Q: The reaction of aqueous \$KMnO_4\$ with \$H_2O_2\$ in acidic conditions gives:

The reaction of aqueous \$KMnO_4\$ with \$H_2O_2\$ in acidic conditions gives:\$3H_2SO_4+ 2KMnO_4 + 5H_2 O_2 \rightarrow 5O_2 + 2MnSO_4 + 8H_2O + K_2SO_4\$The oxidation number of \$Mn\$ in \$MnSO_4\$ is \$+2\$.Hence, option \$B\$ is correct.

Q: The reaction \$H_2S+H_2O_2\rightarrow S+2H_2O\$ describes:

The reaction \$H_2S+H_2O_2\rightarrow S+2H_2O\$ manifests oxidizing nature of \$H_2O_2\$ as it oxidises S from -2 to 0 oxidation states.Hence option C is correct.

Q: (a) \$H_2O_2 + O_3 \rightarrow H_2O + 2O_2\$(b) \$H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2\$Role of hydrogen peroxide in the above reactions is respectively:

(a) \$H_2O_2 + O_3 \rightarrow H_2O + 2O_2\$In this reaction, \$O_3\$ is reduced into \$O^{-2}\$ ion .(b) \$H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2\$In this reaction, \$Ag_2O\$ is reduced to Ag .Hence, \$H_2O_2\$ acts as reducing agent in both the reactions.

Q: In which of the following reactions, \$H_{2}O_{2}\$ acts as a reducing reagent ?

\$HOCl + H_{2}O_{2}\overset{H^+}{\rightarrow}H_{3}O^{+} + Cl^{-} + O_2\$Here, the \$H_2O_2\$ act as the reducing agent, as the \$Cl\$ has \$+1\$ oxidation state of in \$HOCl\$ and after reaction with the \$H_2O_2\$ , in the presence of the acid, it reduces the \$Cl\$ in \$HOCl\$ to \$Cl^{-}\$ having -1 oxidation state. So, as \$Cl\$ oxidation state decreases from the +1 to -1 oxidation state, hence the \$H_2O_2\$ is acting as the reducing agent.

Q: Half litre each of three samples of \$H_{2}O_{2}\$ labelled \$10\$ volume, \$15\$ volume and \$20\$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant \$H_{2}O_{2}\$ solution is:

Volume strength of \$H_{ 2 }{ O }_{ 2 } = 5.6\times \text{N}\$ \$(\text{N} =\$ Normality\$)\$Normality of \$10\$ vol of \$H_{ 2 }{ O }_{ 2 } = \dfrac{10}{5.6}\text{N}\$Normality of \$15\$ vol of \$H_{ 2 }{ O }_{ 2 } = \dfrac{15}{5.6} \text{N}\$Normality of \$20\$ vol of \$H_{ 2 }{ O }_{ 2 } = \dfrac{20}{5.6} \text{N}\$Let \$500\$ ml of each is mixed, then the total volume of mix \$= 1500\$ ml.Also, this is diluted to double its volume, i.e., \$= 3000\$ ml\$\text{N}\times 3000\$ = \$\dfrac { 10 }{ 5.6 } \times 500+\dfrac { 15 }{ 5.6 } \times 500+\dfrac { 20 }{ 5.6 } \times 500\$Normality of \$H_{ 2 }{ O }_{ 2 }\$ mix \$= 1.339\:\text{N}\$Vol. strength of \$H_{ 2 }{ O }_{ 2 } = 1.339 \times 5.6 = 7.5\$.

Q: The volume of \$20\:V-H_{2}O_{2}\$ solution, required to prepare \$5\$ litres of oxygen at STP is:

\$20\$ ml of \${ O }_{ 2 }\$ is prepared by \$1\$ ml of \${ H }_{ 2 }{ O }_{ 2 }\$ at STP.\$1\$ ml of \${ O }_{ 2 }\$ is prepared by \$= \dfrac{1}{20}\$ ml of \${ H }_{ 2 }{ O }_{ 2 }\$.\$5000\$ ml of \${ O }_{ 2 }\$ is prepared by = \$\dfrac { 1 }{ 20 } \times 5000 = 250\$ ml of \${ H }_{ 2 }{ O }_{ 2 }\$.Hence, option B is correct.

