Redox Reactions – JEE Main ChemistryPractice Questions & PYQs

The core of neutralization relies on the law of equivalence, which states that at the point of complete reaction, the number of equivalents of acid must balance exactly with the number of equivalents of base. We calculate the equivalents of any substance by taking the product of its molarity, volume, and n-factor, where the n-factor corresponds to the basicity of the acid or the number of hydroxide ions released by the base.

For the base $\mathrm{M}(\mathrm{OH})_2$, the n-factor is 2, so the total equivalents provided by 30 mL of a 0.05 M solution is calculated as $30 \times 0.05 \times 2$. Because the acid is fully neutralized, these equivalents must match the equivalents of the acid, expressed as $10 \times 0.1 \times n$, where $n$ represents the basicity we aim to find. Equating these gives the relationship $0.1 \times 10 \times n = 30 \times 0.05 \times 2$ which simplifies to $n = 3$. This indicates that the acid 'A' has a basicity of 3.

A disproportionation reaction occurs when a single element in a chemical species simultaneously undergoes both oxidation and reduction, moving from an intermediate oxidation state to both higher and lower states. In the reaction $3 \mathrm{MnO}_4^{2-} + 4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_4^{-} + \mathrm{MnO}_2 + 2 \mathrm{H}_2\mathrm{O}$, the manganese atom in the manganate ion, $\mathrm{MnO}_4^{2-}$, possesses an intermediate oxidation state of 6. Throughout this process, this manganese is oxidized to the 7 state in the permanganate ion, $\mathrm{MnO}_4^{-}$, and concurrently reduced to the 4 state in manganese dioxide, $\mathrm{MnO}_2$. Because the same element is split into higher and lower oxidation states from its intermediate starting point, this reaction exemplifies disproportionation.

The reaction between a weak acid like $\mathrm{CH}_3\mathrm{COOH}$ and a strong base like $\mathrm{NaOH}$ produces a salt that undergoes hydrolysis, resulting in an equivalence point with a pH in the basic range, typically between 8 and 10. An effective indicator for this titration must show a distinct color change within this specific alkaline environment, which necessitates that the indicator exists in its ionized form under basic conditions. Phenolphthalein functions as a weak acid that satisfies these requirements, remaining colourless in acidic solutions while transitioning to its ionized, coloured state as the solution becomes basic.

The volumetric titration process relies on the principle of chemical equivalence, where the number of equivalents of the titrant used equals the number of equivalents of the analyte present. In this specific iodometric procedure, copper(II) ions react with excess iodide to generate iodine, which is subsequently quantified by titration against a sodium thiosulfate solution. Since the stoichiometry dictates that the equivalents of iodine produced correspond exactly to the equivalents of the copper(II) ions initially present, the calculation depends on equating these quantities to the equivalents of the hypo solution used.

Calculating the milliequivalents of the hypo solution involves multiplying its molarity by the volume, yielding $0.02\ \mathrm{M} \times 20\ \mathrm{mL} = 0.4\ \mathrm{meq}$. Because this value must equal the milliequivalents of copper(II) in the $10\ \mathrm{mL}$ sample, and the reduction of $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}^{+}$ involves a single electron transfer per ion, the n-factor is 1, making the millimoles of copper(II) equal to the milliequivalents. Consequently, there are $0.4\ \mathrm{mmol}$ of copper(II) in the sample. Dividing these $0.4\ \mathrm{mmol}$ by the $10\ \mathrm{mL}$ volume of the solution provides a molarity of $0.04\ \mathrm{M}$, which is expressed as $4 \times 10^{-2}\ \mathrm{M}$.

Phenolphthalein functions as a pH-sensitive indicator because its ionization state changes depending on the acidity of the surrounding medium, confirming that it is colorless in acidic solutions and turns pink in basic environments. As a weak acid, it must undergo dissociation in a basic solution to produce the $\mathrm{Ph}^{-}$ conjugate base that gives the liquid its characteristic pink color, represented by the equilibrium $\mathrm{HPh} \rightleftharpoons \mathrm{H}^{+} + \mathrm{Ph}^{-}$. Therefore, the assertion is accurate, but the reasoning is invalid because it falsely claims that the indicator does not dissociate in a basic medium.

