Redox Reactions – JEE Main ChemistryPractice Questions & PYQs

A disproportionation reaction occurs when a single element in a specific oxidation state simultaneously undergoes both oxidation and reduction, resulting in two products where the element exists in both a higher and a lower oxidation state. In the reaction $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$, the copper ion with an oxidation state of +1 is converted into copper in the +2 state and metallic copper with an oxidation state of 0, satisfying the criteria for disproportionation. Similarly, 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, which is in the +6 state, is oxidized to the +7 state in the permanganate ion and reduced to the +4 state in manganese dioxide, making this also a clear case of disproportionation. The other processes, which involve the conversion of multiple starting states to a common product or simple decomposition, do not meet the definition of disproportionation because they do not involve the simultaneous oxidation and reduction of a single species.

Equivalence in redox titrations dictates that the number of equivalents of the oxidizing agent must exactly match the equivalents of the reducing agent. Potassium permanganate reduces in an acidic medium by gaining 5 electrons, establishing an n-factor of 5, while oxalic acid releases 2 electrons during oxidation, establishing an n-factor of 2. Applying the relationship $M_1 V_1 n_1 = M_2 V_2 n_2$ with the provided values leads to the expression $M \times 2 \times 5 = 2 \times 20 \times 2$. Simplification of this equation results in $10 M = 80$, which yields a molarity of 8 M.

An oxidizing agent functions by accepting electrons, which necessitates that the species undergoes reduction to decrease its oxidation state. Within the $\mathrm{N}^{3-}$ ion, nitrogen is already at its lowest possible oxidation state of -3 because it has completed its valence shell octet. As it lacks the ability to accept more electrons, the ion cannot serve as an oxidizing agent.

This reaction relies on the ability of manganese dioxide to function as a powerful oxidizing agent in an acidic environment, which facilitates the oxidation of chloride ions into chlorine gas. This gas is characterized by its distinct greenish-yellow color, and among the provided choices, only $\mathrm{NH}_4\mathrm{Cl}$ contains the chloride ions necessary to participate in this specific redox process. The balanced chemical equation illustrating this transformation is:

$2\mathrm{NH}_4\mathrm{Cl} + \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{SO}_4 \xrightarrow{\Delta} \mathrm{MnSO}_4 + (\mathrm{NH}_4)_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O} + \mathrm{Cl}_2 \uparrow$

The varying chemical behavior of $\mathrm{S}_2\mathrm{O}_3^{2-}$ is determined by the distinct oxidizing strengths of the halogens involved. When reacting with $\mathrm{I}_2$, sulphur is only partially oxidized, shifting its average oxidation state from +2 to +2.5 to form $\mathrm{S}_4\mathrm{O}_6^{2-}$. In contrast, $\mathrm{Br}_2$ functions as a significantly more potent oxidant, driving the oxidation of sulphur all the way to a +6 state in the form of $\mathrm{SO}_4^{2-}$, which confirms that $\mathrm{Br}_2$ is a stronger oxidizing agent than $\mathrm{I}_2$.

Successful identification of the matching pairs depends on correctly aligning each element from the first group with its designated counterpart in the second. Systematic verification establishes that A pairs directly with IV, B is associated with I, C corresponds to II, and D is uniquely linked to III. This set of precise relationships confirms that the sequence (A-IV), (B-I), (C-II), (D-III) provides the valid framework for the items in question.

Redox reactions occurring in acidic media are governed by the specific stoichiometric coefficients present in their balanced chemical equations, which allow us to determine the mole ratios between the reactants. The reaction between permanganate and oxalate ions is represented by the following balanced chemical equation:

$2\mathrm{MnO}_4^{-}+5\mathrm{C}_2\mathrm{O}_4^{2-}+16\mathrm{H}^{+}\rightarrow2\mathrm{Mn}^{2+} +10\mathrm{CO}_2+8\mathrm{H}_2\mathrm{O}$

This equation indicates that 2 moles of $\mathrm{MnO}_4^{-}$ react with 16 moles of $\mathrm{H}^{+}$ ions to achieve completion. By dividing these stoichiometric coefficients to determine the specific requirement for a single mole of the oxidizing agent, we find that 1 mole of $\mathrm{MnO}_4^{-}$ consumes 8 moles of $\mathrm{H}^{+}$ ions to successfully oxidize the oxalate ion into $\mathrm{CO}_2$.

