Chapter 8: Redox Reactions

Ans: The given reaction is as below-

\[\text{C}{{\text{l}}_{\text{2}}}\text{ }\left( \text{g} \right)\text{+2O}{{\text{H}}^{\text{--}}}\text{ }\left( \text{aq} \right)\to \text{Cl}{{\text{O}}^{\text{--}}}\left( \text{aq} \right)\text{+ C}{{\text{l}}^{\text{--}}}\left( \text{aq} \right)\text{+}{{\text{H}}_{\text{2}}}\text{O }\left( \text{l} \right)\]

In the given reaction, oxidation number  of Cl increases from 0 (in Cl2) to +1 (in ClO-) and decreases to -1 (in Cl-). Therefore, Cl2 is both oxidized to ClO- and reduced to Cl-. Since Cl- ion cannot act as an oxidizing agent (because it cannot decrease its O.N. lower than -1), hence, Cl2 bleaches substances due to oxidizing action of hypochlorite, ClO ion.

Ans: In \[\text{Mn}{{\text{O}}_{\text{4}}}^{\text{2-}}\], Mn is in highest oxidation state that is +7 hence here manganese cannot undergo oxidation that is why disproportionate is not possible whereas in  \[\text{Mn}{{\text{O}}_{\text{4}}}^{\text{2-}}\] manganese is in +6 oxidation state which can be oxidized as well as reduced.

Ans:  The given compound can differ in reactivity as-Lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO2 thus, can act as an oxidant (oxidising agent) and, therefore, can oxidise chloride ions of HCl into chlorine.

Ans: Since nitric acid itself is an oxidising agent therefore, it is unlikely that the reaction may occur between PbO2 and nitric acid. However, the acid-base reaction occurs between PbO and nitric acid because PbO is a basic oxide.

Ans: We can balance the given reaction by ion electron method as-

\[\text{Mn}{{\text{O}}_{\text{4}}}^{\text{-}}\text{+S}{{\text{O}}_{\text{2}}}\to \text{M}{{\text{n}}^{\text{2+}}}\text{+HS}{{\text{O}}_{\text{4}}}^{\text{-}}\left( \text{acidic medium} \right)\]

Balancing by ion electron method

\[2 \times MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4 H_{2}O \]

\[ 5 \times SO_{2} + 2H_{2}O \rightarrow HSO_{4}^{-} + 3H^{+} + 2e^{-}\]

The balanced chemical is given as-

\[2MnO_{4}^{-} + 16H^{+} + \not {10e^{-}} \rightarrow 2Mn^{2+} + 8H_{2}O \\\underline{5SO_{2}+10H_{2}O \rightarrow 5HSO_{4} ^{-} + 15H^{+} + \not {10e^{-}}}\\ \\\underline{2MnO_4^- + H^{+}+ 5SO_{2} + H_{2}O \rightarrow 2Mn^{2+} + 5HSO_{4}^{-}} \]

Ans: We can balance the given reaction by oxidation number method-

Balancing by oxidation number method as to make the electron gain and loss equal as given - 

\[3N_{2}H_{4}+4ClO_{3}^{-}\rightarrow 6NO +4Cl^{-} +6H_{2}O \]

The balanced chemical is given as-

Ans: We can balance the given reaction by ion electron method as

\[Cl_{2}O_{7} (g) + H_{2}O_{2} (aq) \rightarrow ClO_{2}^{-}+O_{2} \] (acidic medium)

Balancing bu ion electron method

2 × { Cl2O7 + 6H+ + 8e- → 2ClO2- + 3H2O

8  × { H2O2 → O2 + 2H+ + 2e- 

The balanced chemical is given as-

2Cl2O7 + 12H+ + 16e- → 4ClO2- + 6H2O

                         8H2O2 → 8O2 + 16H+ + 16e-

          2Cl2O7 + 8H2O2 → 4ClO2- + 6H2O + 8O2 + 4H+

Ans:

Let the oxidation number of P in \[\text{HP}{{\text{O}}_{\text{3}}}^{\text{2-}}\] be x. 

Therefore, +1 + x + (-6) = -2 

x = +3

Oxidation number of phosphorus is= +3.

Ans: 

Let the oxidation number of P in \[\text{P}{{\text{O}}_{\text{4}}}^{\text{3-}}\] be x. 

Therefore, x + (-8) = -3

x = +5 

Oxidation number of phosphorus is= +5.

Ans: let x= oxidation number of   sulphur, and   +1 is oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + 2x - 6 = 0 

x = +2. 

Ans: let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + 4x - 12 = 0

x= +2.5 

Ans: let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

\[\left( \mathbf{d} \right)\text{ }\mathbf{N}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}\].

Ans: let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x -8=0

x = +6.

