Redox Reactions

Chemistry Class 11

NCERT

1   Assign oxidation numbers to the underlined elements in each of the following species:$\\$ (a)$NaH_2\underline{P}O_4$$\\$ (b)$NaH\underline{S}O_4$$\\$ (c)$H_4\underline{P}_2O_7$$\\$ (d)$K_2\underline{Mn}O_4$$\\$ (e)$Ca\underline{O}_2$$\\$ (f)$Na\underline{B}H_4$$\\$ (g)$H_2\underline{S}_2O_7$$\\$ (h)$KAl(\underline{S}O_4)_2.12 H_2O$

Solution :

(a)$NaH_2\underline{P}O_4$$\\$ Let the oxidation number of $P$ be $x.$$\\$ We know that,$\\$ Oxidation number of $Na = +1$$\\$ Oxidation number of $H = +1$$\\$ Oxidation number of $O = -2$$\\$ $\Rightarrow \overset{+1}Na \overset{+1}H_ 2 \overset{x}P\overset{-2}O_ 4$ Then, we have$\\$ $1(+1)+2(+1)+1(x)+4(-2)=0$$\\$ $\Rightarrow 1+2+x-8=0$$\\$ $\Rightarrow x=+5$$\\$ Hence, the oxidation number of $P$ is $+5.$$\\$

(b) $NaH \underline{S}O _4$$\\$ $\overset{+1}Na \overset{+1}H \overset{x}S \overset{-2}O _4$$\\$ Then, we have$\\$ $1(+1)+1(+1)+1(x)+4(-2)=0$$\\$ $\Rightarrow 1+1+x-8=0$$\\$ $\Rightarrow x=+6$$\\$ Hence, the oxidation number of $S$ is $+ 6$.

(c)$H_4\underline{P}_2O_7$$\\$ $\overset{+1}H_ 4 \overset{x}P_ 2 \overset{-2}O_ 7$$\\$ Then, we have$\\$ $4(+1)+2(x)+7(-2)=0\\ \Rightarrow 4+2x-14=0\\ \Rightarrow 2x=+10\\ \Rightarrow x=+5$ Hence, the oxidation number of $P$ is $+ 5.$

(d)$K_2\underline{Mn}O_4$$\\$ $\overset{+1}K_ 2\overset{x} Mn\overset{-2}O_ 4$$\\$ Then, we have$\\$ $2(+1)+x+4(-2)=0\\ \Rightarrow 2+x-8=0\\ \Rightarrow x=+6$$\\$ Hence, the oxidation number of $Mn$ is $+ 6.$$\\$ (e)$Ca\underline{O}_2$$\\$ $\overset{+2}Ca\overset{x}O_2$$\\$ Then, we have$\\$ $(+2)+2(x)=0\\ \Rightarrow 2+2x=0\\ \Rightarrow x=-1$$\\$ Hence, the oxidation number of $O$ is $- 1$.

(f)$Na\underline{B}H_4$$\\$ $\overset{+1}Na \overset{x}B\overset{-1}H_4$$\\$ Then,we have$\\$ $1(+1)+1(x)+4(-1)=0\\ \Rightarrow 1+x-4=0\\ \Rightarrow x=+3$$\\$ Hence, the oxidation number of $B$ is $+ 3.$$\\$ (g)$H_2\underline{S}_2O_7$$\\$ $\overset{+1}H_2\overset{x}S_2\overset{-2}O_7$$\\$ Then, we have $\\$ $2(+1)+2(x)+7(-2)=0\\ \Rightarrow 2+2x-14=0\\ \Rightarrow 2x=12\\ \Rightarrow x=+6$$\\$ Hence, the oxidation number of $S$ is $+ 6$.

