Hydrogen peroxide Questions in English

(C) The molecular weight (molar mass) of a substance is defined as the mass of one mole of that substance.
Therefore,the unit of molecular weight is expressed in grams per mole,which is $g \ mol^{-1}$.

(B) The structure of $H_2O_2$ is $H-O-O-H$.
In this molecule,the two oxygen atoms are linked to each other by a single covalent bond ($O-O$ bond).
Therefore,the correct option is $(B)$.

(D) Hydrogen peroxide $(H_2O_2)$ has a non-planar,open-book structure.
In the gas phase,the $O-O$ bond length is $147.5 \text{ pm}$ and the $O-H$ bond length is $95 \text{ pm}$.
The $H-O-O$ bond angle is approximately $94.8^{\circ}$ (often approximated as $97^{\circ}$ in textbooks).
The dihedral angle between the two $H-O-O$ planes is $111.5^{\circ}$ (often approximated as $101^{\circ}$ in some contexts).
Therefore,option $D$ correctly describes the non-planar geometry of $H_2O_2$.

(B) $H_2O_2$ has a non-planar structure.
It adopts an open-book configuration to minimize repulsion between the lone pairs of electrons on the oxygen atoms.

(A) The $H_2O_2$ molecule has a non-planar 'open book' structure in the gas phase.
In this structure,the dihedral angle between the two $O-H$ planes is approximately $111.5^o$.
However,in the solid state,this angle is approximately $90.2^o$.
Given the standard options provided,the closest value representing the dihedral angle in the solid state is $90^o$.

(B) $H_2O_2$ acts as an oxidizing agent and oxidizes $K_4[Fe(CN)_6]$ (potassium ferrocyanide) to $K_3[Fe(CN)_6]$ (potassium ferricyanide) in an acidic medium.
The balanced chemical equation is:
$2K_4[Fe(CN)_6] + H_2SO_4 + H_2O_2 \to 2K_3[Fe(CN)_6] + K_2SO_4 + 2H_2O$
Thus,the correct condition is an acidic solution.

(C) In the given reaction,the oxidation state of sulfur $(S)$ increases from $-2$ in $H_2S$ to $0$ in $S$.
Since $H_2S$ is being oxidized,$H_2O_2$ acts as an oxidizing agent.
Therefore,this reaction represents the oxidizing nature of $H_2O_2$.

(B) $H_2O_2$ can act as an oxidizing agent by gaining electrons (reducing to $H_2O$) and as a reducing agent by losing electrons (oxidizing to $O_2$).
Therefore,it acts as both an oxidizing and a reducing agent.

(D) $H_2O_2$ acts as an oxidant in both acidic and basic media,for example: $2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O$.
It acts as a reductant with 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 to give $H^+$ ions: $H_2O_2 \rightleftharpoons H^+ + HO_2^-$.
Therefore,it exhibits all three properties.

(B) reducing agent is a substance that undergoes oxidation itself while reducing another substance.
In the reaction $Cl_2 + H_2O_2 \to 2HCl + O_2$,the oxidation state of oxygen in $H_2O_2$ increases from $-1$ to $0$ (oxidation),while the oxidation state of chlorine in $Cl_2$ decreases from $0$ to $-1$ (reduction).
Therefore,$H_2O_2$ acts as a reducing agent in this reaction.

(B) The compound $H_2O_2$ (hydrogen peroxide) can act as both an oxidising agent and a reducing agent.
In $H_2O_2$,the oxidation state of oxygen 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 $0$).

(B) The reaction of $BaO_2$ with dilute $H_2SO_4$ is given by:
$BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$
In the products,the most electronegative element is oxygen.
In $H_2O_2$,oxygen is in the peroxide state,so its oxidation state is $-1$.
In $BaSO_4$,oxygen is in the oxide state,so its oxidation state is $-2$.
Therefore,the oxidation states are $-1$ and $-2$.

(A) In hydrogen peroxide $(H_2O_2)$,the oxygen atoms are bonded to each other in a peroxide linkage $(-O-O-)$.
According to the rules for assigning oxidation states,the oxidation state of oxygen in peroxides is always $-1$.

(B) reducing agent is a substance that gets oxidized itself while reducing another substance.
In the reaction $Cl_2 + H_2O_2 \to 2HCl + O_2$,the oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $0$ (oxidation).
Since $H_2O_2$ undergoes oxidation,it acts as a reducing agent for $Cl_2$ (which is reduced from $0$ to $-1$).
In all other options ($A$,$C$,and $D$),$H_2O_2$ acts as an oxidizing agent.

