Hydrogen peroxide Questions in English

(B) The strength of $H_2O_2$ in terms of volume is related to percentage strength by the formula: $\text{Strength (\%)} = \frac{\text{Volume strength}}{5.6}$.
Given volume strength is $10$ volume.
Therefore,$\text{Percentage strength} = \frac{10}{5.6} \approx 1.785\%$.
However,using the standard conversion factor where $1$ volume of $H_2O_2$ corresponds to approximately $0.3035\%$ strength,$10$ volume corresponds to $10 \times 0.3035 = 3.035\%$.
Thus,the percentage strength is nearly $3\%$.

(B) Barium peroxide $BaO_2$ is a true peroxide.
It contains the peroxide ion,$O_2^{2-}$.
It is a white ionic solid obtained by the reaction of barium hydroxide with hydrogen peroxide:
$Ba(OH)_2 + H_2O_2 \rightarrow BaO_2 + 2H_2O$

(D) The peroxide ion is $O_2^{2-}$,which contains an $O-O$ single bond.
Among the given options,$MgO$,$CaO$,and $Li_2O$ are normal oxides containing the $O^{2-}$ ion.
$BaO_2$ is a peroxide of barium,which contains the $O_2^{2-}$ ion.
Therefore,the peroxide bond is present in $BaO_2$.

(D) peroxide is a compound containing an oxygen-oxygen single bond $(-O-O-)$.
$Na_2O_2$ (Sodium peroxide) contains the peroxide ion $(O_2^{2-})$.
$CrO_5$ (Chromium pentoxide) has a butterfly structure with two peroxide linkages.
$H_2SO_5$ (Caro's acid) contains one peroxide linkage $(-O-O-)$.
$PbO_2$ (Lead dioxide) is a lead oxide where lead is in the $+4$ oxidation state. It has a fluorite-type structure and does not contain any peroxide linkage.
Therefore,the correct answer is $PbO_2$.

(B) The reaction between hydrogen peroxide $(H_2O_2)$ and lead sulfide $(PbS)$ is an oxidation reaction.
$PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O$.
Thus,$PbS$ is oxidized to $PbSO_4$ by $H_2O_2$.

(A) peroxide linkage is defined by the presence of an $O-O$ bond where the oxidation state of each oxygen atom is $-1$.
In $Pb_2O_3$,the structure is $Pb^{II}-O-Pb^{IV}=O$,which contains an $O-O$ bond (peroxide linkage).
In $SiO_2$ and $CO_2$,the oxidation state of oxygen is $-2$.
In $PbO_2$,the oxidation state of oxygen is $-2$ (it is a dioxide,not a peroxide).
Therefore,$Pb_2O_3$ contains a peroxide linkage.

(C) The reaction of sodium peroxide $(Na_2O_2)$ with water $(H_2O)$ is as follows:
$2Na_2O_2 + 2H_2O \to 4NaOH + O_2$
In this reaction,sodium hydroxide $(NaOH)$ and oxygen gas $(O_2)$ are formed.
Therefore,the correct option is $(C)$.

(A) Peroxide is $H_2O_2$,where oxygen is in $-1$ oxidation state.
$(B)$ Superoxide is $KO_2$,where oxygen is in $-1/2$ oxidation state.
$(C)$ Dioxide is $PbO_2$,where oxygen is in $-2$ oxidation state.
$(D)$ Suboxide is $C_3O_2$,which contains less oxygen than expected from normal valency.
Therefore,the correct match is $A-4, B-3, C-2, D-1$.

(B) The correct answer is $(B)$.
Dilute $H_2SO_4$ does not react with chloride ions $(Cl^-)$ under normal conditions.
However,it reacts with peroxides (like $BaO_2$) to produce hydrogen peroxide $(H_2O_2)$,which can be further identified by its characteristic reactions,such as the blue coloration with acidified potassium dichromate solution.

