Hydrogen peroxide Questions in English

(D) The reaction of potassium superoxide $(KO_2)$ with water is given by: $2KO_2 + 2H_2O \to 2KOH + H_2O_2 + O_2$.
Thus,the product $(X)$ is hydrogen peroxide $(H_2O_2)$.
$H_2O_2$ has an open book-like structure in the gas phase.
$H_2O_2$ is stabilized by the addition of urea to prevent its decomposition.
$H_2O_2$ is sensitive to light and decomposes into water and oxygen in the presence of light,hence it is stored in wax-lined colored bottles.
Therefore,all the given statements are correct.

(D) From the reaction $A + Na_2S_2O_3 \rightarrow B + Na_2S_4O_6$,$A$ is $I_2$ and $B$ is $NaI$.
From the reaction $C + A \rightarrow B_2H_6 + B + H_2$,where $A = I_2$ and $B = NaI$,$C$ is $NaBH_4$.
$NaBH_4 + I_2 \rightarrow B_2H_6 + NaI + H_2$.
$B_2H_6 + NaH \rightarrow NaBH_4$ $(C)$.
Now evaluating the options:
$A$. $B$ $(NaI)$ is soluble in water and the solution is colourless,not brown.
$B$. Compound $C$ $(NaBH_4)$ contains ionic ($Na^+$ and $[BH_4]^-$) and covalent bonds (within $[BH_4]^-$),not coordinate bonds.
$C$. Aqueous solution of $B$ $(NaI)$ does not give a brown layer in the layer test.
$D$. Compound $B$ $(NaI)$ does not react with starch to give a blue colour. However,if $A$ $(I_2)$ is present,it gives a blue colour with starch. Re-evaluating the reaction: $I_2$ reacts with starch to give a blue colour. Given the options,there might be a typo in the question's premise regarding $B$. Based on standard chemistry,$I_2$ is the species that gives a blue colour with starch.

(B) $1$. Black phosphorus is a highly polymeric form of phosphorus with a layered structure.
$2$. $H_2O_2$ has an open book structure,which is non-planar and polar.
$3$. Chloral $(CCl_3CHO)$ does not have an $-OH$ group,so it cannot form intramolecular hydrogen bonds.
$4$. Isohypophosphoric acid $(H_4P_2O_6)$ has a $P-P$ linkage,but it is a tetrabasic acid,not tribasic.
Therefore,the correct statement is that $H_2O_2$ is polar but not planar.

(D) $H_2O_2$ (Hydrogen peroxide) is a versatile chemical compound.
$1$. It acts as an oxidising agent in both acidic and basic media (e.g.,$2Fe^{2+} + H_2O_2 \rightarrow 2Fe^{3+} + 2OH^-$).
$2$. It acts as a reducing agent in the presence of stronger oxidising agents (e.g.,$2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$).
$3$. $H_2O_2$ is a very weak acid,even weaker than water $(K_a \approx 1.5 \times 10^{-12})$,and it forms hydroperoxides with bases.
$4$. It does not exhibit basic properties.
Therefore,the statement that $H_2O_2$ has basic properties is incorrect.

(D) Hydrogen peroxide $(H_2O_2)$ exhibits all the mentioned properties:
$(i)$ It acts as a mild disinfectant due to its ability to release nascent oxygen.
$(ii)$ It is used as a rocket propellant in high concentrations because its decomposition releases a large amount of energy and oxygen.
$(iii)$ It acts as a bleaching agent for textiles,paper,and hair by oxidizing the pigments.
$(iv)$ It acts as a strong oxidizing agent in both acidic and basic media.

(D) reductant (reducing agent) is a substance that undergoes oxidation itself,meaning the oxidation state of the element (in this case,oxygen in $H_2O_2$) increases.
In the reaction $Cl_2 + H_2O_2 \to 2HCl + O_2$,the oxidation state of oxygen in $H_2O_2$ is $-1$,and it increases to $0$ in $O_2$.
Therefore,$H_2O_2$ acts as a reducing agent in this reaction.
In the other options ($A$,$B$,and $C$),$H_2O_2$ acts as an oxidizing agent because the oxidation state of oxygen decreases from $-1$ to $-2$ (forming $H_2O$).