(D) The decomposition of hydrogen peroxide is represented as: $2H_2O_2 \to 2H_2O + O_2$.
From the stoichiometry,$22.4 \ L$ of $O_2$ at $N.T.P.$ is produced by $68 \ g$ of $H_2O_2$.
For a $10$ volume solution,$10 \ L$ of $O_2$ is produced by $1 \ L$ of $H_2O_2$ solution.
Mass of $H_2O_2$ in $1 \ L$ of solution $= \frac{68}{22.4} \times 10 = 30.35 \ g/L$.
Strength in percentage $(\% w/v) = \frac{\text{mass of solute in } g}{\text{volume of solution in } mL} \times 100$.
Strength $= \frac{30.35 \ g}{1000 \ mL} \times 100 = 3.035\%$.

(A) In the laboratory, $H_2O_2$ is prepared by the action of cold dilute $H_2SO_4$ on barium peroxide $(BaO_2)$.
The chemical reaction is:
$BaO_2 \cdot 8H_2O(s) + H_2SO_4(dil.) \to BaSO_4(s) + H_2O_2(aq) + 8H_2O(l)$

(B) The structure of $H_2O_2$ is non-planar and has an open book-like structure.
In the gas phase,the dihedral angle is $111.5^{\circ}$,and in the solid phase,it is $90.2^{\circ}$.

(C) The reaction of sodium peroxide $(Na_2O_2)$ with dilute sulfuric acid $(H_2SO_4)$ produces hydrogen peroxide $(H_2O_2)$ and sodium sulfate $(Na_2SO_4)$.
The chemical equation is:
$Na_2O_2 + H_2SO_4 \to Na_2SO_4 + H_2O_2$
Therefore,the correct option is $C$.

(D) $Hydrogen$ $peroxide$ $(H_2O_2)$ acts as a reducing agent when it reacts with strong oxidizing agents like $O_3$ or $KMnO_4$ in acidic medium.
However,$H_2O_2$ acts as an oxidizing agent when it reacts with $PbS$ (lead sulphide).
The reaction is:
$PbS(s) + 4H_2O_2(aq) \to PbSO_4(s) + 4H_2O(l)$
In this reaction,$PbS$ is oxidized to $PbSO_4$ and $H_2O_2$ is reduced to $H_2O$.

(C) In the given reaction: $H_2S + H_2O_2 \to S + 2H_2O$.
The oxidation state of sulfur in $H_2S$ changes from $-2$ to $0$ in $S$.
Since the oxidation state of sulfur increases,$H_2S$ is oxidized.
$H_2O_2$ acts as an oxidizing agent because it facilitates the oxidation of $H_2S$ and itself gets reduced to $H_2O$.
Therefore,this reaction manifests the oxidising nature of $H_2O_2$.

(B) The reaction between hydrogen peroxide $(H_2O_2)$ and chlorine $(Cl_2)$ is an oxidation-reduction reaction where $H_2O_2$ acts as a reducing agent.
The balanced chemical equation is:
$H_2O_2 + Cl_2 \to 2HCl + O_2$
Therefore,the products are hydrochloric acid $(HCl)$ and oxygen gas $(O_2)$.

(B) $H_2O_2$ acts as both an oxidising and a reducing agent. In the presence of $PbS$ (lead sulfide),$H_2O_2$ acts as an oxidising agent and oxidises $PbS$ to $PbSO_4$ (lead sulfate).
The chemical equation is: $PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$.

(A) Fenton’s reagent is a solution of hydrogen peroxide $(H_2O_2)$ with ferrous iron (typically iron$(II)$ sulfate,$FeSO_4$) as a catalyst.
It is widely used to oxidize contaminants or waste waters.

(A) The relationship between volume strength and normality of $H_2O_2$ is given by the formula:
Volume strength $= 5.6 \times \text{Normality}$
Given,Normality $= 1.5 \, N$
Therefore,Volume strength $= 5.6 \times 1.5 = 8.4 \, \text{liters}$.

(B) $20$ volume $H_2O_2$ means that $1 \, mL$ of $H_2O_2$ solution produces $20 \, mL$ of $O_2$ gas at $N.T.P$.
Given volume of $H_2O_2$ solution = $15 \, mL$.
Volume of $O_2$ liberated = $(\text{Volume of solution}) \times (\text{Volume strength of } H_2O_2)$.
Volume of $O_2$ liberated = $15 \, mL \times 20 = 300 \, mL$.