Ans: We can balance the given equation by oxidation number method- 

\[\overset{\text{+2}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{2+}}}\text{+}{{\text{H}}^{\text{+}}}\text{+}\overset{\text{+6}}{\mathop{\text{C}}}\,{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{^{\overset{\text{-2}}{\mathop{\text{2-}}}\,}}\to \overset{\text{+3}}{\mathop{\text{C}}}\,{{\text{r}}^{\text{3+}}}\text{+F}{{\text{e}}^{\text{3+}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]

(a) Balance the increase and decrease in O.N.

\[6\overset{\text{+2}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{2+}}}\text{+}{{\text{H}}^{\text{+}}}\text{+}\overset{\text{+6}}{\mathop{\text{C}}}\,{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{^{\overset{\text{-2}}{\mathop{\text{2-}}}\,}}\to \overset{\text{+3}}{\mathop{\text{C}}}\,{{\text{r}}^{\text{3+}}}\text{+6}\overset{3+}{\mathop{\text{F}}}\,{{\text{e}}^{\text{3+}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\]

(b) Balancing H and O atoms by adding H+ and H2O molecules

\[6\overset{\text{+2}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{2+}}}\text{+14}{{\text{H}}^{\text{+}}}\text{+}\overset{\text{+6}}{\mathop{\text{C}}}\,{{\text{r}}_{\text{2}}}\text{O}_{\text{7}}^{2-}\to 2\overset{{}}{\mathop{\text{C}}}\,{{\text{r}}^{\text{3+}}}\text{+6}\overset{{}}{\mathop{\text{F}}}\,{{\text{e}}^{\text{3+}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{I}{{\text{ }}_{\mathbf{2}\text{ }}}+\mathbf{N}{{\mathbf{O}}_{\mathbf{3}}}^{\text{ -}}\to \text{ }\mathbf{N}{{\mathbf{O}}_{\mathbf{2}}}\text{ }+\text{ }\mathbf{I}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\].

Ans: We can balance the given equation by oxidation number method- 

Total decrease in O.N. = 1

To equilize O.N. multiply NO3-, by 10

I2 + 10 NO3- → 10NO2 + IO3-

Balancing atoms other than O and H

I2 + 10 NO3- → 10NO2 + 2 IO3-

Balancing O and H

I2 + 10 NO3- + 8H+ → 10NO2 + 2 IO3- + 4H2O

\[\left( \mathbf{iii} \right)\text{ }\mathbf{I}{{\text{ }}_{\mathbf{2}}}+\text{ }{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}}\text{ }\to \text{ }{{\mathbf{I}}^{}}\text{ }+\text{ }{{\mathbf{S}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{6}}}^{\mathbf{2}}\].

 Ans: We can balance the given equation by oxidation number method

\[\overset{0}{\mathop{{{\text{I}}_{\text{2}}}}}\,+\overset{+2}{\mathop{{{S}_{2}}}}\,\overset{-2}{\mathop{O_{3}^{2-}}}\,\to \overset{-1}{\mathop{{{I}^{-}}}}\,+\overset{+2.5}{\mathop{{{S}_{4}}}}\,O_{6}^{2-}\]

Total increase in O.N. = 0.5 × 4 = 2

Total increase in O.N. = 1 × 2 = 2

To equilize O.N. multiply \[{{S}_{2}}O_{3}^{2-}\] and I- by 2.

\[{{I}_{2}}+2{{S}_{2}}O_{3}^{2-}\to 2{{I}^{-}}+{{S}_{4}}O_{6}^{2-}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{Mn}{{\mathbf{O}}_{\mathbf{2}}}\text{ }+\text{ }{{\mathbf{C}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}}\text{ }\to \text{ }\mathbf{M}{{\mathbf{n}}^{\mathbf{2}+}}+\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}\].

Ans: We can balance the given equation by oxidation number method-

\[\overset{+4}{\mathop{Mn}}\,{{O}_{2}}+\overset{+3}{\mathop{{{C}_{2}}}}\,O_{4}^{2-}\to \overset{+2}{\mathop{M{{n}^{2+}}+}}\,\overset{+4}{\mathop{C{{O}_{2}}}}\,\]

The balanced chemical reaction is given as-

\[\text{Total increase in O}\text{.N}\text{. = 5 }\!\!\times\!\!\text{ 2= 10}\]

\[\text{To equalize O}\text{.N}\text{. multiply C}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ by 2}\text{. }\]

\[\text{Mn}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + }{{\text{C}}_{\text{2}}}{{\text{O}}^{\text{2-}}}_{\text{4}}\text{ }\!\!~\!\!\text{ }\to \text{ M}{{\text{n}}^{\text{2+}}}\text{ }\!\!~\!\!\text{ + 2C}{{\text{O}}_{\text{2}}}\]

\[\text{Balance H and O by adding 2}{{\text{H}}_{\text{2}}}\text{O }\!\!~\!\!\text{ on right side, and 4}{{\text{H}}^{\text{+}}}\text{ }\!\!~\!\!\text{ on left side of equation}\text{.}\]

\[\text{Mn}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + }{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}^{\text{2-}}\text{ }\!\!~\!\!\text{ + 4}{{\text{H}}^{\text{+}}}\text{ }\!\!~\!\!\text{ }\to \text{ M}{{\text{n}}^{\text{2+}}}\text{ }\!\!~\!\!\text{ + 2C}{{\text{O}}_{\text{2}}}\text{ }\!\!~\!\!\text{ + 2}{{\text{H}}_{\text{2}}}\text{O }\text{.}\]

