Redox Reactions – JEE Main ChemistryPractice Questions & PYQs

Potassium belongs to the alkali metal group, which characteristically exhibits an oxidation state of +1 in all its combined states. This property is fixed for all alkali metals and does not vary regardless of whether the metal is part of a standard oxide, a peroxide, or a superoxide. Consequently, in the compounds $\mathrm{K}_2\mathrm{O}$, $\mathrm{K}_2\mathrm{O}_2$, and $\mathrm{KO}_2$, the oxidation number of potassium remains +1 in every case.

The determination of oxidation states relies on the fundamental rule that the algebraic sum of oxidation numbers for all atoms in a neutral molecule is zero. By assigning the established oxidation state of $+1$ to potassium and $-2$ to oxygen, one can calculate the specific oxidation state for each transition metal by solving the resulting linear equation.

For ${K}_{2} {Cr}_{2} {O}_{7}$, the equation $2(+1) + 2x + 7(-2) = 0$ simplifies to $2 + 2x - 14 = 0$, which indicates that $x = 6$. In ${KMnO}_{4}$, the setup $1(+1) + y + 4(-2) = 0$ leads directly to $y = 7$. Finally, in ${K}_{2} {FeO}_{4}$, the balance $2(+1) + z + 4(-2) = 0$ confirms that $z = 6$. Summing these three oxidation states yields $6 + 7 + 6$, which results in a final value of 19.00.

To determine the relative oxidation states of nitrogen in these compounds, we assign oxygen a standard oxidation state of -2 and calculate the charge of the nitrogen atom in each molecule. In $N_2O$, two nitrogen atoms balance one oxygen atom for an oxidation state of +1, while in $NO$, nitrogen carries an oxidation state of +2. In $N_2O_3$, two nitrogen atoms balance three oxygen atoms resulting in a +3 oxidation state, and in $NO_2$, nitrogen balances two oxygen atoms for a +4 oxidation state. Arranging these calculated values from lowest to highest produces the sequence $N_2O < NO < N_2O_3 < NO_2$.

Neutralization reactions depend on the stoichiometric relationship where the number of milliequivalents of the acid equals the number of milliequivalents of the base. Based on the balanced reaction $H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$, each mole of oxalic acid consumes two moles of sodium hydroxide. The product of volume, molarity, and n-factor for both components results in $50 \times 0.5 \times 2 = 25 \times M_{NaOH} \times 1$, which confirms the concentration of the base is $2 \text{ M}$. A $2 \text{ M}$ solution contains $2 \times 40 = 80$ grams of sodium hydroxide per liter, given the molar mass is $40 \text{ g/mol}$. In 50 mL of this solution, the total amount of sodium hydroxide is $80 \times \frac{50}{1000} = 4$ grams. Because 4 grams is not represented among the offered values, none of the options are correct.

Hardness of water is conventionally expressed as the amount of calcium carbonate, $CaCO_3$, equivalent to the concentration of dissolved calcium salts in the sample. Since the concentration of $CaSO_4$ is $10^{-3}$ M, the molarity of calcium ions is $10^{-3}$ moles per liter. By multiplying this molarity by the molar mass of $CaCO_3$, which is 100 grams per mole, we determine the hardness in grams per liter. Converting this concentration into milligrams per liter yields 100 mg/L, which is equal to 100 ppm, while noting that the molar mass of $CaSO_4$ is unnecessary for this calculation.

The fundamental principle of redox stoichiometry requires that the total number of electrons lost by the reducing agent must exactly equal the total number of electrons gained by the oxidizing agent. In the balanced chemical equation $2\mathrm{MnO}_4^{-} + 5\mathrm{C_2O_4}^{2-} + 16\mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}$, the overall reaction involves the transfer of 10 electrons, which corresponds to the production of 10 molecules of $\mathrm{CO}_2$. By distributing the 10 total electrons across the 10 produced molecules of $\mathrm{CO}_2$, it follows that the number of electrons involved per individual molecule of $\mathrm{CO}_2$ is 1.

