $+7$ maximum oxidation state, it is chemically incapable of acting as a reducing agent and consequently does not participate in these reactions.

The reaction rate is the primary factor determining the conditions for these titrations. The reaction between oxalic acid and $\mathrm{KMnO}_4$ in the presence of dilute $\mathrm{H}_2\mathrm{SO}_4$ is kinetically slow at room temperature due to a high activation energy, which necessitates an initial heating to 60 °C to allow the reaction to proceed at a measurable rate. Conversely, the reaction between ferrous ammonium sulphate and $\mathrm{KMnO}_4$ occurs spontaneously and rapidly at room temperature, rendering external heating unnecessary for the procedure.

The distinction in these protocols arises from both catalytic requirements and the need to prevent side reactions. In the titration of oxalic acid, the reaction is autocatalytic, meaning the initial formation of $\mathrm{MnSO}_4$ at higher temperatures produces the $\mathrm{Mn}^{2+}$ ions necessary to catalyze the remainder of the reaction. In contrast, heating a solution of ferrous ammonium sulphate is detrimental because it promotes the unwanted oxidation of $\mathrm{Fe}^{2+}$ ions to $\mathrm{Fe}^{3+}$ ions by atmospheric oxygen, an effect that introduces significant experimental error by altering the stoichiometry of the titration.

In redox stoichiometry, the conservation of matter and electrons dictates that the number of moles of each reactant consumed must align with the molar coefficients of the balanced equation. Given the reaction $5\mathrm{I}^{-} + \mathrm{IO}_3^{-} + 6\mathrm{H}^{+} \longrightarrow 3\mathrm{I}_2 + 3\mathrm{H}_2\mathrm{O}$, the stoichiometry establishes a clear ratio between the reactants and products. A solution containing 200 mL of 0.1 M KI provides 0.02 moles of $\mathrm{I}^{-}$, which according to the 5:1 molar ratio, reacts precisely with 0.004 moles of $\mathrm{KIO}_3$, confirming the first statement. Regarding the equivalent weight of $\mathrm{KIO}_3$, the oxidation state of iodine changes from +5 in the iodate ion to 0 in the diatomic iodine molecule, indicating a change of 5 electrons per iodine atom; thus, the n-factor is 5, making the equivalent weight equal to the molecular weight divided by 5.

Calculations for the remaining scenarios reveal they are incorrect based on the same stoichiometric proportions. For the consumption of acid, 0.02 moles of $\mathrm{I}^{-}$ require 0.024 moles of $\mathrm{H}^{+}$ ions, which, because each mole of $\mathrm{H}_2\mathrm{SO}_4$ yields two moles of $\mathrm{H}^{+}$, necessitates 0.012 moles of $\mathrm{H}_2\mathrm{SO}_4$ rather than the 0.006 moles stated in the options. Similarly, reacting 0.5 L of a 0.1 M KI solution, which contains 0.05 moles of $\mathrm{I}^{-}$, should produce 0.03 moles of $\mathrm{I}_2$ based on the 5:3 stoichiometric ratio, which contradicts the claim that only 0.005 moles of $\mathrm{I}_2$ would be produced.

Redox reactions are systematically categorized based on how the reactants rearrange and interact to form products. By examining the oxidation state shifts and the structural changes occurring in each chemical process, we can determine the specific type of transformation taking place.

In the first reaction, $\mathrm{CH}_{4(g)} + 2 \mathrm{O}_{2(g)} \xrightarrow{\Delta} \mathrm{CO}_{2(g)} + 2 \mathrm{H}_2\mathrm{O}_{(l)}$, the reactants undergo combustion to produce carbon dioxide and water, which aligns with the classification of a combination-type redox process. Regarding the thermal breakdown of sodium hydride, $2 \mathrm{NaH}_{(s)} \xrightarrow{\Delta} 2 \mathrm{Na}_{(s)} + \mathrm{H}_{2(g)}$, a single compound splits into its constituent elements, characterizing it as a decomposition reaction.

When calcium reacts with vanadium(V) oxide, $\mathrm{V}_2\mathrm{O}_{5(s)} + 5 \mathrm{Ca}_{(s)} \xrightarrow{\Delta} 2 \mathrm{V}_{(s)} + 5 \mathrm{CaO}_{(s)}$, the calcium atoms replace the vanadium atoms in the oxide lattice, fitting the criteria for a displacement reaction. Finally, the decomposition of hydrogen peroxide, $2 \mathrm{H}_2\mathrm{O}_{2(aq)} \xrightarrow{\Delta} 2 \mathrm{H}_2\mathrm{O}_{(l)} + \mathrm{O}_{2(g)}$, involves oxygen atoms simultaneously increasing and decreasing in oxidation state to yield water and oxygen gas, which is the definitive characteristic of a disproportionation reaction.

The core principle relies on stoichiometric titration analysis to determine the amount of carbon dioxide consumed by linking the final neutralisation data back to the initial reaction conditions. We begin by calculating the moles of hydrochloric acid used to neutralize the excess base, which is found by multiplying the concentration by the volume as $0.1 \times 0.04 = 0.004$ moles. Based on the chemical equation $\mathrm{Ca}(\mathrm{OH})_2 + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_2 + 2\mathrm{H}_2\mathrm{O}$, the stoichiometric ratio implies the remaining base is $0.004 / 2 = 0.002$ moles.