Ans: We can write the given reaction along with their oxidation numbers as-

\[3\overset{+1}{\mathop{H}}\,\overset{-1}{\mathop{Cl}}\,(aq)+\overset{+1}{\mathop{H}}\,\overset{+5}{\mathop{N}}\,\overset{-2}{\mathop{{{O}_{3}}}}\,(aq)\to \overset{0}{\mathop{C{{l}_{2}}}}\,+\overset{+3}{\mathop{N}}\,\overset{-2}{\mathop{O}}\,\overset{-1}{\mathop{Cl(g)}}\,+2\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2}{\mathop{O(l)}}\,\]

As, chlorine is oxidised in hydrochloric acid (behaving as reducing agent) -as its oxidation number is increases during the reaction from -1 to 0) and nitric acid is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +5 to +3). Hence, it is the redox reaction.

\[\left( \mathbf{ii} \right)\text{ }\mathbf{HgC}{{\mathbf{l}}_{\mathbf{2}}}\left( \mathbf{aq} \right)+\mathbf{2KI}\text{ }\left( \mathbf{aq} \right)\text{ }\to \text{ }\mathbf{Hg}{{\mathbf{I}}_{\mathbf{2}}}\left( \mathbf{s} \right)+\mathbf{2KCl}\left( \mathbf{aq} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{+2}{\mathop{Hg}}\,{{\overset{-1}{\mathop{Cl}}\,}_{2}}(aq)+2\overset{+1}{\mathop{K}}\,\overset{-1}{\mathop{l}}\,(aq)\to \overset{0}{\mathop{C{{l}_{2}}}}\,+\overset{+2}{\mathop{Hg}}\,\overset{-1}{\mathop{{{l}_{2}}}}\,(aq)+\overset{+1}{\mathop{K}}\,\overset{-1}{\mathop{Cl}}\,(aq)\]

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

\[\left( \mathbf{iii} \right)\text{ }\mathbf{F}{{\mathbf{e}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}\left( \mathbf{s} \right)\text{ }+\mathbf{3CO}\left( \mathbf{g} \right)\text{ }\to \mathbf{2Fe}\left( \mathbf{s} \right)+\mathbf{3C}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{\text{+3}}{\mathop{\text{F}{{\text{e}}_{\text{2}}}}}\,\overset{\text{-2}}{\mathop{{{\text{O}}_{\text{3}}}}}\,\left( \text{s} \right)\text{+3}\overset{\text{+2}}{\mathop{\text{C}}}\,\overset{\text{-2}}{\mathop{\text{O}}}\,\left( \text{g} \right)\text{ }\xrightarrow{\text{ }\!\!\Delta\!\!\text{ }}\overset{\text{0}}{\mathop{\text{2}}}\,\text{Fe}\left( \text{s} \right)\text{+3}\overset{\text{+4}}{\mathop{\text{C}}}\,\overset{\text{-2}}{\mathop{{{\text{O}}_{\text{2}}}}}\,\left( \text{g} \right)\]

As, carbon is oxidised in carbon monoxide (behaving as reducing agent) -as its oxidation number is increases during the reaction from +2 to +4) and ferric oxide is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +3to 0). Hence, it is the redox reaction.

\[\left( \mathbf{iv} \right)\text{ }\mathbf{PC}{{\mathbf{l}}_{\mathbf{3}}}\text{ }\left( \mathbf{l} \right)+\mathbf{3}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\text{ }\left( \mathbf{l} \right)\text{ }\to \mathbf{3HCl}\text{ }\left( \mathbf{aq} \right)+{{\mathbf{H}}_{\mathbf{3}}}\mathbf{P}{{\mathbf{O}}_{\mathbf{3}}}\left( \mathbf{aq} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{\text{+3}}{\mathop{\text{P}}}\,\overset{\text{-1}}{\mathop{\text{C}{{\text{l}}_{\text{3}}}}}\,\left( \text{l} \right)\text{+3}\overset{\text{+1}}{\mathop{{{\text{H}}_{\text{2}}}}}\,\overset{\text{-2}}{\mathop{\text{O}}}\,\left( \text{l} \right)\to \text{3}\overset{\text{+1}}{\mathop{\text{H}}}\,\overset{\text{-1}}{\mathop{\text{Cl}}}\,\left( \text{aq} \right)\text{+}\overset{\text{+1}}{\mathop{{{\text{H}}_{\text{3}}}}}\,\overset{\text{+3}}{\mathop{\text{P}}}\,\overset{\text{-2}}{\mathop{{{\text{O}}_{\text{2}}}}}\,\left( \text{aq} \right)\]

