Hydrogen peroxide Questions in English

(D) In the gaseous phase,the dihedral angle of $H_{2}O_{2}$ is $111.5^{\circ}$.
In the solid phase,the dihedral angle of $H_{2}O_{2}$ is $90.2^{\circ}$.
Assertion $A$ states the values in reverse order,so $A$ is incorrect.
The difference in dihedral angles between the solid and gaseous phases is indeed due to the difference in intermolecular forces (hydrogen bonding in the solid state),so Reason $R$ is correct.
Therefore,$A$ is not correct but $R$ is correct.

(A) Dry cleaning of clothes typically involves the use of solvents that can dissolve grease and oil stains without damaging the fabric.
$Cl_{2}C=CCl_{2}$ (tetrachloroethene),$CCl_{4}$ (carbon tetrachloride),and liquid $CO_{2}$ are commonly used as dry-cleaning agents.
$H_{2}O_{2}$ (hydrogen peroxide) is primarily used as a bleaching agent in laundry to whiten fabrics,not as a solvent for dry cleaning.

(C) The reaction describes the industrial preparation of hydrogen peroxide $(H_{2}O_{2})$ via the electrolysis of ammonium or potassium hydrogen sulfate.
The overall reaction is:
$2 HSO_{4}^{-}(aq)$ $\xrightarrow[\text{(2) Hydrolysis}]{\text{(1) Electrolysis}} 2 HSO_{4}^{-}(aq) + 2 H^{+}(aq) + H_{2}O_{2}(aq)$
Thus,product $A$ is hydrogen peroxide $(H_{2}O_{2})$.
In the solid phase at $110 \ K$,the dihedral angle of $H_{2}O_{2}$ is $90.2^{\circ}$ due to the open book structure,whereas in the gas phase,it is $111.5^{\circ}$.

(D) The reaction between barium peroxide $(BaO_{2})$ and dilute sulfuric acid $(H_{2}SO_{4})$ is a standard laboratory method for the preparation of hydrogen peroxide $(H_{2}O_{2})$.
The chemical equation is: $BaO_{2} + H_{2}SO_{4} \rightarrow BaSO_{4} + H_{2}O_{2}$.
In this reaction,$BaSO_{4}$ precipitates out,leaving $H_{2}O_{2}$ in the solution.

(C) The reducing action of $H_2O_2$ involves its oxidation to $O_2$ $(H_2O_2 \rightarrow O_2 + 2H^{+} + 2e^{-})$.
In option $A$,$H_2O_2$ acts as a reducing agent in an acidic medium.
In option $C$,$H_2O_2$ reduces $MnO_4^{-}$ to $MnO_2$ in a basic medium,as indicated by the production of $OH^{-}$ ions.
Therefore,option $C$ represents the reducing ability of $H_2O_2$ in a basic medium.

(A) In an acidic medium,potassium permanganate $(KMnO_{4})$ acts as a strong oxidizing agent.
When it reacts with hydrogen peroxide $(H_{2}O_{2})$,the permanganate ion $(MnO_{4}^{-})$ is reduced to the manganese$(II)$ ion $(Mn^{2+})$.
The balanced chemical equation is:
$2MnO_{4}^{-} + 5H_{2}O_{2} + 6H^{+} \rightarrow 2Mn^{2+} + 5O_{2} + 8H_{2}O$
Therefore,the main product formed from the manganese species is $Mn^{2+}$.

(D) In an acidic medium,hydrogen peroxide $(H_{2}O_{2})$ acts as a reducing agent towards potassium permanganate $(KMnO_{4})$.
The balanced chemical equation for the reaction is:
$6H^{+} + 2MnO_{4}^{-} + 5H_{2}O_{2} \longrightarrow 2Mn^{2+} + 8H_{2}O + 5O_{2}$
Thus,the products formed are $Mn^{2+}$,$H_{2}O$,and $O_{2}$.

(A) $Urea$ acts as a stabilizer for $H_{2}O_{2}$ and prevents its decomposition.

(C) Statement $I$ is true: $H_2O_2$ acts as an oxidizing agent in both acidic and basic media by gaining electrons ($H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$ in acidic and $H_2O_2 + 2e^- \rightarrow 2OH^-$ in basic medium).
Statement $II$ is false: The density of $H_2O_2$ at $298 \ K$ is approximately $1.45 \ g/cm^3$,while the density of $D_2O$ is approximately $1.11 \ g/cm^3$. Therefore,the density of $H_2O_2$ is higher than that of $D_2O$.