Disproportionation occurs when a chemical species simultaneously undergoes oxidation and reduction, such that the same element in a single starting oxidation state ends up in two distinct oxidation states in the products. In the reaction $2\ \mathrm{CuBr} \rightarrow \mathrm{CuBr}_2 + \mathrm{Cu}$, the copper in $\mathrm{CuBr}$ starts with an oxidation state of $+1$. It is simultaneously oxidized to the $+2$ state in $\mathrm{CuBr}_2$ and reduced to the $0$ state in $\mathrm{Cu}$, confirming that this specific transformation satisfies the criteria for a disproportionation reaction.

Oxidation-reduction stoichiometry is based on the principle of equivalence, where the total equivalents of the oxidant must balance the total equivalents of the reductants. The number of equivalents for any species is obtained by multiplying the number of moles by its specific valence factor. For the given mixture of 1 mole each, the total equivalents from the reducing agents are calculated by summing the products for $FeC_2O_4$, $Fe_2(C_2O_4)_3$, $FeSO_4$, and $Fe_2(SO_4)_3$, which yields $1 \times 3 + 1 \times 6 + 1 \times 1 + 1 \times 0 = 10$ equivalents. Since the valence factor of $KMnO_4$ is 5, the moles required can be found by dividing these 10 equivalents by 5, resulting in 2 moles of $KMnO_4$.

A redox reaction is characterized by the simultaneous transfer of electrons between substances, which results in a measurable change in their oxidation states. In the reaction $\mathrm{XeF}_4 + \mathrm{O}_2\mathrm{F}_2 \rightarrow \mathrm{XeF}_6 + \mathrm{O}_2$, the oxidation state of xenon increases from +4 to +6, confirming that it undergoes oxidation. Simultaneously, the oxidation state of oxygen decreases from +1 to 0, indicating that it undergoes reduction, which classifies this process as a redox reaction.

Neutralization occurs when the gram equivalents of the acid equal the gram equivalents of the base, calculated using the relationship $N \times V = \text{Equivalents}$. The base furnishes 0.04 moles of $OH^-$ ions, which corresponds to 0.04 equivalents since the valency factor of the hydroxide ion is 1. For a dibasic acid, the normality is derived by multiplying the molarity by the n-factor of 2; assuming the given concentration 0.1 refers to molarity, the effective normality is $N = 0.1 \times 2 = 0.2$. Substituting these values into the neutralization equation gives $0.2 \times V = 0.04$, which results in $V = 0.2$ L, or 200 mL.

A substance acts as a reducing agent when it undergoes oxidation, a chemical process fundamentally defined by the loss of electrons. By examining the provided reactions, we can see that in equations (b) and (d), $\mathrm{H_2O_2} - 2e^{-} \rightarrow \mathrm{O_2} + 2\mathrm{H}^{+}$ and $\mathrm{H_2O_2} + 2\mathrm{OH}^{-} - 2e^{-} \rightarrow \mathrm{O_2} + 2\mathrm{H_2O}$, the molecule releases electrons, thereby functioning as a reducing agent. Conversely, reactions (a) and (c) involve the gain of electrons, identifying $\mathrm{H_2O_2}$ as an oxidizing agent in those specific instances.

Balancing a redox reaction relies on the principle that the total number of electrons lost during oxidation must equal the total number of electrons gained during reduction. Splitting the process into two half-reactions reveals that permanganate ions are reduced to manganese(II) ions, requiring five electrons, while oxalate ions are oxidized to carbon dioxide, releasing two electrons per ion. To ensure the electron count balances, the reduction half-reaction is multiplied by two and the oxidation half-reaction by five. Adding these balanced halves together accounts for all atoms and charges, confirming the stoichiometric coefficients as $x = 2$, $y = 5$, and $z = 16$, which completes the reaction as:

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

A chemical species exhibits oxidizing behavior by undergoing reduction, which involves a decrease in the oxidation state of one of its atoms as it promotes the oxidation of another substance. Within the reaction $2\mathrm{HI} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{I}_2 + \mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O}$, the sulfur atom in sulfuric acid experiences a reduction in its oxidation number from +6 to +4 as it forms $\mathrm{SO}_2$. Because this reduction process coincides with the iodine in $\mathrm{HI}$ increasing its oxidation state from -1 to 0, sulfuric acid confirms its role as an effective oxidizing agent.