Hydrogen peroxide Questions in English

(C) $H_2O_2$ acts as a reducing agent towards strong oxidizing agents like $Cl_2$.
The chemical reaction is: $H_2O_2 + Cl_2 \longrightarrow 2HCl + O_2$.
In this reaction,$HCl$ is produced,which is a strong acid.
The formation of $HCl$ increases the concentration of $H^+$ ions in the solution.
Consequently,the $pH$ of the resultant solution decreases (becomes less than $6.0$) and oxygen gas $(O_2)$ is evolved.

(D) $i$. $Ca(OH)_2$ (slaked lime) is added to hard water in Clark's method to precipitate calcium and magnesium bicarbonates as carbonates,effectively removing temporary hardness. This statement is correct.
$ii$. $10$ vol $H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $10 \ mL$ of $O_2$ at $STP$. Therefore,$100 \ mL$ of $10$ vol $H_2O_2$ gives $100 \times 10 = 1000 \ mL = 1 \ L$ of $O_2$ at $STP$. This statement is correct.
$iii$. $H_2O_2$ is an unstable liquid and decomposes slowly on exposure to light or on standing. Urea,acetanilide,or phosphoric acid are commonly added in small amounts as stabilizers to prevent its decomposition. This statement is correct.
Thus,all statements $i, ii,$ and $iii$ are correct.

(D) The correct option is $(d)$. Chlorine reacts with water to form hypochlorous acid $(HOCl)$,which is unstable and decomposes to give nascent oxygen $([O])$.
$Cl_2 + H_2O \longrightarrow HCl + HOCl$
$HOCl \longrightarrow HCl + [O]$
This nascent oxygen is responsible for the bleaching and oxidizing properties of chlorine water.

(B) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
According to the definition of '$X$ volume' $H_2O_2$,$1 \ mL$ of $H_2O_2$ solution produces $X \ mL$ of $O_2$ gas at $STP$.
Given that $4 \ mL$ of $H_2O_2$ produces $80 \ mL$ of $O_2$ at $STP$.
Therefore,$1 \ mL$ of $H_2O_2$ produces $\frac{80 \ mL}{4 \ mL} = 20 \ mL$ of $O_2$ at $STP$.
Thus,the value of $X$ is $20$.

(C) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in acidic medium is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$.
From the stoichiometry,$2 \text{ moles of } KMnO_4$ react with $5 \text{ moles of } H_2O_2$.
Moles of $KMnO_4 = \text{Molarity} \times \text{Volume (L)} = 0.4 \times 0.2 = 0.08 \text{ mol}$.
Moles of $H_2O_2$ required = $\frac{5}{2} \times 0.08 = 0.2 \text{ mol}$.
Mass of $H_2O_2$ = $0.2 \times 34 = 6.8 \text{ g}$.
$10$ volume $H_2O_2$ means $1 \text{ mL}$ of solution gives $10 \text{ mL}$ of $O_2$ at $STP$.
The strength of $10$ volume $H_2O_2$ is $\frac{68}{22.4} \times 10 \approx 30.36 \text{ g/L}$ or $3.036 \%$.
Using the relation: $\text{Volume strength} = 5.6 \times \text{Normality}$.
$10 = 5.6 \times N \Rightarrow N = \frac{10}{5.6} = 1.786 \text{ N}$.
Since $n$-factor for $H_2O_2$ is $2$,Molarity $M = \frac{1.786}{2} = 0.893 \text{ M}$.
Equating milliequivalents: $N_1V_1 = N_2V_2$.
$N_{KMnO_4} = 0.4 \times 5 = 2 \text{ N}$.
$2 \times 200 = 1.786 \times V_{H_2O_2} \Rightarrow V_{H_2O_2} = \frac{400}{1.786} \approx 224 \text{ mL}$.

(D) $(i)$ When $KO_2$ reacts with water,it gives $(A)$,$(B)$ and $(C)$ as follows:
$2KO_2 + 2H_2O \rightarrow 2KOH + H_2O_2 + O_2$
Here,$A = KOH$,$B = H_2O_2$,and $C = O_2$.
$(ii)$ When $(B)$,i.e.,$H_2O_2$ reacts with iodine in basic medium,it gives $(C)$,i.e.,$O_2$,as shown below:
$I_2 + H_2O_2 + 2OH^{-} \rightarrow 2I^{-} + 2H_2O + O_2$
Thus,$B$ is $H_2O_2$ and $C$ is $O_2$. Therefore,option $(D)$ is the correct answer.