(A) The molar mass of $H_2O_2$ is $34 \ g/mol$.
The molarity $(M)$ of the solution is given by $M = \frac{\text{mass in } g/L}{\text{molar mass}} = \frac{30.36}{34} = 0.893 \ M$.
The relation between volume strength and molarity is: $\text{Volume strength} = 11.2 \times M$.
$\text{Volume strength} = 11.2 \times 0.893 \approx 10 \text{ volumes}$.

(C) $H_2O_2$ acts as an oxidising as well as a reducing agent in both acidic and alkaline media.

(A) The equivalent weight of a substance is calculated as the molar mass divided by the $n$-factor.
For $H_2O_2$,the molar mass is $2 \times 1 + 2 \times 16 = 34 \ g/mol$.
In redox reactions,$H_2O_2$ acts as an oxidizing or reducing agent,typically involving a change in oxidation state of oxygen from $-1$ to $-2$ or $0$,resulting in an $n$-factor of $2$.
Therefore,the equivalent weight = $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{34}{2} = 17$.

(B) The decomposition of $H_2O_2$ is given by: $2H_2O_2 \rightarrow 2H_2O + O_2$.
$68 \, g$ of $H_2O_2$ produces $22.4 \, L$ of $O_2$ at $N.T.P$.
Therefore,$1 \, L$ of $O_2$ is produced by $\frac{68}{22.4} \, g$ of $H_2O_2$.
For a $20$ volume solution,$1 \, L$ of $H_2O_2$ solution produces $20 \, L$ of $O_2$.
Mass of $H_2O_2$ in $1 \, L$ of solution $= \frac{68}{22.4} \times 20 \approx 60.71 \, g$.
Strength in percentage $(\% \, w/v) = \frac{\text{mass of solute in } g}{\text{volume of solution in } mL} \times 100$.
Strength $= \frac{60.71 \, g}{1000 \, mL} \times 100 \approx 6.07 \%$.
Thus,the strength is approximately $6 \%$.

(C) The correct answer is $(C)$.
Electrolysis of $50\%$ $H_2SO_4$ (sulphuric acid) yields peroxodisulphuric acid $(H_2S_2O_8)$.
This product,upon hydrolysis and subsequent distillation,produces a $30\%$ solution of hydrogen peroxide $(H_2O_2)$.

(A) The volume strength of $H_2O_2$ is defined as the volume of $O_2$ gas (in $mL$) produced at $NTP$ by the decomposition of $1 \ mL$ of $H_2O_2$ solution.
Given that $1 \ mL$ of $H_2O_2$ solution produces $10 \ mL$ of $O_2$ at $NTP$.
Therefore,the solution is $10$ volume $H_2O_2$.

(A) $H_2O_2$ is unstable and decomposes into $H_2O$ and $O_2$.
Substances like $Pt$,$Au$ (Gold),and $MnO_2$ act as catalysts and speed up this decomposition.
Glycerol,phosphoric acid,or acetanilide are added to $H_2O_2$ as stabilizers to prevent or check its decomposition.
Therefore,glycerol does not speed up the decomposition.

(A) $H_2O_2$ acts as both an oxidizing and a reducing agent.
In the reaction with $O_3$,$H_2O_2$ acts as a reducing agent and reduces $O_3$ to $O_2$.
The reaction is: $O_3 + H_2O_2 \to H_2O + 2O_2$.
Since $O_3$ is reduced by $H_2O_2$,it is not oxidized by it.
Conversely,$KI$ is oxidized to $I_2$,$PbS$ is oxidized to $PbSO_4$,and $Na_2SO_3$ is oxidized to $Na_2SO_4$ by $H_2O_2$.

(D) $H_2O_2$ acts as both an oxidizing and a reducing agent.
In the presence of strong oxidizing agents,it acts as a reducing agent.
$KMnO_4/H_2SO_4$,$K_2Cr_2O_7/H_2SO_4$,and $Ag_2O$ are strong oxidizing agents and are reduced by $H_2O_2$.
$Fe^{3+}$ is the highest oxidation state of iron and cannot be further reduced by $H_2O_2$ to $Fe^{2+}$ under standard conditions; instead,$H_2O_2$ can reduce $Fe^{3+}$ only in specific complex environments,but in general redox chemistry,$Fe^{3+}$ is not reduced by $H_2O_2$.

(D) $H_2O_2$ has a higher dielectric constant than $H_2O$,making it a better polar solvent.
However,it is not commonly used as a solvent because of its strong auto-oxidizing ability,which can lead to the decomposition of the solute or the solvent itself.

(A) $H_2O_2$ acts as a very weak acid in aqueous solution.
It dissociates slightly to give $H^+$ and $HO_2^-$ ions.
Therefore,the correct option is $(A)$.