(C) Strength = $\text{Normality} \times \text{Equivalent Weight (EW)}$ of $H_2O_2$
$= 1.5 \ N \times 17 \ g \ eq^{-1} = 25.5 \ g \ L^{-1}$
Decomposition reaction: $2H_2O_2 \rightarrow 2H_2O + O_2$
$2 \times 34 \ g = 68 \ g$ of $H_2O_2$ produces $22.4 \ L$ of $O_2$ at $STP$.
Volume strength = $\frac{11.2 \times \text{Normality}}{2} = 5.6 \times \text{Normality}$
$= 5.6 \times 1.5 = 8.4$
Thus,the volume strength of $1.5 \ N \ H_2O_2$ solution is $8.4$.

(C) The relationship between volume strength and normality is given by the formula: $\text{Normality} = \frac{\text{Volume strength}}{5.6}$.
Given,volume strength = $10$ $V$.
Therefore,$\text{Normality} = \frac{10}{5.6} \approx 1.786 \ \text{N}$.
Rounding to the nearest provided option,the answer is $1.78$.

(D) The decomposition of $H_2O_2$ is represented as: $2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g) + \text{Heat}$.
$1$. It is an exothermic reaction because heat is released during the process.
$2$. It is an example of auto-oxidation (or disproportionation) where the oxygen in $H_2O_2$ (oxidation state $-1$) is both oxidized to $O_2$ $(0)$ and reduced to $H_2O$ $(-2)$.
$3$. It is an example of negative catalysis because substances like acetanilide or glycerol act as negative catalysts to slow down its decomposition.
Therefore,all the given options are correct.

(A) In an acidic medium,$H_2O_2$ acts as a reducing agent towards strong oxidizing agents like $KMnO_4$.
The reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$.
In this reaction,$KMnO_4$ is reduced to $Mn^{2+}$ ions,and the pink color of the permanganate solution disappears.

(C) When an acidified solution of dichromate (chromic acid) is treated with hydrogen peroxide $(H_2O_2)$,a deep blue colored compound,chromium pentoxide $(CrO_5)$,is formed.
The chemical reaction is as follows:
$K_2Cr_2O_7 + H_2SO_4 + 4H_2O_2 \to K_2SO_4 + 2CrO_5 + 5H_2O$
$CrO_5$ is an unstable compound that decomposes to form $Cr^{3+}$ ions in acidic medium.

(B) The industrial preparation of $H_2O_2$ involves the auto-oxidation of $2-$ethylanthraquinol.
When $2-$ethylanthraquinol is dissolved in a mixture of benzene and cyclohexanol and oxidised by air,it produces $2-$ethylanthraquinone and $H_2O_2$ (hydrogen peroxide).
Thus,the product obtained is $H_2O_2$.

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

(B) Let the oxidation number of $O$ be $X$.
In $H_2O_2$,the oxidation number of $H$ is $+1$.
Applying the rule for the sum of oxidation numbers in a neutral molecule:
$2(+1) + 2(X) = 0$
$2 + 2X = 0$
$2X = -2$
$X = -1$
Therefore,the oxidation number of $O$ in $H_2O_2$ is $-1$.

(C) In the reaction $Ag_2O + H_2O_2 \to 2Ag + H_2O + O_2$,the oxidation state of oxygen in $H_2O_2$ changes from $-1$ to $0$ in $O_2$.
Since $H_2O_2$ undergoes oxidation,it acts as a reducing agent.

(C) In the electrolytic production of $H_2O_2$ using $50\% \ H_2SO_4$,the process involves the oxidation of $HSO_4^-$ ions to $H_2S_2O_8$ (peroxydisulfuric acid) at the anode,followed by hydrolysis. Hydrogen gas is liberated at the cathode,not the anode. Therefore,the statement that hydrogen is liberated at the anode is incorrect.

(A) $H_2O_2$ acts as a very weak acid. It is slightly stronger than water.
The dissociation constant for $H_2O_2$ is given by $K_a = 1.5 \times 10^{-12}$.
Therefore,the order of magnitude of the $K_a$ value is $10^{-12}$.