(A) The hydrolysis of a superoxide $(MO_2)$ of a group $1$ element $(M)$ is given by the reaction: $2MO_2 + 2H_2O \longrightarrow 2M^{+} + 2OH^{-} + H_2O_2 + O_2$.
Here,$X$ is $H_2O_2$ and $Y$ is $O_2$.
When $X$ $(H_2O_2)$ reacts with $I_2$ in a basic medium,the reaction is: $I_2 + H_2O_2 + 2OH^{-} \longrightarrow 2I^{-} + 2H_2O + O_2$.
In this reaction,$I_2$ is reduced to $I^{-}$ and $H_2O_2$ is oxidized to $O_2$ $(Y)$.
Thus,$X$ is $H_2O_2$ and $Y$ is $O_2$.

(C) For $H_2O_2$,the relationship between Normality $(N)$ and Molarity $(M)$ is given by: $N = n \times M$,where $n$-factor for $H_2O_2$ is $2$.
$M = \frac{N}{2} = \frac{1.5}{2} = 0.75 \ M$.
$\% \left(\frac{W}{V}\right)$ is defined as the mass of solute in grams present in $100 \ mL$ of solution.
Mass of $H_2O_2$ in $1 \ L$ $(1000 \ mL)$ $= M \times \text{Molar Mass} = 0.75 \times 34 = 25.5 \ g/L$.
Therefore,in $100 \ mL$,the mass is $\frac{25.5}{1000} \times 100 = 2.55 \ g$.
Thus,the $\% \left(\frac{W}{V}\right)$ is $2.55$.

(C) The normality $(N)$ of $H_2O_2$ is related to its molarity $(M)$ by the equation: $N = M \times n$-factor.
For $H_2O_2$,the $n$-factor is $2$,so $M = \frac{N}{2} = \frac{3}{2} = 1.5$ $M$.
The volume strength of $H_2O_2$ is given by the formula: $\text{Volume strength} = M \times 11.2$.
Therefore,$\text{Volume strength} = 1.5 \times 11.2 = 16.8$ $L$.
This is approximately $17$ $L$.

(D) Alkaline formaldehyde solution reacts with $H_2 O_2$ as follows:
$2HCHO + H_2 O_2 \rightarrow 2HCOOH + H_2 \uparrow$
In this reaction,formaldehyde $(HCHO)$ is oxidized to formic acid $(HCOOH)$ and hydrogen gas $(H_2)$ is liberated.

(B) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
According to the definition of "$x$ volume" $H_2O_2$,$1 \ mL$ of $H_2O_2$ solution produces $x \ mL$ of $O_2$ gas at $STP$.
Given that $1 \ mL$ of $H_2O_2$ produces $20 \ mL$ of $O_2$ at $STP$,the solution is $20$ volume $H_2O_2$,so $x = 20$.
The relation between volume strength and $(w/v) \%$ is given by the formula: $\text{Volume strength} = \frac{11.2}{34} \times \text{Concentration in } (w/v) \% \times 2$ (or more simply,$\text{Volume strength} = 5.6 \times \text{Molarity}$ and $\text{Molarity} = \frac{(w/v) \% \times 10}{34}$).
Substituting the values: $20 = \frac{11.2 \times (w/v) \%}{34} \times 2$ is not the standard form. Using $\text{Volume strength} = 5.6 \times \text{Molarity}$,we get $20 = 5.6 \times \text{Molarity}$,so $\text{Molarity} = \frac{20}{5.6} \approx 3.57 \ M$.
Now,$(w/v) \% = \frac{\text{Molarity} \times \text{Molar mass}}{10} = \frac{3.57 \times 34}{10} = 12.138 \%$. However,using the direct conversion factor $1 \text{ volume} = 0.303 \% (w/v)$,we get $20 \times 0.303 = 6.06 \% (w/v)$.

