Hydrogen peroxide Questions in English

(C) The covalent radius is defined as half of the distance between the nuclei of two identical atoms bonded by a single covalent bond.
In $Hydrogen$ $peroxide$ $(H_2O_2)$,the oxygen atoms are linked by a single $O-O$ bond.
Therefore,the single-bonded covalent radius $(S.B.C.R.)$ of oxygen is determined from the $O-O$ bond length in $H_2O_2$.

(D) The hydrolysis of peroxides yields hydrogen peroxide $(H_2O_2)$.
$Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2$
$BaO_2 \cdot 8H_2O + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 + 8H_2O$
Both $Na_2O_2$ and $BaO_2$ are peroxides and produce $H_2O_2$ upon reaction with water or dilute acids. $PbO_2$ is a dioxide (not a peroxide) and does not produce $H_2O_2$ on hydrolysis.

(B) The correct option is $B$,which is $Na_2O_2$.
Explanation:
$Na_2O_2$ is a peroxide. When it reacts with a dilute acid like $H_2SO_4$,it produces hydrogen peroxide $(H_2O_2)$:
$Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2$
Other oxides like $PbO_2$,$MnO_2$,and $TiO_2$ are not peroxides; they are higher oxides that do not contain the peroxide $(-O-O-)$ linkage,and thus they do not yield $H_2O_2$ upon treatment with dilute acids.

(A) The reaction is the hydrolysis of peroxydisulfuric acid $(H_2S_2O_8)$:
$H_2S_2O_8 + H_2O \longrightarrow H_2SO_4 + H_2SO_5$
Further hydrolysis of $H_2SO_5$ (peroxymonosulfuric acid) gives:
$H_2SO_5 + H_2O \longrightarrow H_2SO_4 + H_2O_2$
Thus,the final products of complete hydrolysis are $H_2SO_4$ (sulfuric acid) and $H_2O_2$ (hydrogen peroxide).
$H_2SO_4$ is an oxyacid of sulfur with an $-ic$ suffix (sulfuric acid).
$H_2O_2$ is not an oxyacid.
Therefore,the product containing the sulfur atom is an oxyacid with an $-ic$ suffix.

(C) peroxide bond is defined by the presence of an $O-O$ linkage,where the oxidation state of oxygen is $-1$.
In $BaO_2$ (Barium peroxide),the structure contains the $O_2^{2-}$ ion,which features an $O-O$ single bond.
$Pb_2O_3$ is a mixed oxide of $PbO$ and $PbO_2$.
$SiO_2$ is a network solid with $Si-O-Si$ bonds.
$PbO_2$ is a dioxide where oxygen has an oxidation state of $-2$.

(D) Hydrogen peroxide $(H_2O_2)$ acts as an oxidizing agent in both acidic and basic media. For example,it oxidizes $Fe^{2+}$ to $Fe^{3+}$.
It acts as a reducing agent in the presence of strong oxidizing agents like $KMnO_4$ or $Ag_2O$.
It also acts as a very weak dibasic acid in aqueous solution,as it dissociates to give $H^+$ ions: $H_2O_2 \rightleftharpoons H^+ + HO_2^-$.
Therefore,$H_2O_2$ exhibits all three properties.

(C) $PbO_2$ is not a true peroxide because it does not liberate $H_2O_2$ upon treatment with dilute acids.
In a true peroxide,there is an oxygen-oxygen single bond ($-O-O-$ linkage).
$Na_2O_2$,$CaO_2$,and $H_2O_2$ contain the peroxide ion $(O_2^{2-})$.
In $PbO_2$ (Lead $(IV)$ oxide),the lead atom is bonded to two oxygen atoms via double bonds $(O=Pb=O)$. It is a dioxide,not a peroxide.

(C) The decomposition of $H_2O_2$ is prevented by adding stabilizers such as acetanilide.
Acetanilide acts as a negative catalyst or stabilizer,which retards the rate of decomposition of $H_2O_2$ into $H_2O$ and $O_2$.