(A) The industrial preparation of $H_2O_2$ involves the auto-oxidation of $2$-ethylanthraquinol.
In this process,$2$-ethylanthraquinol is oxidized by air $(O_2)$ to produce $2$-ethylanthraquinone and $H_2O_2$ as a byproduct.
The reaction is: $2$-Ethylanthraquinol $\xrightarrow{O_2/\text{Air}}$ $2$-Ethylanthraquinone + $H_2O_2$.

(A) The decomposition of hydrogen peroxide is given by the equation: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.
$1$ volume of $H_2O_2$ solution gives $10$ volumes of $O_2$ at $STP$.
$2 \times 34 \ g$ of $H_2O_2$ produces $22400 \ mL$ of $O_2$ at $STP$.
So,$22400 \ mL$ of $O_2$ is produced by $68 \ g$ of $H_2O_2$.
Therefore,$1000 \ mL$ of $O_2$ (for $10$ volume) is produced by $(68 \times 1000) / 22400 \approx 3.036 \ g$ of $H_2O_2$.
Since $1000 \ mL$ of $H_2O_2$ solution weighs approximately $1000 \ g$,the percentage concentration is $(3.036 / 1000) \times 100 \approx 3\%$.

(B) The concentration in $g/L$ is calculated as: $\text{Concentration} = \text{Normality} \times \text{Equivalent weight}$.
For $H_2O_2$,the molar mass is $34 \, g/mol$ and the $n$-factor is $2$,so the equivalent weight is $34 / 2 = 17 \, g/eq$.
Given Normality $= 1.5 \, N$ and Volume $= 1 \, L$.
$\text{Mass} = \text{Normality} \times \text{Equivalent weight} \times \text{Volume in } L$.
$\text{Mass} = 1.5 \times 17 \times 1 = 25.5 \, g$.

(D) Currently,$H_2O_2$ is produced industrially by the auto-oxidation of $2$-ethylanthraquinol.
This process involves the catalytic hydrogenation of $2$-ethylanthraquinone to $2$-ethylanthraquinol,followed by oxidation with air to regenerate the quinone and produce $H_2O_2$.

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

(C) The decomposition of $H_2O_2$ is given by: $2H_2O_2(aq) \to 2H_2O(l) + O_2(g)$.
Volume strength is defined as the volume of $O_2$ gas produced at $STP$ by $1 \, mL$ of $H_2O_2$ solution.
Given: $100 \, mL$ of $H_2O_2$ produces $1000 \, mL$ of $O_2$ at $STP$.
Therefore,$1 \, mL$ of $H_2O_2$ produces $\frac{1000}{100} = 10 \, mL$ of $O_2$ at $STP$.
Thus,the volume strength of the $H_2O_2$ sample is $10 \, \text{volume}$.

(A) In a basic medium,$H_2O_2$ acts as a reducing agent towards $I_2$.
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 product formed is iodide ion $(I^-)$.

(C) $Na_2O_2$ (Sodium peroxide) is used as a source of oxygen in submarines and closed spaces and is commercially known as 'Oxone'.

(B) Hydrogen peroxide $(H_2O_2)$ is widely used as a mild bleaching agent for textiles,paper,and hair because it acts as an oxidizing agent.
It releases nascent oxygen which oxidizes the colored impurities to colorless substances.
Therefore,the correct option is $B$.

(B) Hydrogen peroxide $(H_2O_2)$ is widely used as an environmentally friendly bleaching agent for paper and pulp industries.
It replaces chlorine-based bleaching agents,which are harmful to the environment.

(B) The reaction between ozone $(O_3)$ and hydrogen peroxide $(H_2O_2)$ is given by:
$O_3 + H_2O_2 \rightarrow H_2O + 2O_2$
In this reaction,the oxidation state of oxygen in $H_2O_2$ increases from $-1$ to $0$ in $O_2$.
Since $H_2O_2$ undergoes oxidation,it acts as a reducing agent.