(C) The structure of $H_2O_2$ is non-planar.
In the gaseous phase,the dihedral angle is $111.5^{\circ}$.
In the solid phase,due to hydrogen bonding,the dihedral angle decreases to $90.2^{\circ}$.
Therefore,the dihedral angles in gaseous and solid phases are $111.5^{\circ}$ and $90.2^{\circ}$ respectively.

(A) $H_2O_2$ is unstable and decomposes into $H_2O$ and $O_2$ in the presence of light or rough surfaces like glass containing alkali oxides.
To prevent this,it is stored in wax-lined glass or plastic bottles to provide a smooth,inert surface.
Additionally,it must be kept in the dark to avoid photochemical decomposition.

(A) Hydrogen peroxide $(H_2O_2)$ is an unstable compound.
When exposed to light,it undergoes decomposition to form water $(H_2O)$ and oxygen gas $(O_2)$.
The balanced chemical equation for this decomposition is: $2 H_2O_2(aq) \rightarrow 2 H_2O(l) + O_2(g)$.
Therefore,it is stored in dark-colored bottles away from light to prevent this decomposition.

(D) $22.4$ volume $H_2O_2$ means $1 \ mL$ of $H_2O_2$ solution gives $22.4 \ mL$ of $O_2$ at $NTP$.

$2H_2O_2 \longrightarrow 2H_2O + O_2$	$2 \ mol \longrightarrow 1 \ mol$
$68 \ g$	$22400 \ mL$ at $NTP$

$\therefore 22400 \ mL$ of $O_2$ is obtained by $68 \ g$ of $H_2O_2$.
$\therefore 22.4 \ mL$ of $O_2$ is obtained by $\frac{68 \times 22.4}{22400} = 0.068 \ g$ of $H_2O_2$.
Since $1 \ mL$ of $H_2O_2$ solution contains $0.068 \ g$ of $H_2O_2$,
$\therefore 100 \ mL$ of $H_2O_2$ solution contains $0.068 \times 100 = 6.8 \ g$ of $H_2O_2$.
Thus,the strength is $6.8 \%$.

(D) When an acidified solution of $TiO_2$ is treated with $H_2O_2$,an intense yellow-orange colour is obtained due to the formation of pertitanic acid,$H_2TiO_4$.
The chemical reaction is:
$TiO_2 + H_2O_2 \xrightarrow{H_2SO_4} H_2TiO_4$
This reaction is used for the detection of both $Ti(IV)$ and $H_2O_2$.

(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 $H_2S_2O_8$.
$H_2SO_4$ has $0$ peroxy bonds,while $H_2S_2O_8$ (Marshall's acid) has $1$ peroxy bond $(HO_3S-O-O-SO_3H)$.

(C) $H_2O_2$ acts as both an oxidizing and a reducing agent.
$1.$ Oxidizing action:
In acidic medium: $H_2O_2 + 2H^{+} + 2e^{-} \longrightarrow 2H_2O$. Reaction $(I)$ is a standard oxidation of $Fe^{2+}$ to $Fe^{3+}$ by $H_2O_2$ in acidic medium.
$2.$ Reducing action:
In acidic medium: $H_2O_2 \longrightarrow 2H^{+} + O_2 + 2e^{-}$. Reaction $(II)$ is a standard reduction of $MnO_4^{-}$ to $Mn^{2+}$ by $H_2O_2$ in acidic medium.
$3.$ In alkaline medium:
$H_2O_2$ can act as an oxidizing agent: $H_2O_2 + 2e^{-} \longrightarrow 2OH^{-}$. Reaction $(III)$ represents the oxidation of $Fe^{2+}$ to $Fe^{3+}$ in basic conditions.
$H_2O_2$ can act as a reducing agent: $H_2O_2 + 2OH^{-} \longrightarrow 2H_2O + O_2 + 2e^{-}$. Reaction $(IV)$ represents the reduction of $MnO_4^{-}$ to $MnO_2$ in basic conditions.
Since all four reactions represent valid redox processes involving $H_2O_2$ in their respective media,all are feasible.

(A) The balanced chemical equation for the reaction between $KMnO_4$ and $H_2O_2$ in an acidic medium is:
$2 \ KMnO_4 + 3 \ H_2SO_4 + 5 \ H_2O_2 \longrightarrow K_2SO_4 + 2 \ MnSO_4 + 8 \ H_2O + 5 \ O_2$
From the stoichiometry,$2 \ \text{moles of } KMnO_4$ react with $5 \ \text{moles of } H_2O_2$.
Given: $n(KMnO_4) = \text{Molarity} \times \text{Volume} = 0.02 \ M \times 1 \ L = 0.02 \ \text{moles}$.
Therefore,$n(H_2O_2) = \frac{5}{2} \times 0.02 = 0.05 \ \text{moles}$.
Mass of $H_2O_2 = 0.05 \ \text{mol} \times 34 \ g/mol = 1.7 \ g$.
$10 \ \text{vol } H_2O_2$ means $1 \ L$ of $H_2O_2$ solution gives $10 \ L$ of $O_2$ at $STP$.
Since $1 \ L$ of $H_2O_2$ contains $30.3 \ g$ of $H_2O_2$ (as $68 \ g$ of $H_2O_2$ gives $22.4 \ L$ of $O_2$,so $10 \ L$ of $O_2$ corresponds to $\frac{68 \times 10}{22.4} \approx 30.35 \ g$),the concentration is $\frac{30.35 \ g}{1000 \ mL}$.
Volume of $H_2O_2$ solution required $= \frac{1.7 \ g}{30.35 \ g/L} \approx 0.056 \ L = 56 \ mL$.

(B) $H_2O_2$ reacts with disodium hydrogen phosphate $(Na_2HPO_4)$ to form an addition product.
The reaction is as follows:
$H_2O_2 + Na_2HPO_4 \rightarrow Na_2HPO_4 \cdot H_2O_2$
This is a stable addition product.

(D) The oxidising property of $H_2O_2$ is shown when it acts as an oxidising agent,meaning $H_2O_2$ itself gets reduced to $H_2O$ (oxidation state of $O$ changes from $-1$ to $-2$).
In option $D$,$2KI + H_2SO_4 + H_2O_2 \longrightarrow K_2SO_4 + I_2 + 2H_2O$,the oxidation state of $I$ in $KI$ increases from $-1$ to $0$ (oxidation),and the oxidation state of $O$ in $H_2O_2$ decreases from $-1$ to $-2$ (reduction).
Thus,$H_2O_2$ acts as an oxidising agent.

(C) $H_2O_2 + Cl_2 \longrightarrow 2HCl + O_2$
In this reaction,$H_2O_2$ acts as a reducing agent and reduces $Cl_2$ to $HCl$.
The formation of $HCl$ (a strong acid) increases the concentration of $H^+$ ions in the solution.
Consequently,the $pH$ of the resultant solution decreases to a value less than $6.0$,and oxygen gas $(O_2)$ is evolved.

(C) In the given reactions,we analyze the products formed:
$(A)$ $PbO_2 + H_2O_2 \longrightarrow PbO + H_2O + O_2$ (Oxygen gas is formed).
$(B)$ $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \longrightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$ (Oxygen gas is formed).
$(C)$ $PbS + 4H_2O_2 \longrightarrow PbSO_4 + 4H_2O$ (Lead sulphate is a solid,and water is a liquid; no gas is formed).
$(D)$ $Cl_2 + H_2O_2 \longrightarrow 2HCl + O_2$ (Oxygen gas is formed).
Therefore,the reaction in option $C$ does not produce a gaseous product.

(A) reducing agent is a substance that reduces another species and itself gets oxidized. In the reaction $PbO_{2(s)} + H_2O_{2(aq)} \longrightarrow PbO_{(s)} + H_2O_{(l)} + O_{2(g)}$,the oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $0$ (in $O_2$),which indicates oxidation. Therefore,$H_2O_2$ acts as a reducing agent in this reaction. In the other options $(B, C, D)$,$H_2O_2$ acts as an oxidizing agent because it gets reduced to $H_2O$ (oxygen oxidation state changes from $-1$ to $-2$).

