NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

Ans : (a) 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
\( \mathrm{Na} \mathrm{H}_{2} \mathrm{PO}_{4}\)
Then, we have
\( \begin{array}{l}{1(+1)+2(+1)+1(x)+4(-2)=0} \\ {\Rightarrow 1+2+x-8=0} \\ {\Rightarrow x=+5}\end{array}\)
Hence, the oxidation number of P is +5.


(b) \( \stackrel{+1}{N a} H S O_{4}\)
Then, we have
\( \begin{array}{l}{1(+1)+1(+1)+1(x)+4(-2)=0} \\ {\Rightarrow 1+1+x-8=0} \\ {\Rightarrow x=+6}\end{array}\)
Hence, the oxidation number of S is + 6.


(c) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\)
Then, we have
\( \begin{array}{l}{4(+1)+2(x)+7(-2)=0} \\ {\Rightarrow 4+2 x-14=0} \\ {\Rightarrow 2 x=+10} \\ {\Rightarrow x=+5}\end{array}\)
Hence, the oxidation number of P is + 5.


(d) \(\mathrm{K}_{2} \mathrm{MnO}_{4}\)
Then, we have
\( \begin{array}{l}{2(+1)+x+4(-2)=0} \\ {\Rightarrow 2+x-8=0} \\ {\Rightarrow x=+6}\end{array}\)
Hence, the oxidation number of Mn is + 6.


(e) \( \begin{array}{l}{+2 \quad x} \\ {\mathrm{Ca} \mathrm{O}_{2}}\end{array}\)
Then, we have
\( \begin{array}{l}{(+2)+2(x)=0} \\ {\Rightarrow 2+2 x=0} \\ {\Rightarrow x=-1}\end{array}\)


(f) \(9 \mathrm{Na} \mathrm{B} \mathrm{H}_{4}\)
Then, we have
\( \begin{array}{l}{1(+1)+1(x)+4(-1)=0} \\ {\Rightarrow 1+x-4=0} \\ {\Rightarrow x=+3}\end{array}\)
Hence, the oxidation number of B is + 3.


(g) \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7}\)
Then, we have
\( \begin{array}{l}{2(+1)+2(x)+7(-2)=0} \\ {\Rightarrow 2+2 x-14=0} \\ {\Rightarrow 2 x=12} \\ {\Rightarrow x=+6}\end{array}\)
Hence, the oxidation number of S is + 6.




(h ) \( \mathrm{KAl}\left(\underline{\mathrm{SO}}_{4}\right)_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O}\) 
Then, we have
\( \begin{array}{l}{1(+1)+1(+3)+2(x)+8(-2)+24(+1)+12(-2)=0} \\ {\Rightarrow 1+3+2 x-16+24-24=0} \\ {\Rightarrow 2 x=12} \\ {\Rightarrow x=+6}\end{array}\)
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
\( \begin{array}{l}{1(+1)+1(+3)+2(x)+8(-2)=0} \\ {\Rightarrow 1+3+2 x-16=0} \\ {\Rightarrow 2 x=12} \\ {\Rightarrow x=+6}\end{array}\)
Hence, the oxidation number of S is + 6. 

Ans : (a) \( \mathrm{Kl}_{3}\)
In  \( \mathrm{Kl}_{3}\), the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is \( -\frac{1}{3}\) . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3to find the oxidation states.
In a \( \mathrm{Kl}_{3}\) molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a  \( \mathrm{Kl}_{3}\)molecule, the O.N. of the two I atoms forming the l2> molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is as 1.
(b) 
\( \mathrm{H}_{2} \underline{\mathrm{S}_{4}} \mathrm{O}_{6}\)
\( \begin{array}{l}{\text { Now, } 2(+1)+4(x)+6(-2)=0} \\ {\Rightarrow 2+4 x-12=0} \\ {\Rightarrow 4 x=10} \\ {\Rightarrow x=+2 \frac{1}{2}}\end{array}\)
However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

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)  \( \mathrm{Fe}_{3} \mathrm{O}_{4}\)
On taking the O.N. of O as 2, the O.N. of Fe is found to be \( +2 \frac{2}{3}\). 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.


 (d) \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)
\( \begin{array}{l}{2(x)+6(+1)+1(-2)=0} \\ {\text { or, } 2 x+4=0} \\ {\text { or, } x=-2} \\ {\text { Hence, the O.N. of } C \text { is } \hat{a} \epsilon^{u} 2}\end{array}\)


  (e) \( \mathrm{CH}_{3} \mathrm{COOH}\)


\( \begin{array}{l}{2(x)+4(+1)+2(-2)=0} \\ {\text { or, } 2 x=0} \\ {\text { or, } x=0}\end{array}\)


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 +2 and –2 in \( \mathrm{CH}_{3} \mathrm{COOH}\).
 