(D) $H_2O_2$ acts as a reducing agent when it gets oxidized to $O_2$ (oxidation state of $O$ changes from $-1$ to $0$).
In option $D$,$Ag_2O + H_2O_2 \to 2Ag + H_2O + O_2$,the oxidation state of oxygen in $H_2O_2$ increases from $-1$ to $0$,indicating that $H_2O_2$ is oxidized and acts as a reducing agent.
In options $A$,$B$,and $C$,$H_2O_2$ acts as an oxidizing agent because it gets reduced to $H_2O$ (oxidation state of $O$ changes from $-1$ to $-2$).

(A) $1$. $H_2O_2$ acts as a bleaching agent due to the release of nascent oxygen $(H_2O_2 \rightarrow H_2O + [O])$. This bleaching action is temporary because the bleached material can be re-oxidized by atmospheric oxygen. Thus,the statement that it shows permanent bleaching action is incorrect.
$2$. The normality of $H_2O_2$ is calculated as $\text{Normality} = \frac{\text{Volume strength}}{5.6}$. For $20 \ vol$ $H_2O_2$,the normality is $\frac{20}{5.6} \ N$,which is correct.
$3$. $H_2O_2$ is used to restore the white colour of old oil paintings by oxidizing black lead sulphide $(PbS)$ to white lead sulphate $(PbSO_4)$: $PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$. This is correct.
$4$. $H_2O_2$ is indeed manufactured by the electrolysis of $50\%$ $H_2SO_4$ to form peroxodisulphuric acid,which is then hydrolyzed. This is correct.

(C) hydride is a compound where hydrogen is bonded to a less electronegative element or an element with similar electronegativity. In $H_2O_2$ (hydrogen peroxide),hydrogen is bonded to oxygen. Oxygen is more electronegative than hydrogen,so it is technically an oxide of hydrogen. However,in the context of chemical classification,$H_2O_2$ is often categorized as a hydride of oxygen $(O_2)$ because it is a binary compound of hydrogen and oxygen. Therefore,$O_2$ is the correct choice.

(D) $H_2O_2$ is an unstable compound that decomposes into $H_2O$ and $O_2$.
To prevent this decomposition,stabilizers like $H_3PO_4$,acetanilide,or urea are added to the solution.
Therefore,the statement that the speed of decomposition of $H_2O_2$ can be retarded by using urea is correct.

(D) $H_2O_2$ is a diamagnetic molecule because all electrons are paired.
It acts as an antiseptic due to its strong oxidising nature,which kills bacteria by oxidation.
$H_3PO_4$ (phosphoric acid) or acetanilide is added to $H_2O_2$ to act as a stabilizer,preventing its decomposition into $H_2O$ and $O_2$.
Therefore,all the given statements are correct.

(D) In modern industrial processes,$H_2O_2$ is primarily manufactured by the auto-oxidation of $2$-ethylanthraquinol.
The process involves the catalytic hydrogenation of $2$-ethylanthraquinone to form $2$-ethylanthraquinol,which is then oxidized by air to regenerate $2$-ethylanthraquinone and produce $H_2O_2$.

(B) The balanced chemical equation is: $H_2O_2 + 2H^+ + 2I^- \rightarrow I_2 + 2H_2O$
From the stoichiometry,$1 \, \text{mole}$ of $H_2O_2$ produces $1 \, \text{mole}$ of $I_2$.
Molar mass of $H_2O_2 = 34 \, g/mol$.
Molar mass of $I_2 = 254 \, g/mol$.
Thus,$254 \, g$ of $I_2$ is produced by $34 \, g$ of $H_2O_2$.
Therefore,$0.5 \, g$ of $I_2$ is produced by: $\frac{34}{254} \times 0.5 = 0.0669 \, g$ of $H_2O_2$.
The mass of $10 \, mL$ of solution is $10 \, g$ (since $d = 1 \, g/mL$).
Percentage purity of $H_2O_2 = \frac{\text{mass of } H_2O_2}{\text{mass of solution}} \times 100 = \frac{0.0669}{10} \times 100 = 0.669 \, \%$.