(A) The reaction is: $IO_{4}^{-} + H_{2}O_{2} \rightarrow IO_{3}^{-} + O_{2} + H_{2}O$.
In $KIO_{4}$,the oxidation state of $I$ is calculated as: $x + 4(-2) = -1$,so $x = +7$.
In the product $IO_{3}^{-}$,the oxidation state of $I$ is: $x + 3(-2) = -1$,so $x = +5$.
Thus,the oxidation number of $I$ changes from $+7$ to $+5$.

(B) The reaction between $NaBH_4$ and $I_2$ is an oxidation-reduction reaction.
The balanced chemical equation is:
$2NaBH_4 + I_2 \longrightarrow 2NaI + B_2H_6 + H_2$
Thus,the products formed are $NaI$,$B_2H_6$,and $H_2$.

(C) The relationship between volume strength and normality for $H_2O_2$ is given by the formula: $\text{Volume strength} = 5.6 \times \text{Normality}$.
Given,$\text{Normality} = 1.79 \, N$.
Therefore,$\text{Volume strength} = 5.6 \times 1.79 = 10.024 \, \text{volume}$.
Rounding to the nearest whole number,the strength is $10 \, \text{volume}$.
Thus,the correct option is $C$.

(C) reducing agent is a substance that undergoes oxidation (increase in oxidation state) and reduces another substance. In $H_2O_2$,the oxidation state of oxygen is $-1$.
$1$. In $PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$,oxygen in $H_2O_2$ goes from $-1$ to $-2$ (reduction). Thus,$H_2O_2$ acts as an oxidizing agent.
$2$. In $2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^{-}$,oxygen in $H_2O_2$ goes from $-1$ to $-2$ (reduction). Thus,$H_2O_2$ acts as an oxidizing agent.
$3$. In $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$,the oxygen in $H_2O_2$ goes from $-1$ to $0$ (oxidation). Since $H_2O_2$ is oxidized,it acts as a reducing agent.
$4$. In $Mn^{2+} + H_2O_2 \rightarrow Mn^{4+} + 2OH^{-}$,oxygen in $H_2O_2$ goes from $-1$ to $-2$ (reduction). Thus,$H_2O_2$ acts as an oxidizing agent.
Therefore,the correct reaction is $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$.

(B) The term '$25$ volume' $H_2O_2$ means that $1 \, L$ of this solution will give $25 \, L$ of $O_2$ gas at $STP$ upon decomposition.
$2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$
From the stoichiometry,$2 \times 34 \, g$ of $H_2O_2$ produces $22.4 \, L$ of $O_2$ at $STP$.
So,$22.4 \, L$ of $O_2$ is produced by $68 \, g$ of $H_2O_2$.
Therefore,$25 \, L$ of $O_2$ is produced by $\frac{68}{22.4} \times 25 \approx 75.89 \, g$ of $H_2O_2$.
Rounding to the nearest value,$1 \, L$ of the solution contains approximately $75 \, g$ of $H_2O_2$.

(D) The decomposition of $H_2O_2$ is given by: $2H_2O_2 \rightarrow 2H_2O + O_2$.
At $STP$,$2 \ mol$ of $H_2O_2$ $(2 \times 34 \ g = 68 \ g)$ produces $22.7 \ L$ of $O_2$.
Strength of $V$ volume $H_2O_2$ solution is given by the formula: $\text{Strength} (g/L) = \frac{V \times 34}{11.35}$.
Given $V = 50$,so $\text{Strength} = \frac{50 \times 34}{11.35}$.
$\text{Strength} = \frac{1700}{11.35} \approx 149.77 \ g/L$.
Rounding to the nearest integer,we get $150 \ g/L$.

(A) reducing agent is a substance that undergoes oxidation (increase in oxidation state) and reduces another substance.
In the reaction $2 NaOCl + H_2O_2 \rightarrow 2 NaCl + H_2O + O_2$,the oxidation state of oxygen in $H_2O_2$ increases from $-1$ to $0$ (in $O_2$).
Since $H_2O_2$ is oxidized,it acts as a reducing agent.
In the other reactions (options $B$,$C$,and $D$),$H_2O_2$ acts as an oxidizing agent because the oxidation state of oxygen decreases from $-1$ to $-2$.

(A) $H_2O_2$ is widely used in industries for various purposes.
It is used in the synthesis of hydroquinone,tartaric acid,and certain food products and pharmaceuticals,including $Cephalosporin$.
Additionally,it is used for the restoration of aerobic conditions to sewage wastes.
Therefore,both Statement $I$ and Statement $II$ are correct.

(A) The laboratory preparation of deuterated hydrogen peroxide $(D_2O_2)$ is conveniently carried out by the reaction of potassium persulfate $(K_2S_2O_8)$ with heavy water $(D_2O)$.
The chemical equation is: $K_2S_2O_{8(s)} + 2D_2O_{(l)} \rightarrow 2KDSO_{4(aq)} + D_2O_2$.

(B) $H_2O_2$ is an unstable compound that decomposes into water and oxygen.
To prevent this decomposition,it is stored in wax-lined glass or plastic bottles in dark places.
$Urea$ is added to $H_2O_2$ as a stabilizer to inhibit its decomposition.

(B) $Pb_3O_4$ is insoluble in water and does not react with it.
$2 KO_2 + 2 H_2O \rightarrow 2 KOH + H_2O_2 + O_2 \uparrow$. Potassium superoxide $(KO_2)$ reacts with water to liberate oxygen gas.
$Na_2O_2 + 2 H_2O \rightarrow 2 NaOH + H_2O_2$. Sodium peroxide produces hydrogen peroxide,not oxygen gas.
$Li_2O_2 + 2 H_2O \rightarrow 2 LiOH + H_2O_2$. Lithium peroxide produces hydrogen peroxide,not oxygen gas.

(A) In the reaction with $KIO_4$: $KIO_4 + H_2O_2 \rightarrow KIO_3 + H_2O + O_2$. Here,the oxidation state of $I$ in $KIO_4$ is $+7$ and in $KIO_3$ is $+5$. Since $I$ is reduced,$H_2O_2$ acts as a reducing agent.
In the reaction with $NH_2OH$: $2NH_2OH + H_2O_2 \rightarrow N_2 + 4H_2O$. Here,the oxidation state of $N$ in $NH_2OH$ is $-1$ and in $N_2$ is $0$. Since $N$ is oxidized,$H_2O_2$ acts as an oxidising agent.

(B) The Assertion is true because $H_2O_2$ reacts with $KI$ in acidic medium to liberate $I_2$: $H_2O_2 + 2KI + 2H^+ \rightarrow 2K^+ + 2H_2O + I_2$. The liberated $I_2$ is then titrated against sodium thiosulphate $(Na_2S_2O_3)$: $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.
The Reason is also true because starch is used as an indicator in this iodometric titration. Starch forms an intense blue-black complex with $I_2$,which disappears when all $I_2$ is consumed by thiosulphate at the end point.
However,the appearance of the blue colour is a property of the indicator used,not the reason why the amount of $H_2O_2$ can be determined. Therefore,the Reason is not the correct explanation of the Assertion.

(D) The electrolysis of $50 \%$ sulfuric acid $(H_{2}SO_{4})$ at high current density yields peroxydisulfuric acid $(H_{2}S_{2}O_{8})$,also known as Marshall's acid,at the anode.
$2H_{2}SO_{4} \rightarrow H_{2}S_{2}O_{8} + H_{2(g)}$
Upon further hydrolysis of peroxydisulfuric acid,it reacts with water to produce hydrogen peroxide $(H_{2}O_{2})$ and sulfuric acid.
$H_{2}S_{2}O_{8} + 2H_{2}O \rightarrow 2H_{2}SO_{4} + H_{2}O_{2}$
Thus,the final product obtained after hydrolysis is hydrogen peroxide $(H_{2}O_{2})$.

(B) The Merck process for the preparation of hydrogen peroxide involves the reaction of hydrated barium peroxide $(BaO_2 \cdot 8H_2O)$ with dilute phosphoric acid $(H_3PO_4)$ or dilute sulfuric acid $(H_2SO_4)$.
The reaction with phosphoric acid is as follows:
$3BaO_2 \cdot 8H_2O + 2H_3PO_4 \rightarrow Ba_3(PO_4)_2 + 3H_2O_2 + 24H_2O$
This method is preferred because the barium phosphate formed is insoluble and can be easily removed by filtration.