Q: \${ H }_{ 2 }{ O }_{ 2 }\$ solution used for hair bleaching is sold as a solution of approximately 5.0 g \${ H }_{ 2 }{ O }_{ 2 }\$ per 100mL of the solution. The molecular mass of \${ H }_{ 2 }{ O }_{ 2 }\$ is 34. The molarity of this solution is approximately :

Man of \$H_2O_2=5\ g\$molar man of \$H_2O_2=34\$number of mole \$=\dfrac{5}{34}=0.147\$moraity \$\dfrac{0.147}{100\ m.p\times 10^{-3}}=1.47M\simeq1.5M\$

Q: \$H_{2}O_{2}\$ acts as an oxidising agent in:

Because the oxidation state of oxygen in \${ H }_{ 2 }{ O }_{ 2 }\$ is \$-1\$ which can be oxidized to \$'O'\$ oxidation state by removing an \${ e }^{ - }\$ or else can be reduced by adding an \${ e }^{ - }\$ and oxidation state would be reduced to \$-2\$ oxidizing agent means oxidizing others itself getting reduced.Acidic: \${ H }_{ 2 }{ O }_{ 2 }+2{ H }^{ + }+2{ e }^{ - }\rightarrow 2{ H }_{ 2 }O\$.Basic: \${ H }_{ 2 }{ O }_{ 2 }+2{ e }^{ - }\rightarrow 2{ OH }^{ - }\$

Question 1

The reaction of aqueous $KMnO_4$ with $H_2O_2$ in acidic conditions gives:

Accepted Answer

The reaction of aqueous $KMnO_4$ with $H_2O_2$ in acidic conditions gives:$3H_2SO_4+ 2KMnO_4 + 5H_2 O_2 \rightarrow 5O_2 + 2MnSO_4 + 8H_2O + K_2SO_4$The oxidation number of $Mn$ in $MnSO_4$ is $+2$.Hence, option $B$ is correct.

Question 2

The reaction $H_2S+H_2O_2\rightarrow S+2H_2O$ describes:

Accepted Answer

The reaction $H_2S+H_2O_2\rightarrow S+2H_2O$ manifests oxidizing nature of $H_2O_2$ as it oxidises S from -2 to 0 oxidation states.Hence option C is correct.

Question 3

From the following statements regarding $H_2O_2$, choose the incorrect statement?

Accepted Answer

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

Question 4

(a) $H_2O_2 + O_3 \rightarrow H_2O + 2O_2$(b) $H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2$Role of hydrogen peroxide in the above reactions is respectively:

Accepted Answer

(a) $H_2O_2 + O_3 \rightarrow H_2O + 2O_2$In this reaction, $O_3$ is reduced into $O^{-2}$ ion .(b) $H_2O_2 + Ag_2O \rightarrow 2Ag + H_2O + O_2$In this reaction, $Ag_2O$ is reduced to Ag .Hence, $H_2O_2$  acts as reducing agent in both the reactions.

Question 5

In which of the following reactions, $H_{2}O_{2}$ acts as a reducing reagent ?

Accepted Answer

$HOCl + H_{2}O_{2}\overset{H^+}{\rightarrow}H_{3}O^{+} + Cl^{-} + O_2$Here, the $H_2O_2$ act as the reducing agent, as the $Cl$ has $+1$ oxidation state of in $HOCl$ and after reaction with the $H_2O_2$ , in the presence of the acid, it reduces the $Cl$ in $HOCl$ to $Cl^{-}$ having -1 oxidation state. So, as $Cl$ oxidation state decreases from the +1 to -1 oxidation state, hence the $H_2O_2$ is acting as the reducing agent.