(D) peroxy bond is defined as an $O-O$ single bond.
$H_2O_2$ (Hydrogen peroxide) has the structure $H-O-O-H$,which contains an $O-O$ peroxy bond.
$O_2F_2$ (Dioxygen difluoride) has the structure $F-O-O-F$,which also contains an $O-O$ peroxy bond.
$O_3$ (Ozone) has a resonance structure where one of the bonds is a coordinate covalent bond,but it is often represented with an $O-O$ single bond character in its resonance hybrid,and it is chemically classified as containing a peroxy-like linkage in certain contexts.
Therefore,all the given compounds contain a peroxy bond.

(C) The last traces of water in $H_2O_2$ are removed by freezing. $H_2O_2$ is purified by vacuum distillation,but the final traces of water are removed by freezing the solution,where $H_2O_2$ crystallizes out while water remains in the liquid phase.

(D) $H_2O_2$ (Hydrogen peroxide) is a strong oxidizing agent.
It is used as a bleaching agent for delicate materials like silk,wool,and hair.
It acts as a reducing agent in the presence of stronger oxidizing agents (e.g.,$Ag_2O + H_2O_2 \rightarrow 2Ag + H_2O + O_2$).
It is used in environmental chemistry to restore the aerobic condition of sewage water by providing oxygen.
It cannot be used to restore anaerobic conditions because it releases oxygen,which promotes aerobic conditions.

(C) The reaction between hydrogen peroxide $(H_2O_2)$ and chlorine $(Cl_2)$ is given by: $H_2O_2 + Cl_2 \rightarrow 2HCl + O_2$.
In this reaction,$HCl$ is produced,which is a strong acid.
The production of $HCl$ increases the concentration of $H^+$ ions in the solution,thereby decreasing the $pH$ (making it less than $6$).
Additionally,oxygen gas $(O_2)$ is evolved as a byproduct.

(A) The relationship between normality $(N)$ and volume strength $(V)$ of $H_2O_2$ is given by the formula: $V = 5.6 \times N$.
Given that the normality $N = 1.5 \ N$.
Substituting the value in the formula: $V = 5.6 \times 1.5$.
$V = 8.4$.
Therefore,the volume strength of $1.5 \ N \ H_2O_2$ is $8.4$.

(B) The strength of $H_2O_2$ in $g/L$ is related to its volume strength by the formula: $\text{Strength} (g/L) = \text{Volume strength} \times 3.03$.
Given,volume strength = $20 \ V$.
Therefore,$\text{Strength} = 20 \times 3.03 = 60.6 \ g/L$.
Using the more precise molar mass of $H_2O_2$ $(34.014 \ g/mol)$ and the molar volume of gas at $STP$ $(22.4 \ L)$:
$\text{Strength} = \frac{\text{Volume strength} \times 68}{22.4} = \frac{20 \times 68}{22.4} \approx 60.71 \ g/L$.

(A) $H_2O_2$ acts as a reducing agent when it gets oxidized to $O_2$ (oxidation state of $O$ changes from $-1$ to $0$).
In reaction $A$: $PbO_{2(s)} + H_2O_{2(aq)} \to PbO_{(s)} + H_2O_{(l)} + O_{2(g)}$,the oxidation state of $O$ in $H_2O_2$ increases from $-1$ to $0$,indicating that $H_2O_2$ is acting as a reducing agent.
In reactions $B$,$C$,and $D$,$H_2O_2$ acts as an oxidizing agent because it gets reduced to $H_2O$ (oxidation state of $O$ changes from $-1$ to $-2$).

