 # 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?

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.