(h)$KAl(\underline{S}O_4)_2.12 H_2O$$\\$ $\overset{+1}K\overset{3+}Al(\overset{x}S\overset{2-}O_4)_2.12\overset{+1}H_2\overset{-2}O$$\\$ Then , we have$\\$ $1(+1)+1(+3)+2(x)+8(-2)+24(+1)+12(-2)=0\\ \Rightarrow 1+3+2x-16+24-24=0\\ \Rightarrow 2x=12\\ \Rightarrow x=+6$$\\$ Or,$\\$ We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have$\\$ $1(+1)+1(+3)+2(x)+8(-2)=0\\ \Rightarrow 1+3+2x-16=0\\ \Rightarrow 2x=12\\ \Rightarrow x=+6$$\\$ Hence, the oxidation number of $S$ is $+ 6.$

2   What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?$\\$ (a)$K\underline{I}_3$$\\$ (b)$H_2 \underline{S}_4O_6$$\\$ (c)$\underline{Fe}_3O_4$$\\$ (d)$\underline{C}H_3\underline{C}H_2OH$$\\$ (e)$\underline{C}H_3\underline{C}OOH$

Solution :

(a)$K\underline{I}_3$$\\$ In $KI _3$ , the oxidation number (O.N.) of $K$ is $+1$. Hence, the average oxidation number of $I$ is$\dfrac{-1}{3}$ . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of $KI_ 3$ to find the oxidation states.$\\$ In a $KI _3$ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.$\\$ $\overset{+1}{K^+}[\overset{0}I-\overset{0}I\leftarrow \overset{-1}I]$$\\$ Hence, in a $KI _3$ molecule, the O.N. of the two I atoms forming the $I_ 2$ molecule is $0$, whereas the O.N. of the I atom forming the coordinate bond is $-1$.$\\$

(b)$H_2 \underline{S}_4O_6$$\\$ $\overset{+1}H_2 \overset{x}S_4\overset{-2}O_6$$\\$ Now,$2(+1)+4(x)+6(-2)=0$$\\$ $\Rightarrow 2+4x-12=0$$\\$ $\Rightarrow 4x=10$$\\$ $\Rightarrow x=+2\dfrac{1}{2}$$\\$ However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.$\\$ $H-O-\overset{\overset{O}{||}}{\underset{\underset{O}{||}}{S^{+5}}}-\overset{0}S-\overset{0}S-\overset{\overset{O}{||}}{\underset{\underset{O}{||}}{S^{+5}}}-O-H$$\\$ The O.N. of two of the four S atoms is $+5$ and the O.N. of the other two S atoms is $0$.$\\$ (c)$\underline{Fe}_3O_4$$\\$ On taking the O.N. of O as $-2$, the O.N. of Fe is found to be $+2\dfrac{2}{3}.$ However, O.N. cannot be fractional. Here, one of the three Fe atoms exhibits the O.N. of $+2$ and the other two Fe atoms exhibit the O.N. of $+3$.$\\$ $\overset{+2}FeO,\overset{+3}Fe_2O_3$$\\$ (d)$\underline{C}H_3\underline{C}H_2OH$$\\$ $\overset{x}C_2 \overset{+1}H_6 \overset{-2}O$$\\$ $2(x)+6(+1)+1(-2)=0$$\\$ $2x+6-2=0$$\\$ $x=-2$$\\$ The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, $C$ exhibits the oxidation states of $-3$ and $-1$ $\\$

(e)$\underline{C}H_3\underline{C}OOH$$\\$ $\overset{x}C_2\overset{+1}H_4 \overset{-2}O_2$$\\$ $2(x)+4(+1)+2(-2)=0$$\\$ $\Rightarrow 2x+4-4=0$$\\$ $\Rightarrow x=0$$\\$ However, $0$ is average O.N. of $C$. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of $+ 3$ and $-3$ in $CH _3 COOH.$$\\$ $H-\overset{\overset{H}{|}}{\underset{\underset{H}{|}}{C^{-2}}}-\overset{\overset{O}{||}}{C^{-2}}-O-H$$\\$