Question 6

Hydrogen peroxide is prepared in the laboratory by:

Accepted Answer

Hydrogen peroxide can be prepared in the laboratory by the action of ice cold, dilute sulphuric acid on sodium peroxide or hydrated barium peroxide $(BaO_2. 8H_2O)$. A calculated quantity of sodium peroxide is added in small quantities to $20\%$ ice cold solution of sulphuric acid. The reaction is as follows:$Na_2O_2(s)  + H_2SO_4\rightarrow Na_2SO_4(s)  + H_2O_2(l)$Most of the sodium sulphate separates out on cooling as crystals of $Na_2SO_4.10H_2O$. Thus, $30\%$ of hydrogen peroxide containing small amount of sodium sulphate is obtained. Barium peroxide is made into a thin paste and added slowly to ice-cold dilute sulphuric acid. The reaction is as follows:$BaO_2(s)  + H_2SO_4(aq)\rightarrow BaSO_4(s) + H_2O_2(l)$Barium sulphate is allowed to settle down as a precipitate and filtered off. The filtrate obtained is a dilute solution of hydrogen peroxide. Phosphoric acid can be used in place of sulphuric acid. Alternatively, carbon dioxide gas may be passed through hydrated barium peroxide paste in ice cold water, in a slow stream.                      $2BaO_2(s)  +  2H_3PO_4(aq)\rightarrow Ba(PO_4)_2(s) + 3H_2O_2(l)$                  $BaO_2(s) + CO_2 + H_2O_2\rightarrow BaCO_3(s) + H_2O_2(l)$The precipitated barium peroxide or barium carbonate is filtered off which leaves hydrogen peroxide.

Question 7

Half litre each of three samples of $H_{2} O_{2}$ labelled $10$ volume, $15$ volume and $20$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant $H_{2} O_{2}$ solution is:

Accepted Answer

Volume strength of $H_{ 2 }{ O }_{ 2 } = 5.6\times \text{N}$     $(\text{N} =$ Normality$)$Normality of $10$ vol of $H_{ 2 }{ O }_{ 2 } = \dfrac{10}{5.6}\text{N}$Normality of $15$ vol of $H_{ 2 }{ O }_{ 2 } = \dfrac{15}{5.6} \text{N}$Normality of $20$ vol of $H_{ 2 }{ O }_{ 2 } = \dfrac{20}{5.6} \text{N}$Let $500$ ml of each is mixed, then the total volume of mix $= 1500$ ml.Also, this is diluted to double its volume, i.e., $= 3000$ ml$\text{N}\times 3000$ = $\dfrac { 10 }{ 5.6 } \times 500+\dfrac { 15 }{ 5.6 } \times 500+\dfrac { 20 }{ 5.6 } \times 500$Normality of $H_{ 2 }{ O }_{ 2 }$ mix $= 1.339\:\text{N}$Vol. strength of $H_{ 2 }{ O }_{ 2 } = 1.339 \times 5.6 = 7.5$.

Question 8

The volume of $20\:V-H_{2}O_{2}$ solution, required to prepare $5$ litres of oxygen at STP is:

Accepted Answer

$20$ ml of ${ O }_{ 2 }$ is prepared by $1$ ml of ${ H }_{ 2 }{ O }_{ 2 }$ at STP.$1$ ml of ${ O }_{ 2 }$ is prepared by $= \dfrac{1}{20}$ ml of ${ H }_{ 2 }{ O }_{ 2 }$.$5000$ ml of ${ O }_{ 2 }$ is prepared by = $\dfrac { 1 }{ 20 } \times 5000 = 250$ ml of ${ H }_{ 2 }{ O }_{ 2 }$.Hence, option B is correct.

Question 9

${ H }_{ 2 }{ O }_{ 2 }$ solution used for hair bleaching is sold as a solution of approximately 5.0 g ${ H }_{ 2 }{ O }_{ 2 }$ per 100mL of the solution. The molecular mass of ${ H }_{ 2 }{ O }_{ 2 }$ is 34. The molarity of this solution is approximately :

Accepted Answer

Man of $H_2O_2=5\ g$molar man of $H_2O_2=34$number of mole $=\dfrac{5}{34}=0.147$moraity $\dfrac{0.147}{100\ m.p\times 10^{-3}}=1.47M\simeq1.5M$

Question 10

$H_{2}O_{2}$ acts as an oxidising agent in:

Accepted Answer

Because the oxidation state of oxygen in ${ H }_{ 2 }{ O }_{ 2 }$ is $-1$ which can be oxidized to $'O'$ oxidation state by removing an ${ e }^{ - }$ or else can be reduced by adding an ${ e }^{ - }$ and oxidation state would be reduced to $-2$ oxidizing agent means oxidizing others itself getting reduced.Acidic: ${ H }_{ 2 }{ O }_{ 2 }+2{ H }^{ + }+2{ e }^{ - }\rightarrow 2{ H }_{ 2 }O$.Basic: ${ H }_{ 2 }{ O }_{ 2 }+2{ e }^{ - }\rightarrow 2{ OH }^{ - }$

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Properties and Structure of Hydrogen Peroxide

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The reaction of aqueous $K M n O_{4}$ with $H_{2} O_{2}$ in acidic conditions gives:

The reaction $H_{2} S + H_{2} O_{2} \to S + 2 H_{2} O$ describes:

From the following statements regarding $H_{2} O_{2}$ , choose the incorrect statement?

(a) $H_{2} O_{2} + O_{3} \to H_{2} O + 2 O_{2}$
(b) $H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}$

Role of hydrogen peroxide in the above reactions is respectively:

In which of the following reactions, $H_{2} O_{2}$ acts as a reducing reagent ?

Hydrogen peroxide is prepared in the laboratory by:

Half litre each of three samples of $H_{2} O_{2}$ labelled $10$ volume, $15$ volume and $20$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant $H_{2} O_{2}$ solution is:

The volume of $20 V - H_{2} O_{2}$ solution, required to prepare $5$ litres of oxygen at STP is:

$H_{2} O_{2}$ solution used for hair bleaching is sold as a solution of approximately 5.0 g $H_{2} O_{2}$ per 100mL of the solution. The molecular mass of $H_{2} O_{2}$ is 34. The molarity of this solution is approximately :

$H_{2} O_{2}$ acts as an oxidising agent in:

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Hydrogen Peroxide - preparation, Properties, Structure and Uses

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Properties and Structure of Hydrogen Peroxide

Uses of Hydrogen Peroxide

The reaction of aqueous KMnO4​ with H2​O2​ in acidic conditions gives:

The reaction H2​S+H2​O2​→S+2H2​O describes:

From the following statements regarding H2​O2​, choose the incorrect statement?

(a) H2​O2​+O3​→H2​O+2O2​ (b) H2​O2​+Ag2​O→2Ag+H2​O+O2​ Role of hydrogen peroxide in the above reactions is respectively:

In which of the following reactions, H2​O2​ acts as a reducing reagent ?

Hydrogen peroxide is prepared in the laboratory by:

Half litre each of three samples of H2​O2​ labelled 10 volume, 15 volume and 20 volume are mixed and then diluted with equal volume of water. The relative strength of resultant H2​O2​ solution is:

The volume of 20V−H2​O2​ solution, required to prepare 5 litres of oxygen at STP is:

H2​O2​ solution used for hair bleaching is sold as a solution of approximately 5.0 g H2​O2​ per 100mL of the solution. The molecular mass of H2​O2​ is 34. The molarity of this solution is approximately :

H2​O2​ acts as an oxidising agent in:

The reaction of aqueous $K M n O_{4}$ with $H_{2} O_{2}$ in acidic conditions gives:

The reaction $H_{2} S + H_{2} O_{2} \to S + 2 H_{2} O$ describes:

From the following statements regarding $H_{2} O_{2}$ , choose the incorrect statement?

(a) $H_{2} O_{2} + O_{3} \to H_{2} O + 2 O_{2}$
(b) $H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}$

Role of hydrogen peroxide in the above reactions is respectively:

In which of the following reactions, $H_{2} O_{2}$ acts as a reducing reagent ?

Half litre each of three samples of $H_{2} O_{2}$ labelled $10$ volume, $15$ volume and $20$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant $H_{2} O_{2}$ solution is:

The volume of $20 V - H_{2} O_{2}$ solution, required to prepare $5$ litres of oxygen at STP is:

$H_{2} O_{2}$ solution used for hair bleaching is sold as a solution of approximately 5.0 g $H_{2} O_{2}$ per 100mL of the solution. The molecular mass of $H_{2} O_{2}$ is 34. The molarity of this solution is approximately :

$H_{2} O_{2}$ acts as an oxidising agent in: