Redox Reactions – NEET UG ChemistryPractice Questions & PYQs

To determine the oxidation state of $\mathrm{Cr}$ in this molecule, we must recognize that $\mathrm{CrO}_6$ is structurally composed of three peroxide groups, making its formula effectively $\mathrm{Cr}(\mathrm{O}_2)_3$. Within this framework, each peroxo unit, represented as $\mathrm{O}_2^{2-}$, contributes a total charge of -2. Consequently, the three peroxide units impose a combined negative charge of -6 on the compound. Since the entire molecule is electrically neutral, the $\mathrm{Cr}$ atom must therefore be in a +6 oxidation state to achieve charge balance.

Chemical species can coexist in a single mixture only if they are unable to undergo a spontaneous oxidation-reduction reaction with one another. While combinations involving $\mathrm{FeCl}_3$, $\mathrm{HgCl}_2$, or $\mathrm{KI}$ contain agents that readily exchange electrons, $\mathrm{FeCl}_2$ and $\mathrm{SnCl}_2$ are chemically stable together. Because $\mathrm{Sn}^{+2}$ cannot reduce $\mathrm{Fe}^{+2}$, no spontaneous electron transfer occurs between these two compounds, allowing them to remain in the same solution without reacting.

A reducing agent is defined by its ability to facilitate the reduction of another species while undergoing oxidation itself. In the reaction $\mathrm{H}_2\mathrm{O}_2 + \mathrm{O}_3 \rightarrow \mathrm{H}_2\mathrm{O} + 2\mathrm{O}_2$, ozone undergoes reduction, identifying hydrogen peroxide as the reducing agent. Similarly, in the reaction $\mathrm{H}_2\mathrm{O}_2 + \mathrm{Ag}_2\mathrm{O} \rightarrow 2\mathrm{Ag} + \mathrm{H}_2\mathrm{O} + \mathrm{O}_2$, the conversion of silver oxide into metallic silver demonstrates that hydrogen peroxide is once again acting as a reducing agent in both chemical processes.

Identifying the element with the largest shift in oxidation state requires tracking the change in electron distribution for each species involved in the reaction:

$\mathrm{KClO}_3 + (\mathrm{COOH})_2 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{K}_2\mathrm{SO}_4 + \mathrm{KCl} + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$

The chlorine atom in $\mathrm{KClO}_3$ decreases from an oxidation state of +5 to -1 in $\mathrm{KCl}$, reflecting a total change of 6 units. Since carbon atoms in the oxalic acid only shift from +3 to +4 and the other elements show minimal or no change, chlorine undergoes the maximum variation in its oxidation number.

Assigning hydrogen a value of $+1$ and oxygen a value of $-2$ allows for the determination of nitrogen's oxidation state ($x$) by ensuring the total sum of charges equals zero. In hydrazine ($N_2H_4$), the equation $2x + 4(+1) = 0$ yields an oxidation state of $x = -2$, while ammonia ($NH_3$) follows $x + 3(+1) = 0$, resulting in $x = -3$. For hydroxylamine ($NH_2OH$), the sum of components $x + 2(+1) + (-2) + (+1) = 0$ simplifies to $x + 1 = 0$, which gives $x = -1$. Evaluating hydrazoic acid ($N_3H$) requires solving $3x + 1(+1) = 0$, leading to $x = -1/3$. Comparing these values reveals that $-1/3$ is the highest algebraic value among $-1$, $-2$, and $-3$, making $N_3H$ the compound where nitrogen exhibits its highest oxidation state.

In chemical reactions where a single element undergoes simultaneous oxidation and reduction, we refer to the process as a disproportionation reaction. Elemental chlorine, represented as $\mathrm{Cl}_2$, begins with an oxidation state of 0 before reacting with hot, concentrated sodium hydroxide, represented as $\mathrm{NaOH}$. During this reaction, expressed as
$3\mathrm{Cl}_2 + 6\mathrm{NaOH} \rightarrow 5\mathrm{NaCl} + \mathrm{NaClO}_3 + 3\mathrm{H}_2\mathrm{O}$
the chlorine atoms partition into two distinct products. Specifically, the chlorine is reduced to a -1 oxidation state in sodium chloride, $\mathrm{NaCl}$, and simultaneously oxidized to a +5 oxidation state in sodium chlorate, $\mathrm{NaClO}_3$.