3   Justify that the following reactions are redox reactions:$\\$ (a)$ CuO(s) + H _2 (g) \rightarrow Cu(s) + H _2 O(g)$$\\$ (b) $Fe _2 O _3 (s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_ 2 (g)$$\\$ (c) $4BCl _3 (g) + 3LiAlH _4 (s) \rightarrow 2B _2 H _6 (g) + 3LiCl(s) + 3 AlCl _3 (s)$$\\$ (d) $2K(s) + F_ 2 (g) \rightarrow 2K ^+ F ^- (s)$$\\$ (e) $4 NH _3 (g) + 5O _2 (g) \rightarrow 4NO(g) + 6H _2 O(g)$

Solution :

(a) $CuO ( s ) +H _2( g ) \to Cu ( s ) + H _2 O ( g )$$\\$ Let us write the oxidation number of each element involved in the given reaction as:$\\$ $\overset{+2}Cu \overset{-2}O(s)+\overset{0}H_2(g)\to \overset{0}Cu(s)+\overset{+1}H_2 \overset{-2}O(g)$$\\$ Here, the oxidation number of $Cu$ decreases from $+2$ in $CuO$ to $0$ in $Cu$ i.e.,$ CuO$ is reduced to $Cu$. Also, the oxidation number of $H$ increases from $0$ in $H _2$ to $+1$ in $H_ 2 O$ i.e., $H _2$ is oxidized to $_H 2 O$. Hence, this reaction is a redox reaction.$\\$ (b) $Fe _2 O _3 (s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_ 2 (g)$$\\$ Let us write the oxidation number of each element in the given reaction as:$\\$ $\overset{+3}Fe_2 \overset{-2}O_3(s)+3 \overset{+2}C \overset{-2}O(g)\to 2 \overset{0}Fe(s)+3 \overset{+4}C \overset{-2}O_2(g)$$\\$ Here, the oxidation number of $Fe$ decreases from $+3$ in $Fe _2 O _3$ to $0$ in $Fe$ i.e., $Fe _2 O _3$ is reduced to$ Fe$. On the other hand, the oxidation number of $C$ increases from $+2$ in $CO$ to $+4$ in $CO _2$ i.e., $CO$ is oxidized to $CO _2$ . Hence, the given reaction is a redox reaction.$\\$ (c) $4BCl _3 (g) + 3LiAlH _4 (s) \rightarrow 2B _2 H _6 (g) + 3LiCl(s) + 3 AlCl _3 (s)$$\\$ The oxidation number of each element in the given reaction can be represented as:$\\$ $4\overset{+3}B\overset{-1}Cl_3(g)+3\overset{+1}Li\overset{+3}Al \overset{-1}H_4(s)\to 2 \overset{-3}B_2 \overset{+1}H_6(g)+3 \overset{+1}Li \overset{-1}Cl(s)+3 \overset{+3}Al\overset{-1}Cl_3(s)$$\\$ In this reaction, the oxidation number of $B$ decreases from $+3$ in $BCl _3$ to $-3 $ in $B _2 H _6 $ . i.e., $BCl _3$ is reduced to $B _2 H_ 6$ . Also, the oxidation number of $H$ increases from $-1$ in $LiAlH _4$ to $+1$ in $B_ 2 H_ 6$ i.e., $LiAlH _4$ is oxidized to $B _2 H_ 6$ . Hence, the given reaction is a redox reaction.$\\$