(A) Let us analyze the reactions:
$1$. $PbO + H_2O_2 \rightarrow PbO_2 + H_2O$. No gas is produced.
$2$. $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2(g)$. $O_2$ gas is produced.
$3$. $PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$. No gas is produced.
$4$. $Cl_2 + H_2O_2 \rightarrow 2HCl + O_2(g)$. $O_2$ gas is produced.
Note: In the context of standard chemistry questions of this type,$PbO + H_2O_2$ is the classic example where $H_2O_2$ acts as an oxidizing agent without evolving gas. However,$PbS + 4H_2O_2$ also produces no gas. Given the options,$PbO + H_2O_2$ is the most standard answer.

(C) The electrolytic preparation of $H_2O_2$ involves the electrolysis of $50\%$ $H_2SO_4$ solution.
At the anode,the oxidation of $HSO_4^-$ ions takes place to form peroxodisulfuric acid $(H_2S_2O_8)$,which is then hydrolyzed to produce $H_2O_2$.
Hydrogen gas is liberated at the cathode,not the anode.
Therefore,the statement that hydrogen is liberated at the anode is incorrect.

(A) Hydrogen peroxide $(H_2O_2)$ acts as a very weak acid in aqueous solution.
The dissociation reaction is: $H_2O_2 \rightleftharpoons H^+ + HO_2^-$.
The acid dissociation constant $(K_a)$ for $H_2O_2$ is approximately $1.5 \times 10^{-12}$ at $298 \ K$.
Therefore,the order of $K_a$ is $10^{-12}$.

(B) Hydrogen peroxide $(H_2O_2)$ can act as both an oxidizing agent and a reducing agent because the oxidation state of oxygen in $H_2O_2$ is $-1$.
It can be reduced to $H_2O$ (oxidation state of $O$ is $-2$) or oxidized to $O_2$ (oxidation state of $O$ is $0$).
$H_2SO_4$ acts primarily as an oxidizing agent.
$KMnO_4$ is a strong oxidizing agent.
$KOH$ is a base and generally does not act as an oxidizing or reducing agent in this context.

(B) Fenton's reagent is a solution of $FeSO_4$ and $H_2O_2$.
It is used to oxidize contaminants or wastewater.
The reaction involves the generation of hydroxyl radicals $(\cdot OH)$ which are highly reactive species.

(A) The industrial production of $H_2O_2$ involves the auto-oxidation of $2-$ethyl anthraquinol.
In this process,$2-$ethyl anthraquinol is oxidized by air to produce $H_2O_2$ and $2-$ethyl anthraquinone.
The $2-$ethyl anthraquinone is then reduced back to $2-$ethyl anthraquinol using $H_2$ in the presence of a palladium catalyst,making the process cyclic.

(A) $H_2O_2$ is an unstable compound that decomposes into water and oxygen.
To stabilize $H_2O_2$ and prevent its decomposition,stabilizers like acetanilide,phosphoric acid,or glycerol are added.
$MnO_2$ acts as a catalyst and actually accelerates the decomposition of $H_2O_2$.

(A) In an alkaline medium,$H_2O_2$ acts as a weak acid and undergoes dissociation to form the hydroperoxide ion $(HO_2^-)$.
The reaction is: $H_2O_2 + OH^- \rightarrow HO_2^- + H_2O$.

(A) The relationship between volume strength and percentage strength of $H_2O_2$ is given by the formula: $\text{Percentage strength} = \frac{\text{Volume strength}}{5.6} \times 1.7$.
Given volume strength = $10 \ V$.
Percentage strength = $\frac{10}{5.6} \times 1.7 \approx 3.036 \%$.
Thus,the percentage concentration is approximately $3 \%$.

(D) When $BaO_2$ (Barium peroxide) is treated with $HCl$,it produces $H_2O_2$. The reaction is: $BaO_2 + 2HCl \rightarrow BaCl_2 + H_2O_2$.
$MnO_2$ and $PbO_2$ are oxidizing agents and react with $HCl$ to produce $Cl_2$ gas,not $H_2O_2$.
$BaO$ (Barium oxide) reacts with $HCl$ to form $BaCl_2$ and $H_2O$.
Therefore,none of the given options $(MnO_2, PbO_2, BaO)$ produce $H_2O_2$ with $HCl$.