(D) $90 \%$ solution of $H_2O_2$ is cooled with a solid $CO_2 +$ ether bath.
Cooling is continued until crystallization sets in.
The crystals of $H_2O_2$ are separated,melted,and refrozen to achieve higher concentration and remove the last traces of water.

(D) $H_2O_2$ has a non-planar structure.
The $O-O$ bond is at an angle,and the two $O-H$ bonds are in different planes.
This structure is often described as a 'half-open book' structure,where the two pages of the book represent the two $O-H$ planes.

(B) $H_2O_2$ is a pale blue liquid and it can be oxidised by ozone $(O_3 + H_2O_2 \rightarrow H_2O + 2O_2)$.
$H_2O_2$ acts as both an oxidising and a reducing agent.
$H_2O_2$ has a non-planar,open-book structure where the two $OH$ bonds do not lie in the same plane 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 two $OH$ bonds lie in the same plane is false.

(A) Metallic oxides which on treatment with dilute acids produce hydrogen peroxide are called peroxides. All peroxides contain a peroxide ion $(O_2)^{2-}$ having the structure $-O-O-$.
$PbO_2$ does not contain a peroxide ion $(O_2)^{2-}$ and it cannot be classified as a peroxide.
$PbO_2$ contains $O^{2-}$ ions and the oxidation state of $Pb$ is $+4$,whereas $H_2O_2$,$SrO_2$,and $BaO_2$ contain $[O-O]^{2-}$ ions (peroxide ions).
Thus,option $A$ is correct.

(D) $H_2O_2$ acts as an oxidizing agent in both acidic and basic media,for example,$2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O$.
It acts as a reducing agent in the presence of strong oxidizing agents,for example,$2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$.
It also acts as a very weak acid in aqueous solution,dissociating as $H_2O_2 \rightleftharpoons H^+ + HO_2^-$. Therefore,it exhibits all these properties.

(B) The preparation of $H_2O_2$ involves several industrial and laboratory methods:
$1$. The oxidation of $2-$ethyl Anthraquinol is the standard industrial process for producing $H_2O_2$.
$2$. The reaction of $BaO_2 \cdot 8H_2O$ with dilute acid ($H_2SO_4$ or $H_3PO_4$) is a standard laboratory method.
$3$. The complete hydrolysis of peroxodisulphates like $K_2S_2O_8$ yields $H_2O_2$.
$4$. $H_2S_2O_7$ (Oleum) on hydrolysis produces $H_2SO_4$,not $H_2O_2$. Therefore,option $B$ is not a method to prepare $H_2O_2$.

(C) The hydrolysis of peroxo-disulphuric acid $(H_2S_2O_8)$,commonly known as Marshall's acid,produces hydrogen peroxide $(H_2O_2)$ and sulphuric acid $(H_2SO_4)$.
The reaction proceeds in two steps:
$1. H_2S_2O_8 + H_2O \rightarrow H_2SO_4 + H_2SO_5$ (Peroxomonosulphuric acid or Caro's acid)
$2. H_2SO_5 + H_2O \rightarrow H_2SO_4 + H_2O_2$
Overall reaction: $H_2S_2O_8 + 2H_2O \rightarrow 2H_2SO_4 + H_2O_2$

(B) The stability of oxygen species depends on the bond order and the charge density. The oxide ion $(O^{2-})$ is the most stable due to its closed-shell configuration. The peroxide ion $(O_2^{2-})$ has a bond order of $1$,and the superoxide ion $(O_2^-)$ has a bond order of $1.5$. However,in terms of thermodynamic stability in the presence of alkali metals,the order is $Oxide > Peroxide > Superoxide$.

(B) $H_2O_2$ is a pale blue liquid and it can be oxidized by ozone to $O_2$.
$H_2O_2$ acts as both an oxidizing and a reducing agent in different chemical reactions.
Due to its non-planar,open-book structure,the two $OH$ bonds do not lie in the same plane.
Therefore,the statement that the two $OH$ bonds lie in the same plane is false.

(B) $1$. Calculate the molarity $(M)$ of the $H_2O_2$ solution:
$M = \frac{\text{mass of } H_2O_2}{\text{molar mass of } H_2O_2 \times \text{volume of solution in } L}$
$M = \frac{6.8 \, g}{34 \, g/mol \times 0.2 \, L} = \frac{6.8}{6.8} = 1 \, M$
$2$. Use the relation between volume strength and molarity:
$\text{Volume strength} = M \times 11.2$
$\text{Volume strength} = 1 \times 11.2 = 11.2 \, V$
$3$. Re-evaluating the options provided,there seems to be a discrepancy. However,if we consider the calculation $11.2 \, V$,none of the options match exactly. Given the standard context of such problems,if the volume was $0.4 \, L$,the answer would be $5.6 \, V$. Based on the provided values,the calculated result is $11.2 \, V$.

(B) Bleaching agents work by releasing nascent oxygen or by reduction.
$H_2O_2$ acts as a bleaching agent by releasing nascent oxygen: $H_2O_2 \rightarrow H_2O + [O]$.
$O_3$ acts as a bleaching agent by releasing nascent oxygen: $O_3 \rightarrow O_2 + [O]$.
$Cl_2$ in the presence of moisture produces nascent oxygen: $Cl_2 + H_2O \rightarrow 2HCl + [O]$.
$SO_2$ acts as a bleaching agent by reduction,but it is typically used for bleaching delicate materials like wool and silk.
However,in the context of standard chemistry questions regarding common bleaching agents,all the listed options are actually used as bleaching agents.
Given the standard curriculum,if one must be chosen as 'not used' in a specific context,it is often a trick question; however,all four are technically bleaching agents.
Assuming the question implies which one is not a permanent bleaching agent or acts differently,$SO_2$ is a reducing agent,while the others are oxidizing agents.
Since all are bleaching agents,this question is technically flawed,but $SO_2$ is the most distinct due to its reducing nature.

(D) The production of $H_2O_2$ can be achieved through several methods:
$1$. Hydrolysis of $H_2S_2O_8$ (peroxodisulfuric acid): $H_2S_2O_8 + 2H_2O \rightarrow 2H_2SO_4 + H_2O_2$.
$2$. Electrolysis of concentrated $H_2SO_4$: This process yields peroxodisulfuric acid,which upon hydrolysis gives $H_2O_2$.
$3$. Merck's process: This involves the reaction of barium peroxide $(BaO_2)$ with dilute sulfuric acid $(H_2SO_4)$ to produce $H_2O_2$ and barium sulfate $(BaSO_4)$: $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$.
Since all the given options are valid methods for the preparation of $H_2O_2$,the correct answer is $D$.

(D) $H_2O_2$ is highly unstable and decomposes upon heating. Therefore,it cannot be purified by boiling or simple distillation. The last traces of water are removed by fractional crystallization,where the mixture is cooled to allow $H_2O_2$ to crystallize out,leaving the impurities behind.

(A) In a basic medium,$H_2O_2$ acts as a reducing agent and reduces $I_2$ to iodide ions $(I^{-})$.
The balanced chemical equation for the reaction is:
$I_{2(s)} + H_2O_{2(aq)} + 2OH^{-}_{(aq)} \rightarrow 2I^{-}_{(aq)} + 2H_2O_{(l)} + O_{2(g)}$
Thus,the correct product is $I^{-}$.

(C) $H_2O_2$ acts as a reducing agent when it reacts with strong oxidizing agents,where it gets oxidized to $O_2$.
$1$. With $KMnO_4$ in acidic medium: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$.
$2$. With $K_2Cr_2O_7$ in acidic medium: $K_2Cr_2O_7 + 4H_2SO_4 + 3H_2O_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3O_2$.
$3$. With $Ag_2O$: $Ag_2O + H_2O_2 \rightarrow 2Ag + H_2O + O_2$.
$HNO_3$ is an oxidizing agent,but $H_2O_2$ does not typically act as a reducing agent against it in standard conditions; rather,it is often used to reduce higher oxidation states. However,in the context of common textbook examples for $H_2O_2$ as a reducing agent,$I, II,$ and $III$ are the standard reactions.

(C) $1$. The reaction $Ag_2O + H_2O_2 \to 2Ag + H_2O + O_2$ is a correct redox reaction where $H_2O_2$ acts as a reducing agent.
$2$. The reaction $PbS + 4H_2O_2 \to PbSO_4 + 4H_2O$ is a correct redox reaction where $H_2O_2$ acts as an oxidizing agent.
$3$. The reaction $Fe^{2+} + H_2O_2 \xrightarrow{\text{basic}} Fe^{3+} + 2OH^-$ is incorrect because in a basic medium,$Fe^{2+}$ is oxidized to $Fe^{3+}$ by $H_2O_2$,but the stoichiometry and charge balance are not represented correctly as $2Fe^{2+} + H_2O_2 \to 2Fe^{3+} + 2OH^-$.
$4$. The reaction $Mn^{2+} + H_2O_2 \xrightarrow{\text{basic}} Mn^{4+} + 2OH^-$ is a correct representation of the oxidation of $Mn^{2+}$ to $MnO_2$ $(Mn^{4+})$ in basic medium.

(D) An oxidizing agent is a substance that gains electrons and gets reduced in a chemical reaction.
In the reaction $PbS + 4H_2O_2 \to PbSO_4 + 4H_2O$:
$1$. The oxidation state of sulfur $(S)$ in $PbS$ changes from $-2$ to $+6$ in $PbSO_4$. This is an oxidation process.
$2$. The oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $-2$ in $H_2O$. This is a reduction process.
Since $H_2O_2$ is reduced,it acts as an oxidizing agent.
In options $A$,$B$,and $C$,$H_2O_2$ acts as a reducing agent because the oxygen in $H_2O_2$ is oxidized from $-1$ to $0$ (in $O_2$).

(A) $H_2O_2$ acts as a reducing agent when it reacts with strong oxidising agents. In an acidic medium,$H_2O_2$ reduces $MnO_4^-$ to $Mn^{2+}$.
The balanced chemical equation is:
$2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \to K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$
In this reaction,the oxidation state of $Mn$ changes from $+7$ to $+2$,and $H_2O_2$ is oxidized to $O_2$.

(D) Correct option: $D$. Hydrogen peroxide $(H_{2}O_{2})$ acts as both an oxidising and a reducing agent in both acidic and basic media.
$H_{2}O_{2}$ as an oxidant:
In acidic medium: $H_{2}O_{2} + 2H^{+} + 2e^{-} \longrightarrow 2H_{2}O$
In basic medium: $H_{2}O_{2} + 2e^{-} \longrightarrow 2OH^{-}$
$H_{2}O_{2}$ as a reductant:
In acidic medium: $H_{2}O_{2} \longrightarrow O_{2} + 2H^{+} + 2e^{-}$
In basic medium: $H_{2}O_{2} + 2OH^{-} \longrightarrow 2H_{2}O + O_{2} + 2e^{-}$

(C) The relationship between volume strength and molarity $(M)$ of $H_2O_2$ is given by the formula: $\text{Volume strength} = 11.35 \times M$.
Given $M = 1 \, M$.
Therefore,$\text{Volume strength} = 11.35 \times 1 = 11.35$.

(A) The decomposition reaction of $H_2O_2$ is: $2H_2O_2 \to 2H_2O + O_2 \uparrow$
From the stoichiometry,$2 \times 34 \ g = 68 \ g$ of $H_2O_2$ produces $22.4 \ L$ of $O_2$ at $STP$.
Therefore,$11.2 \ L$ of $O_2$ is produced by $\frac{68 \ g}{22.4 \ L} \times 11.2 \ L = 34 \ g$ of $H_2O_2$.
This means $34 \ g$ of $H_2O_2$ is present in $1 \ L$ $(1000 \ mL)$ of the solution.
Strength in $g/L = 34 \ g/L$.
Percentage strength = $\frac{\text{Strength in } g/L}{10} = \frac{34}{10} = 3.4\%$.

(D) Perhydrol is the commercial name for a $30\%$ (by weight) aqueous solution of hydrogen peroxide $(H_2O_2)$. It is used as a laboratory reagent and in various industrial applications.

(D) First,calculate the number of moles of $H_2O_2$: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{3.4 \ g}{34 \ g/mol} = 0.1 \ mol$.
Next,calculate the molarity $(M)$ of the solution: $M = \frac{n}{V(L)} = \frac{0.1 \ mol}{112 \times 10^{-3} \ L} = \frac{0.1}{0.112} \approx 0.8928 \ M$.
The relationship between volume strength and molarity is: $\text{Volume strength} = M \times 11.2$.
Therefore,$\text{Volume strength} = \frac{0.1}{0.112} \times 11.2 = \frac{0.1 \times 11.2}{0.112} = \frac{1.12}{0.112} = 10 \ V$.

(B) In $H_2O_2$,the $O-O$ bond length is $1.48 \, \mathring{A}$.
In $O_2F_2$,due to the high electronegativity of fluorine,the $O-F$ bond has more $p$-character,which according to Bent's Rule,increases the $s$-character in the $O-O$ bond.
Greater $s$-character leads to a shorter bond length.
Thus,the $O-O$ bond length in $O_2F_2$ is $1.22 \, \mathring{A}$.
Therefore,the bond lengths for $H_2O_2$ and $O_2F_2$ are $1.48 \, \mathring{A}$ and $1.22 \, \mathring{A}$ respectively.

(C) $H_2S_2O_8 + H_2O \to H_2SO_4 + H_2SO_5$
The hydrolysis of $H_2S_2O_8$ (peroxodi-sulphuric acid) yields one mole of $H_2SO_4$ (sulphuric acid) and one mole of $H_2SO_5$ (peroxomono-sulphuric acid).

(D) $H_2O_2$ is thermally unstable and it decomposes easily according to the reaction: $H_2O_{2(l)} \to H_2O_{(l)} + \frac{1}{2}O_{2(g)}$.
Its decomposition is catalysed by trace amounts of impurities,such as alkali metals present in the glass of the vessel,which makes it unsuitable as a stable solvent.

(B) To determine the acidity,we compare the dissociation constants or the stability of the conjugate bases.
$H_2O$ is a very weak acid with a $pK_a$ of approximately $15.7$.
$H_2O_2$ is a stronger acid than $H_2O$ due to the inductive effect of the oxygen atom,with a $pK_a$ of approximately $11.6$.
$CO_2$ in water forms carbonic acid $(H_2CO_3)$,which is a stronger acid than $H_2O_2$,with a $pK_a$ of approximately $6.35$.
Therefore,the increasing order of acidity is $H_2O < H_2O_2 < CO_2$.

(D) $A: 2OF \xrightarrow{\text{Dimerisation}} F-O-O-F$ (contains peroxy linkage).
$B: H_4P_2O_8 + H_2O \rightarrow 2H_3PO_4 + H_2O_2$ (contains peroxy linkage in $H_2O_2$).
$C: 2Na + \text{excess } O_2 \xrightarrow{\Delta} Na_2O_2$ ($Na^+ O^--O^- Na^+$,contains peroxy linkage).
Since all the given reactions produce compounds containing a peroxy linkage,the correct answer is $D$.

(B) $H_2O_2$ is a peroxide. For its preparation,we require a peroxide source.
$1$. $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$ (This is a standard method).
$2$. $PbO_2$ is a dioxide,not a peroxide. The reaction $PbO_2 + H_2SO_4 \rightarrow PbSO_4 + H_2O + \frac{1}{2}O_2$ does not produce $H_2O_2$.
$3$. Aerial oxidation of $2$-ethyl anthraquinol is the industrial process for $H_2O_2$ production.
$4$. Electrolysis of $(NH_4)_2SO_4$ produces ammonium peroxodisulphate,which on hydrolysis yields $H_2O_2$.

(A) The reaction of chlorine with hydrogen peroxide $(H_2O_2)$ is an oxidation-reduction reaction where $Cl_2$ acts as an oxidizing agent.
The chemical equation is: $H_2^{18}O_2 + Cl_2 \rightarrow 2HCl + ^{18}O_2$.
In this reaction,the oxygen atoms in $H_2^{18}O_2$ are released as molecular oxygen $(^{18}O_2)$,while chlorine is reduced to chloride ions $(Cl^-)$ in the form of $HCl$.

(B) Peroxymonosulfuric acid,with the chemical formula $H_2SO_5$,is commonly known as $Caro's$ acid.

(C) The reaction of sodium peroxide $(Na_2O_2)$ with water $(H_2O)$ produces sodium hydroxide $(NaOH)$ and oxygen gas $(O_2)$.
The balanced chemical equation is: $2Na_2O_2 + 2H_2O \to 4NaOH + O_2$.

(C) $H_2S_2O_7$ (Oleum) contains an $S-O-S$ linkage and does not contain a peroxo $(O-O)$ linkage.
$CF_3CO_3H$ (Peroxytrifluoroacetic acid),$H_2S_2O_8$ (Peroxydisulfuric acid),and $H_2SO_5$ (Peroxymonosulfuric acid) all contain peroxo linkages.