The fundamental principle for this redox titration is that at the equivalence point, the number of milliequivalents of the oxidizing agent must equal the number of milliequivalents of the reducing agent. In this reaction, potassium permanganate acts as the oxidizing agent in an acidic environment, where the permanganate ion gains five electrons, giving it a valence factor of 5. Conversely, Mohr's salt acts as the reducing agent because the ferrous iron is oxidized to ferric iron, which corresponds to a valence factor of 1.

Applying the law of equivalence with the relationship $M \times V \times n_F = M \times V \times n_F$, we substitute the known values into the equation $0.01 \times V \times 5 = 0.05 \times 20 \times 1$. Solving for the volume $V$ reveals that 20 mL of the potassium permanganate solution was consumed during the titration. Since the burette initially contained 50 mL, subtracting the consumed volume from the initial amount leaves 30 mL remaining in the burette.

A disproportionation reaction occurs when the same element within a reactant is simultaneously oxidized and reduced, resulting in products where that element exhibits both higher and lower oxidation states. In the reaction
$\mathrm{MnO}_4^{-} + 4\,\mathrm{H}^{+} + 3\,\mathrm{e}^{-} \rightarrow \mathrm{MnO}_2 + 2\,\mathrm{H}_2\,\mathrm{O}$
the manganese atom solely undergoes reduction as its oxidation state decreases from +7 to +4 through the gain of electrons. Because this mechanism lacks the necessary simultaneous oxidation of manganese, it is classified as a reduction process rather than a disproportionation reaction.

The normality of a mixture depends on the net number of equivalents remaining after the acid and base neutralize each other. The milliequivalents of $\mathrm{H}_2\mathrm{SO}_4$ are calculated as $100 \times 0.1 \times 2 = 20$, while the milliequivalents of $\mathrm{NaOH}$ are $50 \times 0.1 \times 1 = 5$. Subtracting the base equivalents from the acid equivalents leaves $15$ milliequivalents of excess $\mathrm{H}_2\mathrm{SO}_4$. Dividing this value by the total volume of $150\ \mathrm{mL}$ results in the final normality calculation:

$N = \frac{15}{150} = 0.1\ \mathrm{N}$

This equates to $1 \times 10^{-1}\ \mathrm{N}$, meaning the required integer is 1.

Sodium hydride, denoted as $NaH$, behaves exclusively as a reducing agent because the hydrogen atom exists in the $-1$ oxidation state, meaning it cannot accept additional electrons to be reduced further. Given that sodium is present in the $+1$ state, the characterization of $NaH$ as an oxidising agent is chemically incorrect. In the case of pyridine, the lone pair of electrons on the nitrogen atom resides in an $sp^2$ hybrid orbital that does not participate in the aromatic $\pi$ system, allowing it to remain localized and available to accept a $H^{+}$ ion. This availability of the electron pair is exactly what defines pyridine as a basic substance.

Determining the change in oxidation state involves finding the numerical difference between the initial and final oxidation states of the central atom. Within the permanganate ion $\mathrm{MnO}_4^{-}$, manganese occupies a +7 oxidation state, which transitions to +2 upon forming the $\mathrm{Mn}^{2+}$ ion. This absolute difference of 5 corresponds exactly to the gain of five electrons required during the reduction process.

When dichromate ions encounter a basic solution, they undergo a transformation into chromate ions ($\mathrm{CrO}_{4}^{2-}$). To find the oxidation state of chromium in this species, we assign each oxygen atom an oxidation number of -2 and calculate the value $x$ using the equation $x + 4(-2) = -2$. This algebraic calculation results in $x = +6$, showing that the chromium atom maintains an oxidation state of 6 throughout the process.