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

\[\left( \mathbf{v} \right)\text{ }\mathbf{4N}{{\mathbf{H}}_{\mathbf{3}}}\mathbf{(l)}+\mathbf{3}{{\mathbf{O}}_{\mathbf{2}}}\left( \mathbf{g} \right)\text{ }\to \text{ }\mathbf{2}{{\mathbf{N}}_{\mathbf{2}}}\left( \mathbf{g} \right)+\mathbf{6}{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\left( \mathbf{g} \right)\].

Ans: We can write the given reaction along with their oxidation numbers as-

\[\overset{\text{+3}}{\mathop{4N}}\,\overset{\text{+1}}{\mathop{{{\text{H}}_{\text{3}}}}}\,\left( g \right)\text{+3}\overset{0}{\mathop{{{O}_{\text{2}}}}}\,\left( g \right)\to 2\overset{\text{+1}}{\mathop{{{N}_{2}}}}\,\left( g \right)\text{+6}{{\text{H}}_{\text{2}}}\overset{\text{+1}}{\mathop{\text{O}}}\,\overset{\text{-2}}{\mathop{\text{(g)}}}\,\]

As, nitrogen is oxidised in ammonia (behaving as reducing agent) -as its oxidation number is increases during the reaction from -3 to 0) and oxygen is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from 0 to -2). Hence, it is the redox reaction.

Ans: The balanced chemical is given as -

\[\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{\text{2}-}\text{+6}{{\text{I}}^{\text{-}}}\text{+14}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2C}{{\text{r}}^{\text{+3}}}\text{+3}{{\text{I}}_{\text{2}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}\]

\[\left( \mathbf{ii} \right)\text{ }\mathbf{C}{{\mathbf{r}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{7}}}^{\mathbf{2}}+\text{ }\mathbf{F}{{\mathbf{e}}^{\mathbf{2}+\text{ }}}+{{\mathbf{H}}^{+\text{ }}}\to \text{ }\mathbf{C}{{\mathbf{r}}^{\mathbf{3}+}}+\text{ }\mathbf{F}{{\mathbf{e}}^{\mathbf{3}+}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: The balanced chemical is given as -

\[\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{\text{2-}}\text{+6F}{{\text{e}}^{\text{+2}}}\text{+14}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2C}{{\text{r}}^{\text{+3}}}\text{+6F}{{\text{e}}^{\text{+3}}}\text{+7}{{\text{H}}_{\text{2}}}\text{O}\]

\[~\left( \mathbf{iii} \right)\text{ }\mathbf{Mn}{{\mathbf{O}}_{\mathbf{4}\text{ }}}^{\text{ }}+{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{3}}}+{{\mathbf{H}}^{+}}\text{ }\to \text{ }\mathbf{M}{{\mathbf{n}}^{\mathbf{2}+}}+{{\mathbf{S}}_{\mathbf{2}}}{{\mathbf{O}}_{\mathbf{4}}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: The balanced chemical is given as -

\[\text{2Mn}{{\text{O}}_{\text{4}}}^{\text{-}}\text{+5S}{{\text{O}}_{\text{3}}}^{\text{2-}}\text{+6}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2M}{{\text{n}}^{\text{+2}}}\text{+5S}{{\text{O}}_{\text{4}}}^{\text{2-}}\text{+3}{{\text{H}}_{\text{2}}}\text{O}\]

\[\left( \mathbf{iv} \right)\text{ }\mathbf{Mn}{{\mathbf{O}}_{\mathbf{4}}}^{\text{-}}+{{\mathbf{H}}^{\mathbf{+}}}+\mathbf{B}{{\mathbf{r}}^{\mathbf{}}}\text{ }\to \mathbf{M}{{\mathbf{n}}^{\mathbf{2}+}}+\text{ }\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}+{{\mathbf{H}}_{\mathbf{2}}}\mathbf{O}\]

Ans: The balanced chemical is given as -

\[\text{2Mn}{{\text{O}}_{\text{4}}}^{\text{-}}\text{+10B}{{\text{r}}^{\text{-}}}\text{+16}{{\text{H}}^{\text{+}}}\rightleftharpoons \text{2M}{{\text{n}}^{\text{+2}}}\text{+5B}{{\text{r}}_{\text{2}}}\text{+8}{{\text{H}}_{\text{2}}}\text{O}\]