(D) Heating $Pb(NO_3)_2$ gives $PbO$,$NO_2$,and $O_2$.
Heating $KClO_3$ gives $KCl$ and $O_2$.
Heating $HgO$ gives $Hg$ and $O_2$.
$MnO_2$ is a stable oxide and does not decompose to give $O_2$ upon simple heating; it is often used as a catalyst for the decomposition of $KClO_3$.

(D) The chemical reaction is: $BaO_2 + H_2SO_4 \to BaSO_4 + H_2O_2$.
The most electronegative element in the products is oxygen $(O)$.
In $BaSO_4$,the oxidation state of oxygen is $-2$.
In $H_2O_2$,the oxidation state of oxygen is $-1$.

(B) peroxide contains the peroxide ion,$O_2^{2-}$,which features an $O-O$ single bond.
In $BaO_2$,the oxidation state of $Ba$ is $+2$,so the $O_2$ unit must have a charge of $-2$,which corresponds to the peroxide ion $(O_2^{2-})$.
$KO_2$ is a superoxide $(O_2^-)$.
$MnO_2$ and $NO_2$ are dioxides where oxygen is in the $-2$ oxidation state.

(C) On prolonged standing,hydrogen peroxide undergoes auto-oxidation or disproportionation as follows:
$2H_2O_2 \rightarrow 2H_2O + O_2$
Light acts as a catalyst for this decomposition reaction.
Hence,hydrogen peroxide is always stored in black bottles,which cuts off the light and suppresses the auto-oxidation process.

(B) $1$. Calcium bicarbonate $(Ca(HCO_3)_2)$ is responsible for temporary hardness,not permanent hardness. So,option $A$ is incorrect.
$2$. Hydrogen peroxide $(H_2O_2)$ has a non-planar structure and all its electrons are paired,making it diamagnetic. So,option $B$ is correct.
$3$. The Calgon method uses sodium hexametaphosphate $(Na_6P_6O_{18})$ to sequester $Ca^{2+}$ and $Mg^{2+}$ ions,not anion exchange. So,option $C$ is incorrect.
$4$. Since only option $B$ is correct,option $D$ is incorrect.

(B) The structure of $H_2O_2$ is not planar; it adopts an open book structure to minimize repulsion between the lone pairs on the oxygen atoms.
In the gas phase,the $O-O-H$ bond angle is $101.5^{\circ}$,and the $O-O$ single bond distance is $1.48 \mathring{A}$.
This non-planar geometry is essential for the stability of the molecule.

(C) $H_2O_2$ (Hydrogen peroxide) is a light-sensitive compound. It undergoes decomposition into water and oxygen in the presence of light: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$. To prevent this photochemical decomposition,it is stored in dark-colored (black) bottles. Additionally,it is often stored in wax-lined glass bottles to prevent the catalytic decomposition caused by rough glass surfaces.

(B) Old lead paintings often contain lead pigments that react with atmospheric $H_2S$ to form black lead sulfide $(PbS)$.
$H_2O_2$ acts as an oxidizing agent and converts the black $PbS$ into white lead sulfate $(PbSO_4)$.
The chemical reaction is: $PbS(s) + 4H_2O_2(aq) \rightarrow PbSO_4(s) + 4H_2O(l)$.

(C) In the reaction $H_2S + H_2O_2 \to S + 2H_2O$,the oxidation state of sulfur in $H_2S$ changes from $-2$ to $0$ (oxidation).
Simultaneously,the oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $-2$ (reduction).
Since $H_2O_2$ undergoes reduction,it acts as an oxidizing agent.
Therefore,the reaction manifests the oxidizing nature of $H_2O_2$.
Hence,option $C$ is correct.

(D) reducing agent is a substance that loses electrons and undergoes oxidation.
In reaction $(II)$,$H_2O_2 \to O_2 + 2H^{+} + 2e^-$,$H_2O_2$ loses electrons and is oxidized,thus acting as a reducing agent.
In reaction $(IV)$,$H_2O_2 + 2OH^{-} \to O_2 + 2H_2O + 2e^-$,$H_2O_2$ loses electrons and is oxidized,thus acting as a reducing agent.
In reactions $(I)$ and $(III)$,$H_2O_2$ gains electrons and is reduced,acting as an oxidizing agent.
Therefore,$H_2O_2$ acts as a reducing agent in reactions $(II)$ and $(IV)$.