A chemical species can undergo a disproportionation reaction only if the central atom exists in an intermediate oxidation state, enabling it to act as both an oxidizing and a reducing agent simultaneously. Evaluating the provided list, $\mathrm{H}_2\mathrm{O}_2$, $\mathrm{ClO}_3^{-}$, $\mathrm{P}_4$, $\mathrm{Cl}_2$, $\mathrm{Cu}^{+1}$, and $\mathrm{NO}_2$ all fulfill this condition because their respective elements can reach both higher and lower oxidation states. Conversely, $\mathrm{Ag}$, $\mathrm{F}_2$, and $\mathrm{K}^{+}$ are excluded because they lack the necessary range of oxidation states to undergo such a reaction, resulting in a total of 6 valid species.

Disproportionation reactions represent a process where the same species undergoes both oxidation and reduction, necessitating a careful balance of atoms and charge to ensure electron conservation. We begin by noting that chlorine in $\mathrm{Cl}_2$ starts with an oxidation state of $0$, and then shifts to $-1$ in $\mathrm{Cl}^-$ and $+1$ in $\mathrm{ClO}^-$. This redox process requires that one chlorine atom accepts an electron while another releases one, which provides the stoichiometric basis for the entire reaction.

Balancing the mass and charge in the equation $\mathrm{Cl}_2 + \mathrm{OH}^- \rightarrow \mathrm{ClO}^- + \mathrm{Cl}^- + \mathrm{H}_2\mathrm{O}$ requires setting coefficients that satisfy the conservation laws. Assigning $a=1$ for $\mathrm{Cl}_2$ implies the production of $1$ molecule of $\mathrm{ClO}^-$ and $1$ ion of $\mathrm{Cl}^-$, fixing $c=1$ and $d=1$. To satisfy the oxygen and hydrogen balance along with the net charge, $2$ hydroxide ions are needed on the reactant side, establishing $b=2$, which produces $1$ water molecule on the product side. Therefore, the stoichiometric coefficients $a=1, b=2, c=1$, and $d=1$ correctly balance the reaction.

The inert pair effect, which describes the tendency of outermost $s$-electrons to remain unbonded, significantly influences the stability of oxidation states for heavier $p$-block elements, causing lower oxidation states to become more stable than higher ones as one descends a group. Because of this, thallium and lead prefer their lower oxidation states of $\mathrm{Tl}^{+}$ and $\mathrm{Pb}^{2+}$ over the higher $\mathrm{Tl}^{3+}$ and $\mathrm{Pb}^{4+}$, driving these higher-state ions to act as oxidizing agents by readily accepting electrons. Among the species provided ($\mathrm{Sn}^{4+}, \mathrm{Sn}^{2+}, \mathrm{Pb}^{2+}, \mathrm{Tl}^{3+}, \mathrm{Pb}^{4+}, \mathrm{Tl}^{+}$), only $\mathrm{Tl}^{3+}$ and $\mathrm{Pb}^{4+}$ demonstrate this behavior, resulting in a total of 2 ions that function as oxidizing agents.

To determine the oxidation state of an element within a neutral molecule, we assign standard values of 1 for hydrogen and -2 for oxygen, establishing that the algebraic sum of these states must equal zero. In bromic acid, $\mathrm{HBrO}_3$, the calculation $1 + x + 3(-2) = 0$ identifies the oxidation state of bromine as 5. Similarly, for perbromic acid, $\mathrm{HBrO}_4$, the equation $1 + x + 4(-2) = 0$ reveals that the oxidation state of bromine is 7. The sum of these two values is $5 + 7 = 12$.

Determining oxidation states involves balancing the sum of individual atomic charges with the overall charge of the chemical species, where neutral ligands such as carbonyl provide no charge contribution. In $\mathrm{Fe}(\mathrm{CO})_5$, the iron center is $0$ because the carbon monoxide ligands are neutral. The vanadyl ion $\mathrm{VO}^{2+}$ contains an oxygen atom with a $-2$ charge, requiring an oxidation state of $+4$ for vanadium to satisfy the overall $2+$ charge. For tungsten trioxide $\mathrm{WO}_3$, the three oxygen atoms contribute a total of $-6$, which dictates that tungsten must be in the $+6$ oxidation state to maintain neutrality. The total sum of these values is calculated as $0 + 4 + 6 = 10$.

The oxidation state of a central metal atom is determined by ensuring the total sum of the individual oxidation numbers in a neutral molecule equals zero. In the case of ammonium phosphomolybdate, which has the chemical formula $(\mathrm{NH}_4)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$, we focus on the $\mathrm{MoO}_3$ component to calculate the oxidation state of molybdenum. Assigning oxygen its common oxidation state of $-2$ and letting $x$ represent the oxidation state of molybdenum, we establish the equation $x + 3(-2) = 0$. Solving this equation results in $x - 6 = 0$, which confirms that the oxidation state of molybdenum is $+6$.

Chemical reactions must adhere to the Law of Conservation of Mass and Charge, ensuring that the number of atoms for every element and the total net charge are balanced on both sides. Since the reactant $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ possesses two chromium atoms, the product side must generate $2\mathrm{Cr}^{3+}$ ions, which confirms that $Y$ is 2.

The seven oxygen atoms provided by the dichromate ion necessitate the formation of $7\mathrm{H}_2\mathrm{O}$ molecules, meaning $Z$ must be 7. These seven water molecules contain 14 hydrogen atoms, requiring 14 $\mathrm{H}^{+}$ ions on the reactant side to satisfy the mass balance, thereby setting $X$ to 14. Combining these coefficients, the sum $X + Y + Z = 23$ is obtained.

This redox process is governed by the principle that total electron gain must equal total electron loss, which is directly determined by the $n$-factor of the participating species. In this reaction, $\mathrm{IO}_3^{-}$ undergoes a reduction where the oxidation state of iodine changes from $+5$ to $0$, resulting in an $n$-factor of $5$, whereas $\mathrm{I}^{-}$ is oxidized with a change of $-1$ to $0$, corresponding to an $n$-factor of $1$. To maintain electrical neutrality, these reactants must be present in a molar ratio of $1:5$, ensuring the total electrons transferred are balanced. Because the equation requires $2$ moles of $\mathrm{IO}_3^{-}$, we must utilize $2 \times 5$ moles of $\mathrm{I}^{-}$ to satisfy this requirement, which identifies $x$ as $10$.

Neutralization between a monobasic acid and sodium hydroxide follows a one-to-one stoichiometric ratio, implying the moles of acid must exactly match the moles of base. Since the base solution has a concentration of 0.24 M and a volume of 25 mL, the millimoles of base used is $0.24 \times 25 = 6$. Consequently, we require 6 millimoles of the acid. Multiplying these millimoles by the molar mass of 24.2 g/mol gives a total mass of $6 \times 24.2 = 145.2$ mg.

To determine the volume of this mass, we use the given density of 1.21 g/mL. Expressing the acid mass in grams, we have 0.1452 g. Using the relation $V = \frac{\text{mass}}{\text{density}}$, we calculate the volume as:
$V = \frac{0.1452}{1.21} = 0.12$
This result corresponds to $12 \times 10^{-2}$ mL.

Redox titrations involve the transfer of electrons, which requires indicators to be sensitive to the change in the oxidation potential of the solution, rather than the $\mathrm{pH}$. Conversely, acid-base titrations are governed by the transfer of protons, meaning their indicators function by detecting changes in the concentration of hydrogen ions, or $\mathrm{pH}$. Because the statements provided reverse these fundamental underlying principles, both assertions are incorrect.

The reduction of the permanganate ion (\text{MnO}_4^-$) \in an alkaline medium involves the conversion of manganese from the$+7$oxidation state to the$+4$oxidation state found \in manganese dioxide (\text{MnO}_2$). This chemical change is described by the following half-reaction:
$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3\text{e}^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$
Because the oxidation state of manganese decreases by 3 units during this transformation, the process necessitates the gain of 3 electrons.

Chemical reactions that involve a simultaneous transfer of electrons, characterized by changes in the oxidation states of the reacting species, are classified as redox reactions. In this specific transformation, gold changes its oxidation state from +1 in the complex $[\mathrm{Au}(\mathrm{CN})_2]^-$ to 0 in metallic $\mathrm{Au}$, while zinc shifts from an oxidation state of 0 in $\mathrm{Zn}(\mathrm{s})$ to +2 in the complex $[\mathrm{Zn}(\mathrm{CN})_4]^{2-}$. Because zinc effectively replaces gold within the cyanide complex, this process serves as a classic example of a displacement reaction occurring alongside the redox process.

The chemical structure of hypophosphoric acid, represented by the formula $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6$, is characterized by a direct phosphorus-phosphorus single bond linking two equivalent phosphorus centers. Because this P-P bond contributes zero to the oxidation states of the phosphorus atoms involved, the oxidation number for each phosphorus is determined by the oxygen and hydrogen atoms attached to it. By applying standard oxidation states of -1 to each oxygen in the hydroxyl groups and -2 to the double-bonded oxygen atom, we set up the equation $x + 2(-1) + (-2) = 0$ for each phosphorus, which identifies the oxidation state as +4.