The oxidation number of an atom within a polyatomic ion is calculated by ensuring the sum of all oxidation states equals the net charge of the species, keeping in mind that oxygen generally carries an oxidation state of -2. For the phosphate ion $\mathrm{PO}_4^{3-}$, setting the oxidation state of phosphorus to $x$ results in the equation $x + 4(-2) = -3$, which solves to $x = +5$. Applying the same principle to the sulfate ion $\mathrm{SO}_4^{2-}$, the equation $y + 4(-2) = -2$ demonstrates that the oxidation state of sulfur is $y = +6$. Finally, for the dichromate ion $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, assigning the oxidation state $z$ to chromium leads to the equation $2z + 7(-2) = -2$, which simplifies to $2z = 12$, confirming that the oxidation state of chromium is $z = +6$.

In an acidic environment, permanganate ($\mathrm{MnO_4^-}$) functions as an oxidizing agent, while ferrous oxalate ($\mathrm{FeC_2O_4}$) acts as the reducing agent. Success in this problem requires balancing the total electrons lost by the reducing agent with the total electrons gained by the oxidizing agent.

Within each mole of ferrous oxalate, the $\mathrm{Fe^{2+}}$ ion oxidizes to $\mathrm{Fe^{3+}}$, losing 1 electron. Simultaneously, the $\mathrm{C_2O_4^{2-}}$ ion oxidizes to two molecules of $\mathrm{CO_2}$, as each carbon atom undergoes an oxidation state change from $+3$ to $+4$, releasing a total of 2 electrons. This results in a combined loss of 3 electrons per mole of $\mathrm{FeC_2O_4}$. During the reduction process, $\mathrm{MnO_4^-}$ follows the reaction:
$\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5e^- \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$
This indicates that each mole of $\mathrm{MnO_4^-}$ gains 5 electrons. By equating the electron exchange, where the moles of $\mathrm{MnO_4^-}$ multiplied by 5 equals 1 mole of $\mathrm{FeC_2O_4}$ multiplied by 3, the calculation reveals that 0.6 moles of $\mathrm{MnO_4^-}$ are required for the oxidation.

Oxidation and reduction are defined by changes in the oxidation numbers of atoms throughout a chemical process. In the reaction $\mathrm{H}_2 \mathrm{O} + \mathrm{Br}_2 \rightarrow \mathrm{HOBr} + \mathrm{HBr}$, the bromine atoms in $\mathrm{Br}_2$ possess an initial oxidation number of 0. Within the products, part of the bromine is converted to $\mathrm{HOBr}$ where the oxidation number increases to +1, while the other part is converted to $\mathrm{HBr}$ where the oxidation number decreases to -1. Because the bromine undergoes both an increase and a decrease in its oxidation state, it is simultaneously oxidized and reduced.

To determine the oxidation state of sulfur in these anions, we rely on the principle that the sum of the oxidation states of all atoms in a polyatomic ion must equal its net charge. By assigning the oxidation state of oxygen as $-2$, we can treat the oxidation state of sulfur as an unknown variable, $x$, and set up an algebraic equation for each species. For the sulfite ion $\mathrm{SO}_{3}^{2-}$, the calculation $x + 3(-2) = -2$ simplifies to $x = +4$. For the dithionite ion $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}$, the expression $2x + 4(-2) = -2$ leads to $2x = +6$, resulting in $x = +3$. Finally, in the dithionate ion $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$, we solve the equation $2x + 6(-2) = -2$ to find $2x = +10$, which gives $x = +5$. Comparing these values, the oxidation states follow the ascending order $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} < \mathrm{SO}_{3}^{2-} < \mathrm{S}_{2} \mathrm{O}_{6}^{2-}$.