(d) $2K(s) + F_ 2 (g) \rightarrow 2K ^+ F ^- (s)$$\\$ The oxidation number of each element in the given reaction can be represented as:$\\$ $2\overset{0}K(s)+\overset{0}F_2(g) \to 2\overset{+1}{k^+} \overset{-1}{F^-}(s)$$\\$ In this reaction, the oxidation number of $K$ increases from $0$ in $K$ to $+1 $ in $KF$ i.e., $K$ is oxidized to $KF$. On the other hand, the oxidation number of $F$ decreases from $0$ in $F_ 2$ to $- 1$ in $KF$ i.e., $F _2$ is reduced to $KF.$$\\$ Hence, the above reaction is a redox reaction.$\\$ (e) $4 NH _3 (g) + 5O _2 (g) \rightarrow 4NO(g) + 6H _2 O(g)$$\\$ The oxidation number of each element in the given reaction can be represented as:$\\$ $4\overset{-3} N\overset{+1}H _3 (g) + 5\overset{0}O _2 (g) \rightarrow 4\overset{+2}N\overset{-2}O(g) + 6\overset{+1}H _2 \overset{-2}O(g)$$\\$ Here, the oxidation number of $N$ increases from $-3$ in $NH _3$ to $+2$ in $NO$. On the other hand, the oxidation number of $O_ 2$ decreases from $0$ in $O _2$ to $-2$ in $NO$ and $H _2 O$ i.e., $O _2$ is reduced.$\\$ Hence, the given reaction is a redox reaction.

4   Fluorine reacts with ice and results in the change:$\\$ $H _2 O(s) + F_ 2 (g) \to HF(g) + HOF(g)$$\\$ Justify that this reaction is a redox reaction.

Solution :

Let us write the oxidation number of each atom involved in the given reaction above its symbol as:$\\$ $\overset{+1}H_2\overset{-2}O+\overset{0}F_2\to \overset{+1}H\overset{-1}F+\overset{+1}H\overset{-2}O\overset{+1}F$$\\$ Here, we have observed that the oxidation number of $F$ increases from $0$ in $F _2$ to $+1$ in $HOF.$$\\$ Also, the oxidation number decreases from $0$ in $F _2$ to $-1$ in $HF$. $\\$Thus, in the above reaction,$ F$ is both oxidized and reduced. Hence, the given reaction is a redox reaction.$\\$

5   Calculate the oxidation number of sulphur, chromium and nitrogen in $H_ 2 SO_ 5 , Cr_ 2 O _7^{ 2 - } $ and $NO_ 3^{-}$ Suggest structure of these compounds. Count for the fallacy

Solution :

(i)$\overset{+1}H_2\overset{x}S\overset{-2}O_5$$\\$ $2(+1)+1(x)+5(-2)=0$$\\$ $\Rightarrow 2+x-10=0$$\\$ $\Rightarrow x=+8$$\\$ However, the O.N. of S cannot be $+8$. S has six valence electrons. Therefore, the O.N. of S cannot be more than $+6.$$\\$ The structure of $H _2 SO_ 5$ is shown as follows:$\\$ $\overset{+1}H-\overset{-2}O-\overset{\overset{O^{-2}}{||}}{\underset{\underset{O_{-2}}{||}}{S^x}}-\overset{-1}O-\overset{-1}O-\overset{+1}H$ $\\$ $2(H)+1(S)+3(O)+2(O$ in peroxy linkage)$\\$ Now,$2(+1)+1(x)+3(-2)+2(-1)=0$$\\$ $\Rightarrow 2+x-6-2=0$$\\$ $\Rightarrow x=+6$$\\$ Therefore, the O.N. of S is $+6.$$\\$ (ii)$\overset{x}Cr_2 \overset{2-}{O_7^{2-}}$$\\$ $2(x)+7(-2)=-2$$\\$ $\Rightarrow 2x-14=-2$$\\$ $\Rightarrow x=+6$$\\$ Here, there is no fallacy about the O.N. of $Cr$ in $Cr_ 2 O _7^{ 2 -}$$\\$ The structure of $Cr_ 2 O_ 7^{ 2 -} $ is shown as follows:$\\$ $\overset{2-}O=\overset{\overset{\overset{2-}{O}}{||}}{\underset{\underset{\underset{1-}{O^-}}{||}}{Cr^{+6}}}-\overset{2-}{O}-\overset{\overset{\overset{2-}{O}}{||}}{\underset{\underset{\underset{1-}{O^-}}{||}}{Cr^{+6}}}-\overset{2-}O$$\\$ Here, each of the two $Cr$ atoms exhibits the O.N. of $+6.$$\\$ (iii)$\overset{x}N\overset{2-}{O_3^{-}}$$\\$ $1(x)+3(-2)=-1$$\\$ $\Rightarrow x-6=-1$$\\$ $\Rightarrow x=+5$$\\$ Here, there is no fallacy about the O.N. of N in $NO _3 ^{-}$ . The structure of $NO _3^{-}$ is shown as follows:$\\$