Redox titrations rely on the fundamental principle that the number of equivalents of the oxidizing agent must equal the number of equivalents of the reducing agent at the equivalence point. Because $\mathrm{KMnO}_4$ functions with an n-factor of 5 in an acidic medium and oxalic acid dihydrate reacts with an n-factor of 2, the relationship between their concentrations and volumes is expressed by the equation $\frac{10 \times 0.05}{1000} \times 5 = \frac{10 \times M}{1000} \times 2$. Solving for $M$ identifies the molarity of the oxalic acid solution as 0.125 mol/L.

Determining the strength of the solution in grams per liter requires multiplying the molarity by the molar mass of oxalic acid dihydrate, which is 126 g/mol. Multiplying 0.125 by 126 yields 15.75 g/L. To express this in the specified format, the final value becomes $1575 \times 10^{-2}$ g/L.

In neutral molecules, the sum of all oxidation numbers must equal zero, with hydrogen typically assigned +1 and oxygen -2. For $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$, the balance is defined by the following equation:
$4(+1) + 2x + 7(-2) = 0$
Solving this yields $2x - 10 = 0$, so $x = +5$. Similarly, for $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_5$, the calculation follows the same principle:
$4(+1) + 2x + 5(-2) = 0$
This simplifies to $2x - 6 = 0$, resulting in $x = +3$. For the final compound, $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_6$, the expression is:
$4(+1) + 2x + 6(-2) = 0$
This simplifies to $2x - 8 = 0$, confirming that $x = +4$.

Determining the oxidation state of nitrogen involves setting the oxidation state of oxygen to -2 and ensuring that the sum of the oxidation numbers across all atoms equals the total charge on the species. Applying this calculation, $\mathrm{NO}$ yields $x + (-2) = 0$, resulting in +2 for nitrogen. In $\mathrm{NO}_2$, the calculation is $x + 2(-2) = 0$, which gives a state of +4, while for $\mathrm{N}_2\mathrm{O}$, we solve $2x + (-2) = 0$ to find that nitrogen is +1. Finally, for the nitrate ion $\mathrm{NO}_3^{-}$, the equation $x + 3(-2) = -1$ leads to an oxidation state of +5, confirming the decreasing order of oxidation states as $\mathrm{NO}_3^{-} > \mathrm{NO}_2 > \mathrm{NO} > \mathrm{N}_2\mathrm{O}$.

The determination of $\mathrm{Na}_2\mathrm{CO}_3$ in a mixture with $\mathrm{NaOH}$ relies on the specific neutralization stages indicated by phenolphthalein and methyl orange. Phenolphthalein signals the completion of $\mathrm{NaOH}$ neutralization and the conversion of $\mathrm{Na}_2\mathrm{CO}_3$ into $\mathrm{NaHCO}_3$. The subsequent addition of methyl orange completes the conversion of $\mathrm{NaHCO}_3$ into $\mathrm{NaCl}$, $\mathrm{H}_2\mathrm{O}$, and $\mathrm{CO}_2$. Because each mole of $\mathrm{Na}_2\mathrm{CO}_3$ consumes one mole of $\mathrm{HCl}$ to form $\mathrm{NaHCO}_3$ and another to reach the final endpoint, the volume of acid used in the second stage is directly proportional to the amount of $\mathrm{Na}_2\mathrm{CO}_3$ in the original sample.

Given that the second titration requires 1.5 mL of 0.1 N $\mathrm{HCl}$, the equivalents of $\mathrm{HCl}$ used are calculated as:
$\text{Equivalents} = 0.1 \times \frac{1.5}{1000} = 1.5 \times 10^{-4}$
Since the n-factor for $\mathrm{Na}_2\mathrm{CO}_3$ is 1, its equivalent weight equals its molar mass of 106 g/mol. Multiplying the number of equivalents by the equivalent weight yields the mass of $\mathrm{Na}_2\mathrm{CO}_3$:
$\text{Mass} = 1.5 \times 10^{-4} \times 106 = 0.0159 \text{ g}$
Determining the weight percentage in the 0.4 g mixture involves the following calculation:
$\text{Weight \%} = \frac{0.0159}{0.4} \times 100 = 3.975\%$
Rounding 3.975% to the nearest integer gives 4%.