(C) The decomposition of $H_2O_2$ is prevented by adding stabilizers such as acetanilide.
Acetanilide acts as a negative catalyst or stabilizer that retards the rate of decomposition of $H_2O_2$.

(C) $H_2O_2$ is an unstable liquid that decomposes into water and oxygen $(2H_2O_2 \rightarrow 2H_2O + O_2)$ when exposed to light.
To prevent this photochemical decomposition,it is stored in dark-colored (black) bottles.

(D) $H_2O_2$ is used as an aerating agent in the production of sponge rubber,as an antichlor to remove excess chlorine from fabrics,and for restoring the white colour of old lead paintings by converting black $PbS$ to white $PbSO_4$. Therefore,none of the given statements are wrong.

(A) In the given reaction,$H_2O_2$ undergoes oxidation by losing electrons $(2e^-)$ to form $O_2$ and $H^{+}$ ions.
Since $H_2O_2$ is undergoing oxidation,it acts as a reducing agent.
Therefore,the correct option is $A$.

(A) The structure of hydrogen peroxide $(H_2O_2)$ is non-planar.
It possesses an open book structure where the two $O-H$ bonds are in different planes,resulting in a dihedral angle of $111.5^{\circ}$ in the gas phase.

(C) When $H_2O_2$ is added to an acidified solution of potassium dichromate $(K_2Cr_2O_7)$ in the presence of ether,a deep blue color is observed in the ethereal layer.
This is due to the formation of chromium pentoxide $(CrO_5)$,which is soluble in ether.
The chemical reaction is:
$K_2Cr_2O_7 + H_2SO_4 + 4H_2O_2 \to K_2SO_4 + 2CrO_5 (\text{Blue}) + 5H_2O$

(A) The acid dissociation constant $(K_a)$ for hydrogen peroxide $(H_2O_2)$ is approximately $1.55 \times 10^{-12}$.
Therefore,the order of magnitude is $10^{-12}$.

(D) $H_2O_2$ acts as an oxidising agent in both acidic and alkaline media.
In acidic medium: $H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$.
In alkaline medium: $H_2O_2 + 2e^- \rightarrow 2OH^-$.

(D) In the gaseous phase,the $H_2O_2$ molecule has a non-planar structure.
The $H-O-O$ bond angle is $97^o$ in the gaseous phase.
Therefore,the correct option is $(D)$.

(B) The balanced chemical equation for the decomposition of $H_2O_2$ is:
$2H_2O_2 \rightarrow 2H_2O + O_2$
From the stoichiometry:
$2 \times 34 \ g$ of $H_2O_2$ produces $22400 \ mL$ of $O_2$ at $STP$.
$68 \ g$ of $H_2O_2$ produces $22400 \ mL$ of $O_2$.
Therefore,the volume of $O_2$ produced from $0.68 \ g$ of $H_2O_2$ is:
$\text{Volume} = \frac{22400 \ mL \times 0.68 \ g}{68 \ g} = 224 \ mL$.

(D) Hydrogen peroxide $(H_2O_2)$ acts as a weak acid,stronger than water $(H_2O)$.
It acts as an oxidising agent in both acidic and basic media (e.g.,$2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O$).
It also acts as a reducing agent towards strong oxidising agents (e.g.,$2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2$).
Therefore,all the given statements are correct.

(D) $MnO_2$,$PbO_2$,and $BaO$ do not produce $H_2O_2$ when reacted with $HCl$.
$MnO_2$ reacts with $HCl$ to produce $MnCl_2$,$Cl_2$,and $H_2O$.
$PbO_2$ reacts with $HCl$ to produce $PbCl_2$,$Cl_2$,and $H_2O$.
$BaO$ reacts with $HCl$ to produce $BaCl_2$ and $H_2O$.
$H_2O_2$ is typically prepared by the action of acids on hydrated barium peroxide $(BaO_2 \cdot 8H_2O)$,not on $BaO$ or other oxides listed.

(B) Strength of the solution is given by the formula: $\text{Strength} = \text{Normality} \times \text{Equivalent mass}$.
The equivalent mass of $H_2O_2$ is $\frac{\text{Molar mass}}{n\text{-factor}} = \frac{34}{2} = 17 \ g \ eq^{-1}$.
Given,$\text{Normality} = 1.5 \ N$ and $\text{Volume} = 1 \ L$.
$\text{Strength} = 1.5 \times 17 = 25.5 \ g \ L^{-1}$.
Since the volume is $1 \ L$,the amount of $H_2O_2$ present is $25.5 \ g$.