6   Write the formulae for the following compounds:$\\$ (a) Mercury(II) chloride$\\$ (b) Nickel(II) sulphate$\\$ (c) Tin(IV) oxide$\\$ (d) Thallium(I) sulphate$\\$ (e) Iron(III) sulphate$\\$ (f) Chromium(III) oxide

Solution :

(a) Mercury (II) chloride:$\\$ $HgCl _2$ $\\$ (b) Nickel (II) sulphate:$\\$ $NiSO _4$ $\\$ (c) Tin (IV) oxide:$\\$ $SnO_ 2$ $\\$ (d) Thallium (I) sulphate:$\\$ $Tl _2 SO _4$ $\\$ (e) Iron (III) sulphate:$\\$ $Fe _2 (SO _4 )_ 3$ $\\$ (f) Chromium (III) oxide:$\\$ $Cr_ 2 O _3$

7   Suggest a list of the substances where carbon can exhibit oxidation states from $-4$ to $+4$ and nitrogen from $-3$ to $+5.$

Solution :

The substances where carbon can exhibit oxidation states from $-4$ to $+4$ are listed in the following table.$\\$ $\begin{array}{|c|c|}\hline \text{Substance}& \text{O.N. of carbon}\\ \hline CH_ 2 Cl_ 2&0 \\ \hline ClC \equiv CCl &+1\\ \hline HC \equiv CH&-1\\ \hline CHCl _3 , CO &+2\\ \hline CH _3 Cl&-2\\ \hline Cl _3 C - CCl_ 3&+3\\ \hline H _3 C - CH _3&-3\\ \hline CCl _4 , CO _2& +4\\ \hline CH _4 &-4\\ \hline \end{array}$ $\\$ The substances where nitrogen can exhibit oxidation states from $-3$ to $+5$ are listed in the following table.$\\$ $\begin{array}{|c|c|} \hline \text{Substance}& \text{O.N. of nitrogen}\\ \hline N_2&0\\ \hline N _2 O &+1\\ \hline N_ 2 H_ 2 &-1\\ \hline NO& +2\\ \hline N _2 H_ 4& -2\\ \hline N _2 O _3& +3\\ \hline NH _3 &-3\\ \hline NO _2&+4\\ \hline N _2 O _5&+5\\ \hline \end{array}$ $\\$

8   While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Solution :

In sulphur dioxide $(SO_ 2 )$, the oxidation number $(O.N.)$ of $S$ is $+4$ and the range of the $O.N.$ that $S$ can have is from $+6 $ to $-2.$$\\$ Therefore, $SO _2$ can act as an oxidising as well as a reducing agent.$\\$ In hydrogen peroxide $(H _2 O _2 )$, the $O.N.$ of $O $ is $-1$ and the range of the $O.N.$ that $O$ can have is from $0$ to $-2. O$ can sometimes also attain the oxidation numbers $+1$ and $+2.$ $\\$ Hence, $H _2 O_ 2$ can act as an oxidising as well as a reducing agent.$\\$ In ozone $(O _3 )$, the $O.N.$ of $O$ is zero and the range of the $O.N.$ that $O$ can have is from $0$ to $-2.$ $\\$ Therefore, the $O.N.$ of $O $ can only decrease in this case. Hence, $O _3$ acts only as an oxidant. In nitric acid $(HNO _3 )$, the $O.N.$ of $N$ is $+5$ and the range of the $O.N.$ that $N$ can have is from $+5$ to $-3.$ Therefore, the $O.N.$ of $N $ can only decrease in this case. $\\$Hence, $HNO _3$ acts only as an oxidant.