Since the carbon dioxide reacted with half the initial calcium hydroxide and the excess base was neutralized by the acid, the moles of $\mathrm{CO}_2$ must be equal to the moles of the excess base, which is 0.002 moles. Knowing this molar amount, we apply the molar volume of a gas at standard temperature and pressure, where one mole occupies 22.4 liters. Multiplying the moles of gas by this factor gives $0.002 \times 22.4 = 0.0448$ liters, which is equivalent to 44.8 cubic centimeters. Rounding this result to the nearest integer provides the final volume of 45.

In acidic medium, $\mathrm{KMnO}_4$ acts as an oxidizing agent where manganese is reduced from the $+7$ oxidation state in $\mathrm{MnO}_4^-$ to the $+2$ oxidation state in $\mathrm{Mn}^{2+}$, resulting in a difference of $X = (+7) - (+2) = 5$. The acetate ion test with neutral ferric chloride produces a brown-red precipitate of basic ferric acetate, often represented by the complex $[\mathrm{Fe}_3(\mathrm{OH})_2(\mathrm{CH}_3\mathrm{COO})_6]^+$, wherein the iron remains in the $\mathrm{Fe}^{3+}$ oxidation state. The ground state electronic configuration of the iron atom is $[Ar] 3d^6 4s^2$, and the $\mathrm{Fe}^{3+}$ ion is formed by losing three electrons to achieve a configuration of $[Ar] 3d^5$, which contains $5$ electrons in the $d$-subshell, yielding $Y = 5$. Summing these calculated values, $X + Y = 5 + 5 = 10$.

The balancing of the dichromate half-reaction in an acidic medium relies on the principles of mass and charge conservation. Dichromate ions, $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, undergo reduction where each chromium atom transitions from an oxidation state of $+6$ to $+3$, requiring a total of $6\mathrm{e}^{-}$ to form $2\mathrm{Cr}^{3+}$ ions. Because the reactant contains seven oxygen atoms, seven $\mathrm{H}_2\mathrm{O}$ molecules must be produced on the right side, which necessitates the presence of $14\mathrm{H}^{+}$ ions on the reactant side to balance the hydrogen atoms. Putting these coefficients into the half-reaction yields the balanced equation:
$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$
Comparing this result to the given template where the reaction is $\mathrm{Cr}_2\mathrm{O}_7^{2-} + X\mathrm{H}^{+} + Y\mathrm{e}^{-} \rightarrow 2A + Z\mathrm{H}_2\mathrm{O}$, we identify $X = 14$, $Y = 6$, $Z = 7$, and $A = \mathrm{Cr}^{3+}$.

The permanganate ion functions as a potent oxidizing agent in an alkaline solution, undergoing reduction to form manganese dioxide. As it accepts electrons from the iodide ion, the iodide is pushed to a higher oxidation state. Specifically, the reaction between $\mathrm{MnO}_4^-$ and $\mathrm{I}^-$ in this basic environment proceeds such that the iodide is oxidized to the iodate ion:

$2\mathrm{MnO}_4^{-} + \mathrm{H}_2\mathrm{O} + \mathrm{I}^{-} \rightarrow 2\mathrm{MnO}_2 + 2\mathrm{OH}^{-} + \mathrm{IO}_3^{-}$

The reaction between potassium dichromate, potassium chloride, and concentrated sulfuric acid is a classic chemical procedure known as the chromyl chloride test, which results in the formation of chromyl chloride ($CrO_2Cl_2$) as the primary chromium-containing product. By applying standard oxidation rules to this neutral molecule, we assign oxygen an oxidation state of -2 and chlorine an oxidation state of -1. Solving for the unknown oxidation state $x$ of chromium using the mathematical expression $x + 2(-2) + 2(-1) = 0$ confirms that the chromium atom maintains an oxidation state of +6.

Oxidation-reduction reactions rely on balancing the exchange of electrons between reactants in a basic medium. The reduction half-reaction involves permanganate ions gaining electrons to form manganese dioxide, which requires water molecules and produces hydroxide ions to account for the change in oxidation states and charge balance: $2\,\mathrm{MnO}_4^{-} + 4\,\mathrm{H}_2\mathrm{O} + 6\,\mathrm{e}^{-} \rightarrow 2\,\mathrm{MnO}_2 + 8\,\mathrm{OH}^{-}$ To satisfy the electron exchange, the oxidation half-reaction consumes iodide ions, which lose electrons to form iodine: $6\,\mathrm{I}^{-} \rightarrow 3\,\mathrm{I}_2 + 6\,\mathrm{e}^{-}$ By adding these two balanced half-reactions and eliminating the electrons, we arrive at the complete stoichiometric equation: $2\,\mathrm{MnO}_4^{-} + 6\,\mathrm{I}^{-} + 4\,\mathrm{H}_2\mathrm{O} \rightarrow 3\,\mathrm{I}_2 + 2\,\mathrm{MnO}_2 + 8\,\mathrm{OH}^{-}$ This balanced equation confirms that the integer coefficient $z$ for the hydroxide ions is 8.

The brown ring test produces the coordination complex $[\mathrm{Fe}(\mathrm{H_2O})_5(\mathrm{NO})]^{2+}$. In this structure, the five water molecules function as neutral ligands, and the nitrosyl ligand is treated as $\mathrm{NO}^+$ to account for the specific electronic environment of the central metal. By balancing the sum of the oxidation states of these ligands with the total complex charge of +2, we determine that the oxidation number of the iron central atom is +1.