Ans : (a) \(\mathrm{CuO}_{(s)}+\mathrm{H}_{2(g)} \)longrightarrow \(\mathrm{Cu}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(g)}\)
Let us write the oxidation number of each element involved in the given reaction as:
\( \stackrel{+2}{\mathrm{Cu}} \mathrm{O}_{(s)}+\stackrel{\theta}{\mathrm{H}_{2(\mathrm{g})}}\) -> \( \mathrm{Cu}_{(s)}+\mathrm{H}_{2} \mathrm{O}_{(g)}\)
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 \( \mathrm{H}_{2}\) to +1 in \( \mathrm{H}_{2} \mathrm{O}\) i.e., \( \mathrm{H}_{2}\) is oxidized to \( \mathrm{H}_{2} \mathrm{O}\). Hence, this reaction is a redox reaction.


(b) \( \mathrm{Fe}_{2} \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(\mathrm{g})} \longrightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(\mathrm{g})}\)
Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in \( \mathrm{Fe}_{2} \mathrm{O}_{3}\) to 0 in Fe i.e.,  \( \mathrm{Fe}_{2} \mathrm{O}_{3}\) is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in \( \mathrm{CO}_{2}\) i.e., CO is oxidized to  \( \mathrm{CO}_{2}\). Hence, the given reaction is a redox reaction.


(c)  \( 4 \mathrm{BCl}_{3}(\mathrm{g})+3 \mathrm{LiAlH}_{4}(\mathrm{s}) \rightarrow 2 \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_{3}(\mathrm{s})\)
The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in \( \mathrm{BCl}_{3}\) to –3 in \( \mathrm{B}_{2} \mathrm{H}_{6}\). i.e., \( \mathrm{BCl}_{3}\) is reduced to\( \mathrm{B}_{2} \mathrm{H}_{6}\). Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in \( \mathrm{B}_{2} \mathrm{H}_{6}\) i.e., LiAlH4 is oxidized to \( \mathrm{B}_{2} \mathrm{H}_{6}\). Hence, the given reaction is a redox reaction.


(d) \( 2 \mathrm{K}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{K}+\mathrm{F}-\)
The oxidation number of each element in the given reaction can be represented as:





(e) \( 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)
The oxidation number of each element in the given reaction can be represented as:
Here, the oxidation number of N increases from –3 in \( \mathrm{NH}_{3}\) to +2 in NO. On the other hand, the oxidation number of \( \mathrm{O}_{2}\)  decreases from 0 in \( \mathrm{O}_{2}\) to –2 in NO and \( \mathrm{H}_{2} \mathrm{O}\)  i.e., \( \mathrm{O}_{2}\) is reduced. Hence, the given reaction is a redox reaction. 

Ans : Let us write the oxidation number of each atom involved in the given reaction above its symbol as:



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. 

Ans : \( \begin{array}{l}{\text { (i) } \mathrm{H}_{2} \mathrm{SO}_{5}} \\ {2(+1)+1(x)+5(-2)=0} \\ {\Rightarrow 2+x-10=0} \\ {\Rightarrow x=+8}\end{array}\)
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 \( \mathrm{H}_{2} \mathrm{SO}_{5}\) is shown as follows:

\( \begin{array}{l}{\text { Now, } 2(+1)+1(x)+3(-2)+2(-1)=0} \\ {\Rightarrow 2+x-6-2=0} \\ {\Rightarrow x=+6}\end{array}\)
Therefore, the O.N. of S is +6.
(ii)\( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)
\( \begin{array}{l}{2(x)+7(-2)=-2} \\ {\Rightarrow 2 x-14=-2} \\ {\Rightarrow x=+6}\end{array}\)
Here, there is no fallacy about the O.N. of Cr in\( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)
The structure of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)is shown as follows:

Here, each of the two Cr atoms exhibits the O.N. of +6.
(iii) \( \mathrm{NO}_{3}^{-}\)
\( \begin{array}{l}{1(x)+3(-2)=-1} \\ {\Rightarrow x-6=-1} \\ {\Rightarrow x=+5}\end{array}\)
Here, there is no fallacy about the O.N. of N in \( \mathrm{NO}_{3}^{-}\)
The structure of  \( \mathrm{NO}_{3}^{-}\)is shown as follows:

The N atom exhibits the O.N. of +5.