9   Consider the reactions:$\\$ (a) $6 CO _2 (g) + 6H _2 O(l) \to C _6 H _{12 }O _6 (aq) + 6O_ 2 (g)$ $\\$ (b) $O _3 (g) + H_ 2 O _2 (l) \to H_ 2 O(l) + 2O_ 2 (g)$ $\\$ Why it is more appropriate to write these reactions as:$\\$ (a) $6CO _2 (g) + 12H _2 O(l) \to C _6 H _{12} O_ 6 (aq) + 6H _2 O(l) + 6O_ 2 (g)$ $\\$ (b) $O _3 (g) + H_ 2 O_ 2 (l) \to H_ 2 O(l) + O_ 2 (g) + O_ 2 (g)$ $\\$ Also suggest a technique to investigate the path of the above (a) and (b) redoxreactions.

Solution :

(a)The process of photosynthesis involves two steps.$\\$ $\text{Step 1:}$ $\\$ $H _2 O$ decomposes to give $H_ 2$ and $O_ 2 .$ $\\$ $2 H _2 O ( l ) \to 2 H _2( g ) + O _2( g )$ $\\$ $\text{Step 2:}$ $\\$ The $H_ 2$ produced in step 1reduces $CO_ 2$ , thereby producing glucose $(C _6 H _{12} O _6 )$ and $H _2 O$.$\\$ $6 CO _2( g ) + 12 H _2( g ) \to C_ 6 H _{12} O_ 6( s ) + 6 H _2 O ( l )$ $\\$ Now, the net reaction of the process is given as:$\\$ $2 H _2 O ( l ) \to 2 H_ 2( g ) +O _2( g ) ]* 6$ $\\$ $\dfrac{6 CO _2( g ) + 12 H _2( g ) \to C _6 H _{12} O _6( s ) + 6 H _2 O ( l )}{6 CO _2( g ) + 12 H_ 2 O ( l ) \to C _6 H _{12} O _6( g ) + 6 H _2 O ( l ) +6 O _2( g )}$ $\\$ It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.$\\$ The path of this reaction can be investigated by using radioactive $H _2 O_{ 18}$ in place of $H _2 O.$ $\\$ (b) $O_ 2$ is produced from each of the two reactants $O_ 3$ and $H _2 O _2 $. For this reason, $O _2$ is written twice.$\\$ The given reaction involves two steps. First, $O _3 $ decomposes to form $O _2 $ and $O$. In the second step, $H _2 O _2$ reacts with the $O$ produced in the first step, thereby producing $H_ 2 O$ and $O _2 .$$\\$ $O _3( g ) \to O _2( g ) + O ( g )$ $\\$ $\dfrac{H_ 2 O _2( l ) + O ( g ) \to H_ 2 O ( l ) +O _2( g )}{H _2 O _2( l ) + O _3( g ) \to H_ 2 O ( l ) + O_ 2( g ) +O _2( g )} $ $\\$ The path of this reaction can be investigated by using $H _2 O_ 2 ^{18}$ or$ O _3^{ 18}$ .

10   The compound $AgF _2$ is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?

Solution :

The oxidation state of $Ag$ in $AgF_ 2$ is $+2$. But, $+2$ is an unstable oxidation state of $Ag.$$\\$ Therefore, whenever $AgF_ 2$ is formed, silver readily accepts an electron to form $Ag^ + $. This helps to bring the oxidation state of $Ag$ down from $+2 $ to a more stable state of $+1$. As a result, $AgF _2$ acts as a very strong oxidizing agent.