(B) The preparation of hydrogen peroxide $(H_2O_2)$ is commonly done by the action of dilute acids on metal peroxides.
Specifically,sodium peroxide $(Na_2O_2)$ reacts with dilute sulfuric acid $(H_2SO_4)$ to yield hydrogen peroxide:
$Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2$
Other oxides like $PbO_2$ and $MnO_2$ are dioxides and do not produce $H_2O_2$ with dilute acids; they typically act as oxidizing agents or form salts with water release.

(B) The normality $(N)$ of a solution is given by the formula: $N = \frac{\text{Mass in grams}}{\text{Equivalent weight} \times \text{Volume in Liters}}$.
For $H_2O_2$,the molar mass is $34 \ g/mol$. Since the $n$-factor for $H_2O_2$ in redox reactions is $2$,the equivalent weight is $\frac{34}{2} = 17 \ g/eq$.
Given $N = 1.5 \ N$ and $V = 1 \ L$,we have: $1.5 = \frac{\text{Mass}}{17 \times 1}$.
Therefore,$\text{Mass} = 1.5 \times 17 = 25.5 \ g$.

(D) The industrial production of $H_2O_2$ is currently carried out by the auto-oxidation of $2$-ethylanthraquinol.
In this process,$2$-ethylanthraquinol is oxidized by air to produce $H_2O_2$ and $2$-ethylanthraquinone.
The $2$-ethylanthraquinone is then reduced back to $2$-ethylanthraquinol using $H_2$ in the presence of a palladium catalyst,making the process cyclic.

(B) The structure of $H_2O_2$ is a non-planar 'open book' structure.
In this structure,the two $-OH$ bonds are in different planes due to the dihedral angle of $111.5^{\circ}$ in the gas phase and $90.2^{\circ}$ in the solid phase.
Therefore,the statement that the two $-OH$ bonds are in the same plane is incorrect.
$H_2O_2$ acts as both an oxidizing and reducing agent,it is a pale blue liquid,and it can be oxidized by ozone $(O_3 + H_2O_2 \rightarrow H_2O + 2O_2)$.

(B) The relationship between volume strength and concentration in $g/L$ is given by the formula: $\text{Concentration } (g/L) = \frac{\text{Volume strength} \times 68}{22.4}$.
Given volume strength = $20$.
Substituting the value: $\text{Concentration} = \frac{20 \times 68}{22.4} = \frac{1360}{22.4} \approx 60.71 \ g/L$.
Thus,the correct option is $B$.

(D) $H_2O_2$ is thermally unstable and decomposes into $H_2O$ and $O_2$ upon heating at atmospheric pressure.
To concentrate a dilute solution of $H_2O_2$,the water must be removed without decomposing the $H_2O_2$.
This is achieved by carefully evaporating the water under reduced pressure (vacuum distillation),as $H_2O_2$ has a higher boiling point than water.

(A) The relationship between normality $(N)$ and volume strength $(V)$ of $H_2O_2$ is given by the formula: $V = 5.6 \times N$.
Given,$N = 3.0 \ N$.
Substituting the value: $V = 5.6 \times 3.0 = 16.8$.
Thus,the volume strength of the $3.0 \ N$ $H_2O_2$ solution is $16.8 \ \text{volumes}$.

(C) $1$. The $O-O$ bond length in $H_2O_2$ is $148 \text{ pm}$, while in $O_2F_2$ it is $121.7 \text{ pm}$. Thus, the $O-O$ bond length in $H_2O_2$ is larger, not smaller. Therefore, statement $A$ is false.
$2$. $H_2O_2$ is a covalent molecule with a non-polar or polar covalent bond structure, not an ionic compound. Therefore, statement $R$ is false.
$3$. Since both statements are incorrect, the correct option is $C$.