A redox reaction is fundamentally characterized by a change in the oxidation numbers of the reacting elements, as electrons are transferred between species. The reaction of dinitrogen with dioxygen at 2000 K, represented as $\mathrm{N}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{NO}$, exemplifies this because nitrogen is oxidized from 0 to +2 and oxygen is reduced from 0 to -2. In contrast, the formation of ozone, the acid-base neutralization between $\mathrm{H}_2\mathrm{SO}_4$ and $\mathrm{NaOH}$, and the double displacement reaction involving the complex $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]\mathrm{Cl}_3$ and $\mathrm{AgNO}_3$ all involve reactant rearrangements without any shifts in oxidation states, thereby excluding them from being classified as redox reactions.

Normality is defined as the number of gram equivalents per liter of solution. With a solution volume of 100 mL at a concentration of 0.1 N, the total amount of solute corresponds to 0.01 gram equivalents, derived by multiplying the normality by the volume in liters.

Gram equivalents are calculated by dividing the mass of the solute by its equivalent weight. The compound $\mathrm{Na}_2\mathrm{CO}_3 \cdot x\mathrm{H}_2\mathrm{O}$ has a molar mass of $106 + 18x$, calculated from the atomic weights of its constituents. Because the salt dissociates into ions carrying a total charge of 2 per formula unit, the $n$-factor is 2, making the equivalent weight $\frac{106 + 18x}{2}$, which simplifies to $53 + 9x$.

Equating the gram equivalents to the mass divided by the equivalent weight produces the following relationship:
$0.01 = \frac{1.43}{53 + 9x}$
Multiplying both sides by the denominator gives $53 + 9x = 143$. Subtracting 53 from both sides results in $9x = 90$, which identifies $x$ as 10.

Whether a chemical species functions as an oxidizing agent, a reducing agent, or both depends on the range of possible oxidation states for its central atom. Compounds such as $H_2O_2$, $H_2SO_3$, and $HNO_2$ contain central atoms capable of both increasing and decreasing their oxidation state, which enables them to act as either oxidizing or reducing agents depending on the specific reaction conditions. In contrast, the phosphorus in $H_3PO_4$ has already reached its highest possible oxidation state, meaning it cannot lose more electrons and therefore cannot act as a reducing agent, limiting its role strictly to that of an oxidizing agent.

Titration of dilute $\mathrm{HCl}$ with aqueous $\mathrm{NaOH}$ is a neutralization process occurring spontaneously at room temperature, represented by the equation $\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}$, meaning a Bunsen burner is entirely unnecessary for the procedure. Because titrations demand high precision, a measuring cylinder is also inappropriate as it lacks the accuracy provided by essential volumetric glassware such as a pipette and burette. The experimental setup instead utilizes a clamp for support, a porcelain tile for visualizing the phenolphthalein endpoint, and distilled water for rinsing, making a Bunsen burner and measuring cylinder the only items not required for the process.

The core principle governing this reaction is the law of equivalence, which dictates that the number of gram equivalents of the oxidizing agent must equal the number of gram equivalents of the reducing agent. Potassium dichromate ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$) participates in the acidic medium with an n-factor of 6, while ferrous oxalate ($\mathrm{FeC}_2\mathrm{O}_4$) behaves as the reducing agent with an n-factor of 3, derived from the oxidation of iron from $+2$ to $+3$ and two carbon atoms from $+3$ to $+4$.

The molar mass of $\mathrm{FeC}_2\mathrm{O}_4$ is calculated as $56 + 2(12) + 4(16) = 144 \mathrm{g \, mol^{-1}}$. With a sample mass of $0.288 \mathrm{g}$, the moles of ferrous oxalate present equal $\frac{0.288}{144} = 0.002 \mathrm{mol}$. Multiplying this by its n-factor of 3 yields $0.006$ gram equivalents. Since the equivalents of potassium dichromate are defined by the product of its molarity, n-factor, and volume in liters, the relationship $0.02 \times 6 \times V = 0.006$ holds true. Solving for $V$ gives $0.05 \mathrm{L}$, or $50.00 \mathrm{mL}$.