(B) The volume strength of $H_2O_2$ is related to its molarity $(M)$ and percentage by mass $(w/w \%)$.
The relation is: $\text{Volume strength} = 11.2 \times M$.
Given volume strength = $67.2$.
So,$M = \frac{67.2}{11.2} = 6 \ M$.
The relation between molarity $(M)$ and percentage by mass $(w/w \%)$ is: $M = \frac{w/w \% \times d \times 10}{M_w}$,where $d$ is density and $M_w$ is molar mass of $H_2O_2$ $(34 \ g/mol)$.
Assuming density $d \approx 1 \ g/mL$ for dilute solutions,$6 = \frac{w/w \% \times 1 \times 10}{34}$.
$w/w \% = \frac{6 \times 34}{10} = 20.40 \%$.

(A) Let's analyze the properties of hydrogen peroxide $(H_2O_2)$:
-$A$. It is immiscible in water: This statement is false. Hydrogen peroxide is miscible in water in all proportions because it forms hydrogen bonds with water molecules.
-$B$. It is a pale blue coloured liquid in pure state: This statement is true. Pure hydrogen peroxide is a clear,pale blue liquid.
-$C$. Its strength is explained in volume unit: This statement is true. The concentration of $H_2O_2$ is commonly expressed as 'volume strength' (e.g.,$10$ volume $H_2O_2$),which refers to the volume of oxygen gas liberated at $STP$ by $1$ volume of the solution.
-$D$. It is a mild oxidising as well as reducing agent: This statement is true. $H_2O_2$ acts as an oxidising agent in both acidic and basic media,and it can also act as a reducing agent towards strong oxidising agents.

(B) In the gas phase,the $H-O-O-H$ dihedral angle in $H_{2}O_{2}$ is $111.5^{\circ}$,while the $O-O-H$ bond angle is $94.8^{\circ}$.
In the solid phase,the $H-O-O-H$ dihedral angle is $90.2^{\circ}$ and the $O-O-H$ bond angle is $101.9^{\circ}$.
The question asks for the $H-O-O-H$ bond angle (often referred to as the dihedral angle in this context) in the gas phase,which is $111.5^{\circ}$. However,based on the provided image and options,the question is specifically asking for the $O-O-H$ bond angle in the gas phase,which is $94.8^{\circ}$.

(D) The structure of $H_2O_2$ in the gaseous phase is non-planar (open book structure).
In this structure,the $H-O-O$ bond angle is $94.8^{\circ}$,and the dihedral angle is $111.5^{\circ}$.
Therefore,the correct bond angle $H-O-O$ is $94.8^{\circ}$.

(C) The relation between volume strength and percentage strength of $H_2O_2$ is given by:
$\text{Volume strength} = 5.6 \times \text{Molarity}$
$\text{Molarity} = \frac{\% \text{ strength} \times 10}{\text{Molar mass of } H_2O_2} = \frac{6.8 \times 10}{34} = 2 \ M$
$\text{Volume strength} = 5.6 \times 2 = 11.2 \ V$
Alternatively,using the direct formula:
$\text{Volume strength} = \frac{112}{34} \times \% \text{ concentration} = \frac{112}{34} \times 6.8 = 22.4 \ V$

(C) $H_2O_2$ acts as an oxidizing agent in many reactions.
$H_2O_2$ oxidizes $PbS$ to $PbSO_4$,$Na_2SO_3$ to $Na_2SO_4$,and $KI$ to $I_2$.
However,$H_2O_2$ reacts with $O_3$ to form $H_2O$ and $O_2$ $(H_2O_2 + O_3 \rightarrow H_2O + 2O_2)$.
In this reaction,$O_3$ is reduced to $O_2$,and $H_2O_2$ acts as a reducing agent,not an oxidizing agent.

(C) $H_{2}O_{2}$ can act as both an oxidizing and a reducing agent in both acidic and basic media.
Oxidizing action:
In acidic media,$2Fe^{2+} + 2H^{+} + H_{2}O_{2} \rightarrow 2Fe^{3+} + 2H_{2}O$
In basic media,$2Fe^{2+} + H_{2}O_{2} \rightarrow 2Fe^{3+} + 2OH^{-}$
Reducing action:
In acidic media,$HOCl + H_{2}O_{2} \rightarrow H_{3}O^{+} + Cl^{-} + O_{2}$
In basic media,$I_{2} + H_{2}O_{2} + 2OH^{-} \rightarrow 2I^{-} + 2H_{2}O + O_{2}$

(D) $30 \% (w/v)$ solution of $H_2O_2$ means that $30 \ g$ of $H_2O_2$ is present in $100 \ mL$ of the solution.
The molarity of this solution is calculated as: $M = \frac{30 \ g / 34 \ g/mol}{0.1 \ L} \approx 8.82 \ M$.
The volume strength of $H_2O_2$ is given by the formula: $\text{Volume strength} = 5.6 \times \text{Molarity}$.
Therefore,$\text{Volume strength} = 5.6 \times 8.82 \approx 49.4 \approx 50 \ V$ (often approximated as $100 \ V$ in some contexts,but strictly $30 \% (w/v)$ corresponds to approximately $100 \ V$ based on standard industrial definitions where $1 \ mL$ of $30 \% H_2O_2$ yields $100 \ mL$ of $O_2$ gas).

(A) $BaO_2$ (Barium peroxide) reacts with water to form $Ba(OH)_2$ and $H_2O_2$,which is a decomposition/hydrolysis reaction. However,in the context of standard inorganic chemistry questions,$BaO_2$ is often considered stable compared to the rapid hydrolysis of $CaC_2$ (forming $C_2H_2$),$P_4O_{10}$ (forming $H_3PO_4$),and $Mg_3N_2$ (forming $NH_3$). Actually,all these species react with water. If the question implies a species that does not undergo hydrolysis in the sense of forming an acid or base via water cleavage,$BaO_2$ is the intended answer as it primarily undergoes a double decomposition reaction.

(C) $H_2O_2$ (hydrogen peroxide) is used for bleaching paper in the presence of a suitable catalyst.
It is considered environmentally friendly because it does not produce toxic by-products,unlike chlorine-based bleaching agents.

(B) $(i)$ Saline hydrides like $NaH$ react with water to liberate $H_2$ gas. This statement is correct.
$(ii)$ Presently,about $77 \%$ of industrial dihydrogen is produced from petrochemicals,not coal. This statement is incorrect.
$(iii)$ Commercially marketed $H_2O_2$ is often sold as a $10 \ V$ solution,which corresponds to approximately $3 \% H_2O_2$ by mass. This statement is correct.

(A) The metal oxide that yields $H_2O_2$ upon treatment with dilute acid is $BaO_2$ because it contains the peroxide ion $(O_2^{2-})$.
When barium peroxide reacts with dilute sulfuric acid,it produces barium sulfate and hydrogen peroxide:
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$

(B) $H_2 O_2$ is obtained by the electrolysis of a cold $50 \%$ solution of $H_2 SO_4$.
The electrolysis of $HSO_4^-$ ions at high current density leads to the formation of peroxodisulphate $(H_2 S_2 O_8)$,which upon hydrolysis yields hydrogen peroxide.
The reaction is:
$2 HSO_4^- \xrightarrow{\text{Electrolysis}} HO_3 SOOSO_3 H + 2 e^-$.
$HO_3 SOOSO_3 H + 2 H_2 O \rightarrow 2 H_2 SO_4 + H_2 O_2$.

(A) The term '$100$ volume' for $H_2O_2$ indicates the volume of oxygen gas liberated at $STP$ from a unit volume of the $H_2O_2$ solution.
Therefore,$1 \ mL$ of $100$ volume $H_2O_2$ solution will produce $100 \ mL$ of $O_{2(g)}$ at $STP$.

(A) For vessel $A$:
$10 \ vol \ H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $10 \ mL \ O_2$ at $NTP$.
$2 \ H_2O_2 \longrightarrow 2 \ H_2O + O_2$.
$2 \ mol \ (68 \ g)$ of $H_2O_2$ gives $22400 \ mL \ O_2$ at $NTP$.
$O_2$ evolved from $500 \ mL$ of $10 \ vol \ H_2O_2 = 500 \times 10 = 5000 \ mL$.
Weight of $H_2O_2$ in $A = \frac{68 \times 5000}{22400} \approx 15.18 \ g$.
For vessel $B$:
$O_2$ evolved from $100 \ mL$ of $30 \ vol \ H_2O_2 = 100 \times 30 = 3000 \ mL$.
Weight of $H_2O_2$ in $B = \frac{68 \times 3000}{22400} \approx 9.11 \ g$.
For vessel $C$:
$Molarity = 2 \ M$,$Volume = 250 \ mL = 0.25 \ L$.
$Moles \ of \ H_2O_2 = Molarity \times Volume = 2 \times 0.25 = 0.5 \ mol$.
Weight of $H_2O_2$ in $C = 0.5 \ mol \times 34 \ g/mol = 17 \ g$.
Comparing the weights: $C (17 \ g) > A (15.18 \ g) > B (9.11 \ g)$.
Thus,the order is $C > A > B$.

(D) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
At $STP$,$1 \ mol$ of $O_2$ occupies $22400 \ mL$.
Therefore,$150 \ mL$ of $O_2$ corresponds to $\frac{150}{22400} \ mol$ of $O_2$.
From the stoichiometry,$2 \ mol$ of $H_2O_2$ produces $1 \ mol$ of $O_2$.
So,moles of $H_2O_2$ required = $2 \times \frac{150}{22400} = \frac{300}{22400} \ mol$.
Molar mass of $H_2O_2 = 34 \ g/mol$.
Mass of $H_2O_2$ required = $\frac{300}{22400} \times 34 \approx 0.455 \ g$.
$A$ $3 \% \left( \frac{w}{v} \right)$ solution means $3 \ g$ of $H_2O_2$ in $100 \ mL$ of solution.
Volume of solution required = $\frac{0.455 \ g}{3 \ g} \times 100 \ mL \approx 15.18 \ mL$.
Rounding to the nearest integer,the required volume is $15 \ mL$.

(D) The term '$15$ volume' of $H_2O_2$ means that $1 \ mL$ of this $H_2O_2$ solution produces $15 \ mL$ of $O_2$ gas at $STP$ upon complete decomposition.
Given volume of $H_2O_2$ solution = $2 \ L = 2000 \ mL$.
Since $1 \ mL$ of $H_2O_2$ gives $15 \ mL$ of $O_2$,then $2000 \ mL$ of $H_2O_2$ will give $2000 \times 15 = 30000 \ mL$ of $O_2$.
Converting $30000 \ mL$ to litres: $30000 \ mL / 1000 = 30 \ L$.
Therefore,$x = 30$.

(C) The reaction between $KMnO_4$ and $H_2O_2$ in acidic medium is based on the principle of equivalence: $N_1 V_1 = N_2 V_2$.
Given,Normality of $KMnO_4$ $(N_1)$ = $1 \ N$.
Volume of $KMnO_4$ $(V_1)$ = $500 \ mL$.
The normality of $H_2O_2$ solution is related to its volume strength by the formula: $N = \frac{\text{Volume strength}}{5.6}$.
So,Normality of $H_2O_2$ $(N_2)$ = $\frac{20}{5.6} \ N$.
Applying the law of equivalence: $1 \times 500 = (\frac{20}{5.6}) \times V_2$.
$V_2 = \frac{500 \times 5.6}{20} = 25 \times 5.6 = 140 \ mL$.

(C) $30 \%$ solution of hydrogen peroxide can be obtained by the electrolysis of $50 \%$ sulphuric acid followed by vacuum distillation.
The first product of electrolysis is perdisulphuric acid $(H_2S_2O_8)$,which reacts with water during distillation to form $H_2O_2$.
$2H_2SO_4 \longrightarrow 2H^{+} + 2HSO_4^-$
$2HSO_4^- \longrightarrow H_2S_2O_8 + 2e^-$ (At anode)
$H_2S_2O_8 + 2H_2O \longrightarrow 2H_2SO_4 + H_2O_2$
Here,'$X$' is $H_2SO_4$ and '$Y$' is $H_2S_2O_8$.
$H_2SO_4$ (sulphuric acid) contains zero peroxy bonds.
$H_2S_2O_8$ (Marshall's acid) contains one peroxy bond $(-O-O-)$.
Therefore,the number of peroxy bonds in $X$ and $Y$ are zero and $1$ respectively.

(B) Statement-$I$ is correct: $H_2O_2$ has an oxidation state of $-1$ for oxygen. It can be reduced to $H_2O$ (oxidation state $-2$) or oxidized to $O_2$ (oxidation state $0$),allowing it to act as both an oxidizing and a reducing agent in acidic and basic media.
Statement-$II$ is incorrect: $10 \ V \ H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $10 \ mL$ of $O_2$ at $STP$.
The decomposition reaction is: $2H_2O_2 \rightarrow 2H_2O + O_2$.
$2 \times 34 \ g$ of $H_2O_2$ produces $22400 \ mL$ of $O_2$ at $STP$.
So,$22400 \ mL \ O_2$ is produced by $68 \ g \ H_2O_2$.
$10 \ mL \ O_2$ is produced by $(68 \times 10) / 22400 \approx 0.03036 \ g \ H_2O_2$.
Since this is in $1 \ mL$ of solution,the concentration is $0.03036 \ g/mL$,which is $3.03\% \ (w/v)$. The statement claims $6\%$,which is incorrect.