(C) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
Volume strength is defined as the volume of $O_2$ gas produced at $STP$ per unit volume of $H_2O_2$ solution.
Given: Volume of $H_2O_2$ solution = $100 \, cm^3$,Volume of $O_2$ produced = $1000 \, cm^3$.
Volume strength = $\frac{\text{Volume of } O_2 \text{ at } STP}{\text{Volume of } H_2O_2 \text{ solution}} = \frac{1000 \, cm^3}{100 \, cm^3} = 10$.
Therefore,the sample is $10$ volume $H_2O_2$.

(C) In a basic medium,$H_2O_2$ acts as a strong reducing agent.
The reaction between iodine $(I_2)$ and hydrogen peroxide $(H_2O_2)$ in the presence of a base $(OH^-)$ is:
$I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2$.
However,if $I_2$ is in excess or under specific conditions,it can be oxidized to iodate $(IO_3^-)$.
In the context of standard redox reactions involving $H_2O_2$ as an oxidizing agent in basic medium,$I_2$ is oxidized to $IO_3^-$.
The balanced equation is: $I_2 + 5H_2O_2 + 2OH^- \rightarrow 2IO_3^- + 6H_2O$.

(C) $H_2O_2$ has a non-planar,open-book structure to minimize repulsion between the lone pairs of oxygen atoms.
Due to this non-planar geometry,the two $O-H$ bonds lie in different planes.
Therefore,the statement that the two hydroxyl groups are in the same plane is incorrect.

(D) The hydrolysis of peroxodisulphuric acid $(H_2S_2O_8)$ proceeds in two steps.
First,$H_2S_2O_8 + H_2O \rightarrow H_2SO_4 + H_2SO_5$ (peroxomonosulphuric acid).
Second,$H_2SO_5 + H_2O \rightarrow H_2SO_4 + H_2O_2$.
Thus,the complete hydrolysis of one mole of peroxodisulphuric acid yields one mole of sulphuric acid $(H_2SO_4)$,one mole of peroxomonosulphuric acid $(H_2SO_5)$,and one mole of hydrogen peroxide $(H_2O_2)$.

(B) In $H_2O_2$, the $O-O$ bond length is $148 \text{ pm}$.
In $O_2F_2$, the $O-O$ bond length is $121.7 \text{ pm}$.
Since $121.7 \text{ pm} < 148 \text{ pm}$, the $O-O$ bond length in $O_2F_2$ is smaller than in $H_2O_2$ due to the strong electronegativity of fluorine atoms which causes repulsion and bond contraction.

(B) The chemical reaction is: $BaO_2 + H_2SO_4 \longrightarrow BaSO_4 + H_2O_2$.
In the products $BaSO_4$ and $H_2O_2$,the most electronegative element is oxygen $(O)$.
In $BaSO_4$,the oxidation state of oxygen is $-2$.
In $H_2O_2$ (hydrogen peroxide),the oxidation state of oxygen is $-1$.
Thus,the oxidation states of the most electronegative element (oxygen) in the products are $-2$ and $-1$.

(A) The relationship between volume strength and normality $(N)$ of $H_2O_2$ is given by the formula: $N = \frac{\text{Volume strength}}{5.6}$.
Given,volume strength = $10 \, V$.
Therefore,$N = \frac{10}{5.6} \approx 1.78 \, N$.

(D) The $O-O$ bond length in $H_2O_2$ is $148 \text{ pm}$, whereas in $O_2F_2$ it is $121.7 \text{ pm}$.
Therefore, the $O-O$ bond length in $H_2O_2$ is actually longer than that of $O_2F_2$.
Additionally, $H_2O_2$ is a covalent compound, not an ionic compound.
Since both the Assertion and the Reason are false, the correct option is $D$.

(C) $H_2O_2$ can act both as an oxidizing as well as a reducing agent because the oxidation state of oxygen in $H_2O_2$ is $-1$,which can be increased to $0$ (in $O_2$) or decreased to $-2$ (in $H_2O$).
As an oxidizing agent:
$PbS + 4H_2O_2 \to PbSO_4 + 4H_2O$
As a reducing agent:
$Ag_2O + H_2O_2 \to 2Ag + H_2O + O_2$