(D) The relationship between volume strength and normality of $H_2O_2$ is given by the formula: $\text{Volume strength} = 5.6 \times \text{Normality}$.
Given,volume strength = $20$.
Therefore,$\text{Normality} = \frac{\text{Volume strength}}{5.6} = \frac{20}{5.6} \approx 3.571 \ N$.
Thus,the correct option is $D$.

(D) The hydrolysis of sodium superoxide $(NaO_2)$ proceeds according to the following chemical equation:
$2NaO_2 + 2H_2O \rightarrow 2NaOH + H_2O_2 + O_2$
From the reaction,it is evident that the products formed are sodium hydroxide $(NaOH)$,hydrogen peroxide $(H_2O_2)$,and oxygen gas $(O_2)$.
Therefore,the correct options are $A$,$B$,and $C$.

(A) Perhydrol is a $30\% \ w/v$ $H_2O_2$ solution,which corresponds to $100$ volume $H_2O_2$.
This means $1 \ mL$ of perhydrol produces $100 \ mL$ of $O_2$ at $STP$.
The reaction for conversion is: $2SO_2 + O_2 \rightarrow 2SO_3$.
From the stoichiometry,$2 \ L$ of $SO_2$ requires $1 \ L$ of $O_2$ for complete conversion.
Volume of $O_2$ required = $1 \ L = 1000 \ mL$.
Therefore,volume of perhydrol required = $\frac{1000 \ mL}{100} = 10 \ mL$.

(B) The volume of $O_2$ gas evolved at $STP$ is calculated using the formula: $\text{Volume of } O_2 = \text{Volume of } H_2O_2 \text{ solution} \times \text{Volume strength}$.
$(A) 2 \text{ mL} \times 100 = 200 \text{ mL}$
$(B) 500 \text{ mL} \times 30 = 15000 \text{ mL}$
$(C) 1000 \text{ mL} \times 10 = 10000 \text{ mL}$
$(D) 100 \text{ mL} \times 20 = 2000 \text{ mL}$
Comparing the results,option $(B)$ yields the highest volume of $O_2$.

(D) When $H_{2}O_{2}$ reacts with an acidified solution of $K_{2}Cr_{2}O_{7}$,it forms chromium pentoxide $(CrO_{5})$.
The reaction is: $K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 4H_{2}O_{2} \rightarrow K_{2}SO_{4} + 5H_{2}O + 2CrO_{5}$.
$CrO_{5}$ is unstable in aqueous solution but is stabilized in ether,which gives a characteristic blue color to the ethereal layer.

(A) $10 \text{ V } H_{2}O_{2}$ means $1 \text{ L}$ of this solution will produce $10 \text{ L}$ of $O_{2}$ at $STP$.
$2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$
$68 \text{ g}$ of $H_{2}O_{2}$ produces $22.4 \text{ L}$ of $O_{2}$ at $STP$.
Therefore,$22.4 \text{ L}$ of $O_{2}$ is obtained from $68 \text{ g}$ of $H_{2}O_{2}$.
$10 \text{ L}$ of $O_{2}$ will be obtained from $H_{2}O_{2} = \frac{68}{22.4} \times 10 = 30.36 \text{ g}$.
$1000 \text{ mL}$ of the given solution contains $30.36 \text{ g}$ of $H_{2}O_{2}$.
$100 \text{ mL}$ of the given solution contains $\frac{30.36 \times 100}{1000} = 3.036 \text{ g}$ of $H_{2}O_{2}$.
Thus,the percentage strength of $H_{2}O_{2}$ is approximately $3$.

(C) The blackening of oil paintings is caused by the formation of $PbS$ (lead sulfide) due to atmospheric $H_2S$.
This can be restored by the action of $H_2 O_2$ (hydrogen peroxide),which oxidizes the black $PbS$ into white $PbSO_4$ (lead sulfate).
The chemical reaction is:
$PbS + 4H_2 O_2 \rightarrow PbSO_4 + 4H_2 O$

(B) The relationship between volume strength and normality is given by the formula: $\text{Volume strength} = 5.6 \times \text{Normality}$.
Given,volume strength $= 30$.
Substituting the value: $30 = 5.6 \times N$.
Therefore,$N = \frac{30}{5.6} \approx 5.357 \ N$.
Comparing this with the given options,the closest value is $5.336 \ N$.