(C) Perhydrol is a trade name for a $30\%$ solution of hydrogen peroxide $({H_2}{O_2})$ by weight.
$10$ volume ${H_2}{O_2}$ corresponds to a $3\%$ solution.
Since $3\%$ ${H_2}{O_2}$ is $10$ volume,then $30\%$ ${H_2}{O_2}$ is $10 \times 10 = 100$ volume.
Therefore,the volume strength of perhydrol is $100$.

(B) $H_2O_2$ acts as both an oxidizing and a reducing agent.
In the case of $Fe^{3+}$,it is already in its highest oxidation state $(+3)$ for iron in common compounds,and $H_2O_2$ cannot reduce it to $Fe^{2+}$.
Conversely,$H_2O_2$ reduces $Ag_2O$ to $Ag$,and it reduces acidified $KMnO_4$ and $K_2Cr_2O_7$ to $Mn^{2+}$ and $Cr^{3+}$ respectively.

(D) The reaction between $AgNO_3$ and $NaOH$ produces a brown precipitate of silver oxide:
$2AgNO_{3(aq)} + 2NaOH_{(aq)} \to Ag_2O_{(s)} + H_2O_{(l)} + 2NaNO_{3(aq)}$
(Brown ppt.)
When $H_2O_2$ is added to this brown precipitate,it acts as a reducing agent and reduces $Ag_2O$ to metallic silver,which appears as a black precipitate:
$Ag_2O_{(s)} + H_2O_{2(aq)} \to 2Ag_{(s)} + H_2O_{(l)} + O_{2(g)}$
(Black ppt.)
Therefore,the black precipitate is finely divided metallic silver $(Ag)$.

(B) Atomic hydrogen reacts with oxygen to give almost pure hydrogen peroxide.
$2[H] + O_2 \to H_2O_2$

(C) When $CO_2$ is bubbled through a cold pasty solution of barium peroxide $(BaO_2)$ in water,$H_2O_2$ is obtained.
The chemical reaction is: $BaO_2 + CO_2 + H_2O \to BaCO_3 + H_2O_2$.
Barium carbonate $(BaCO_3)$ being insoluble is filtered off.
This method of preparation of hydrogen peroxide is known as Merck's process.

(A) $PbO_2$ (Lead dioxide) is a dioxide where lead is in the $+4$ oxidation state,and it does not contain the peroxide ion $(O_2^{2-})$.
$H_2O_2$,$SrO_2$,and $BaO_2$ are peroxides as they contain the $( - O - O - )$ linkage.

(C) In reaction $I$: The oxidation state of oxygen in $H_2O_2$ is $-1$. It changes to $0$ in $O_2$. Since the oxidation state increases,$H_2O_2$ acts as a reducing agent.
In reaction $II$: The oxidation state of oxygen in $H_2O_2$ is $-1$. It changes to $0$ in $O_2$. Since the oxidation state increases,$H_2O_2$ acts as a reducing agent.
Therefore,$H_2O_2$ acts as a reducing agent in both reactions.

(D) reducing agent is a substance that undergoes oxidation by losing electrons.
In the reaction $H_2O_2 + 2OH^- \rightarrow O_2 + 2H_2O + 2e^-$,the oxidation state of oxygen in $H_2O_2$ increases from $-1$ to $0$ in $O_2$.
Since $H_2O_2$ loses electrons and is oxidized,it acts as a reducing agent.

(C) . It has to be stored in plastic or wax-lined glass bottles in the dark to prevent its decomposition. Thus,the statement is correct.
$B$. It has to be kept away from dust as it decomposes in the presence of dust. Thus,the statement is correct.
$C$. Hydrogen peroxide can act as both an oxidizing and a reducing agent. For example,in an acidic medium,it oxidizes ferrous ions to ferric ions: $2 Fe^{2+} + H_2O_2 + 2 H^{+} \rightarrow 2 Fe^{3+} + 2 H_2O$. In an acidic medium,it also reduces $HOCl$ to chloride ions: $HOCl + H_2O_2 \rightarrow H^{+} + Cl^{-} + H_2O + O_2$. Thus,the statement that it acts only as an oxidizing agent is incorrect.
$D$. It decomposes on exposure to light to form water and oxygen: $2 H_2O_2 \rightarrow 2 H_2O + O_2$. Thus,the statement is correct.