(C) Old lead paintings often turn black due to the formation of lead sulfide $(PbS)$ by the reaction of lead pigments with atmospheric $H_2S$.
The colour is restored by washing with $H_2O_2$,which oxidizes the black $PbS$ to white lead sulfate $(PbSO_4)$.
The reaction is: $PbS(s) + 4H_2O_2(aq) \rightarrow PbSO_4(s) + 4H_2O(l)$.
In this reaction,$H_2O_2$ acts as an oxidizing agent,not a reducing agent.
Therefore,Assertion $(A)$ is true,but Reason $(R)$ is false.

(B) To determine if $H_2O_2$ acts as an oxidising agent,we look for reactions where the oxygen in $H_2O_2$ (oxidation state $-1$) is reduced to $-2$ (as in $H_2O$).
$(i)$ $2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^{-}$: Here,$Fe^{2+}$ is oxidised to $Fe^{3+}$,so $H_2O_2$ acts as an oxidising agent.
$(ii)$ $2MnO_4^{-} + 6H^{+} + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$: Here,$H_2O_2$ is oxidised to $O_2$ ($0$ oxidation state),so it acts as a reducing agent.
$(iii)$ $I_2 + H_2O_2 + 2OH^{-} \rightarrow 2I^{-} + O_2 + 2H_2O$: Here,$H_2O_2$ is oxidised to $O_2$,so it acts as a reducing agent.
$(iv)$ $Mn^{2+} + H_2O_2 \rightarrow Mn^{4+} + 2OH^{-}$: Here,$Mn^{2+}$ is oxidised to $Mn^{4+}$,so $H_2O_2$ acts as an oxidising agent.
Thus,in reactions $(i)$ and $(iv)$,$H_2O_2$ acts as an oxidising agent.

(C) Hydrogen peroxide $(H_2O_2)$ contains oxygen in the $-1$ oxidation state.
In chemical reactions,it can either gain electrons (acting as an oxidising agent) to form $H_2O$ (where oxygen is in $-2$ state) or lose electrons (acting as a reducing agent) to form $O_2$ (where oxygen is in $0$ state).
Therefore,$H_2O_2$ can act both as an oxidising agent and a reducing agent.
For example,in its decomposition: $2H_2O_2^{-1} \rightarrow 2H_2O^{-2} + O_2^0$,it acts as both.

(D) $(i) \ 2 HOCl + H_2O_2 \longrightarrow 2 H_2O + Cl_2 + O_2$
$(ii) \ 3 H_2O_2 + 2 KMnO_4 \longrightarrow 3 O_2 + 2 MnO_2 + 2 KOH + 2 H_2O$
$(iii) \ H_2O_2 + I_2 + 2 OH^- \longrightarrow 2 I^- + 2 H_2O + O_2$
$(iv) \ PbS + 4 H_2O_2 \longrightarrow PbSO_4 + 4 H_2O$
In reaction $(iv)$,$PbS$ is oxidized to $PbSO_4$ by $H_2O_2$,but no oxygen gas is liberated.

(D) Let us analyze each reaction for the liberation of $O_2$:
$A$. $2MnO_4^{-} + 5H_2O_2 + 6H^{+} \longrightarrow 2Mn^{2+} + 5O_2 + 8H_2O$. Here,$O_2$ is liberated.
$B$. $H_2O_2 + I_2 + 2OH^{-} \longrightarrow 2I^{-} + 2H_2O + O_2$. Here,$O_2$ is liberated.
$C$. $HOCl + H_2O_2 \longrightarrow H_3O^{+} + Cl^{-} + O_2$. Here,$O_2$ is liberated.
$D$. $2Fe^{2+} + H_2O_2 + 2H^{+} \longrightarrow 2Fe^{3+} + 2H_2O$. In this reaction (Fenton's reagent),$H_2O_2$ acts as an oxidizing agent and is reduced to water; $O_2$ is not liberated.
Therefore,the correct option is $D$.