(A) In an alkaline medium,hydrogen peroxide $(H_2O_2)$ acts as an oxidizing agent. It oxidizes manganese $(II)$ hydroxide $(Mn(OH)_2)$ to manganese $(IV)$ oxide hydrate or manganese dioxide ($MnO_2 \cdot H_2O$ or $MnO(OH)_2$).
The balanced chemical equation is:
$Mn(OH)_2 + H_2O_2 \to MnO(OH)_2 + H_2O$

(D) The hydrolysis of peroxides yields $H_2O_2$.
$Na_2O_2 + 2H_2O \rightarrow 2NaOH + H_2O_2$.
Superoxides like $KO_2$ react with water to produce $H_2O_2$ along with $O_2$ and $KOH$.
$2KO_2 + 2H_2O \rightarrow 2KOH + H_2O_2 + O_2$.
Therefore,both $Na_2O_2$ and $KO_2$ produce $H_2O_2$ upon hydrolysis.

(B) $1$. Calcium bicarbonate $(Ca(HCO_3)_2)$ causes 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 calcium and magnesium ions,not anion exchange. So,option $C$ is incorrect.
$4$. Since only option $B$ is correct,option $D$ is incorrect.

(D) $H_2SO_4$ is a strong acid and a dehydrating agent.
To dry a substance,the drying agent must not react with it.
$CaO$ is a basic oxide and will react with $H_2SO_4$ to form $CaSO_4$ and $H_2O$.
$P_2O_5$ is an acidic oxide but it reacts with $H_2SO_4$ to form phosphoric acid and sulfur trioxide.
$CaCl_2$ also reacts with $H_2SO_4$ to form $HCl$ gas.
Therefore,none of the given substances can be used to dry $H_2SO_4$.

(B) $KMnO_4$ acts as a strong oxidizing agent in acidic medium and $H_2O_2$ acts as a reducing agent.
The reaction is: $2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 5O_2 + 8H_2O$.
In this reaction,the manganate$(VII)$ ions $(MnO_4^-)$ are reduced to manganese$(II)$ ions $(Mn^{2+})$,which causes the decolourisation of the purple solution.

(D) $H_2O_2$ (Hydrogen peroxide) shows acidic nature as it ionizes in water to give $H^+$ ions.
$H_2O_2$ can act as an oxidising agent by gaining electrons (e.g.,$H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$).
$H_2O_2$ can act as a reducing agent by losing electrons (e.g.,$H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$).
Therefore,all the given statements are correct.

(D) The $-O-O-$ bond is characteristic of peroxides.
$BaO_2$ (Barium peroxide) contains the peroxide ion $O_2^{2-}$,which has an $-O-O-$ bond.
$Na_2O_2$ (Sodium peroxide) also contains the peroxide ion $O_2^{2-}$,which has an $-O-O-$ bond.
$CrO_5$ (Chromium pentoxide) has a butterfly structure containing two peroxide linkages $(-O-O-)$.
Therefore,all the given compounds contain $-O-O-$ bonds.

(C) The electrolysis of a concentrated solution of $KHSO_4$ or $NH_4HSO_4$ yields peroxodisulphate ($K_2S_2O_8$ or $(NH_4)_2S_2O_8$) at the anode.
The reaction is: $2KHSO_4 \rightarrow K_2S_2O_8 + H_2$.
This peroxodisulphate is then subjected to hydrolysis to produce $H_2O_2$.
The reaction is: $K_2S_2O_8 + 2H_2O \rightarrow 2KHSO_4 + H_2O_2$.