(a) Ion electron method Write the skeleton equation for the given reaction. $\mathrm{MnO}_4^{-}(a q)+\mathrm{SO}_2(g) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{HSO}_4^{-}(a q)$

Find out the elements which undergo change in O.N.

Divide the given skeleton into two half equations.

Reduction half equation : $\mathrm{MnO}_4^{-}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})$

Oxidation half equation: $\mathrm{SO}_2(\mathrm{~g}) \longrightarrow \mathrm{HSO}_4^{-}(a q)$

To balance reduction half equation

In acidic medium, balance H and O -atoms

$$\mathrm{MnO}_4^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{H}_2 \mathrm{O}(l)$$

To balance the complete reaction

$$\begin{aligned} & 2 \mathrm{MnO}_4(a q)+16 \mathrm{H}^{+}(a q)+10 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_2 \mathrm{O}(l) \\ & 5 \mathrm{SO}_2(g)+10 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 5 \mathrm{HSO}_4^{-}(a q)+15 \mathrm{H}^{+}(a q)+10 e^{-} \end{aligned}$$

$2 \mathrm{MnO}_4^{-}(a q)+5 \mathrm{SO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}^{+}(a q) \rightarrow 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{HSO}_4^{-}(a q)$

(b) Oxidation number method Write the skeleton equation for the given reaction.

$$\mathrm{N}_2 \mathrm{H}_4(l)+\mathrm{ClO}_3^{-}(a q) \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}^{-}(g)$$

O.N. increases by 4 per N -atom

Multiply NO by 2 because in $\mathrm{N}_2 \mathrm{H}_4$ there are 2 N atoms

$$\mathrm{N}_2 \mathrm{H}_4(l)+\mathrm{ClO}_3^{-}(a q) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}^{-}(a q)$$

Total increase in O.N. of $\mathrm{N}=2 \times 4=8$ ( $8 \mathrm{e}^{-}$ lost) Total decrease in $\mathrm{O} . \mathrm{N}$. of $\mathrm{Cl}=1 \times 6=6\left(6 e^{-}\right.$gain $)$

Therefore, to balance increase or decrease in O.N. multiply $\mathrm{N}_2 \mathrm{H}_4$ by $3,2 \mathrm{NO}$ by 3 and $\mathrm{ClO}_3^{-}, \mathrm{Cl}^{-}$ by 4

$$3 \mathrm{~N}_2 \mathrm{H}_4(l)+4 \mathrm{ClO}_3^{-}(a q) \longrightarrow 6 \mathrm{NO}(g)+4 \mathrm{Cl}^{-}(a q)$$

Balance O and H -atoms by adding $6 \mathrm{H}_2 \mathrm{O}$ to RHS

$$3 \mathrm{~N}_2 \mathrm{H}_4(l)+4 \mathrm{ClO}_3^{-}(a q) \longrightarrow 6 \mathrm{NO}(g)+4 \mathrm{Cl}^{-}(a q)+6 \mathrm{H}_2 \mathrm{O}(l)$$

(c) Ion electron method Write the skeleton equation for the given reaction.

$$\mathrm{Cl}_2 \mathrm{O}_7(g)+\mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \longrightarrow \mathrm{ClO}_2^{-}(\mathrm{aq})+\mathrm{O}_2(g)$$

Find out the elements which undergo a change in O.N.

Divide the given skeleton equation into two half equations.

Reduction half equation : $\mathrm{Cl}_2 \mathrm{O}_7 \longrightarrow \mathrm{ClO}_2^{-}$

Oxidation half equation : $\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{O}_2$

To balance the reduction half equation

$$\mathrm{Cl}_2 \mathrm{O}_7(g)+6 \mathrm{H}^{+}(a q)+8 \mathrm{e}^{-} \longrightarrow 2 \mathrm{ClO}_2^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l)$$

To balance the oxidation half equation

$$\mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \longrightarrow \mathrm{O}_2(g)+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$$

To balance the complete reaction

$$\begin{aligned} \mathrm{Cl}_2 \mathrm{O}_7(g)+6 \mathrm{H}^{+}(a q)+8 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{ClO}_2^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \\ 4 \mathrm{H}_2 \mathrm{O}_2(a q) & \longrightarrow \mathrm{O}_2(g)+8 \mathrm{H}^{+}(a q)+8 e^{-} \end{aligned}$$

$$ \mathrm{Cl}_2 \mathrm{O}_7(g)+4 \mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow 2 \mathrm{ClO}_2^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l)+4 \mathrm{O}_2(g)+2 \mathrm{H}^{+}+(a q)$$

This represents the balanced redox reaction.

(a) Suppose that the $\mathrm{O} . \mathrm{N}$. of P in $\mathrm{HPO}_3^{2-}$ be $x$.

$$\begin{aligned} \text{Then,}\quad 1+x+3(-2) & =-2 \\ \text{or,}\quad x+1-6 & =-2 \\ \text{or,}\quad x & =+3 \end{aligned}$$

(b) Suppose that the O.N. of P in $\mathrm{PO}_4^{3-}$ be $x$.

Then, $x+4(-2)=-3$

or, $x-8=-3$

or, $x=+5$

The oxidation number of each sulphur atom in the following compounds are given below

(a) $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ Let us consider the structure of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$.

There is a coordinate bond between two sulphur atoms. The oxidation number of acceptor S -atom is -2 . Let, the oxidation number of other S -atom be $x$.

$$\mathrm{\mathop {2( + 1)}\limits_{For\,Na} + \mathop {3 \times ( - 2)}\limits_{For\,O - atoms} + x + \mathop {1( - 2) = 0}\limits_{For\,coordinate\,S - atom}}$$

$x=+6$

Therefore, the two sulphur atoms in $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ have -2 and +6 oxidation number.

(b) $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ Let us consider the structure of $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$.

In this structure, two central sulphur atoms have zero oxidation number because electron pair forming the S-S bond remain in the centre. Let, the oxidation number of (remaining S-atoms) S-atom be $x$.

$$\mathrm{\mathop {2( + 1)}\limits_{For\,Na} + \mathop {6\,( - 2)}\limits_{For\,O} + 2x + \mathop {2(0) = 0}\limits_{}}$$

$2-12+2 x=0$ or $x=+\frac{10}{2}=+5$

Therefore, the two central S-atoms have zero oxidation state and two terminal S-atoms have +5 oxidation state each.

(c) $\mathrm{Na}_2 \mathrm{SO}_3$ Let the oxidation number of S in $\mathrm{Na}_2 \mathrm{SO}_3$ be $x$.

$$2(+1)+x+3(-2)=0 \text { or } x=+4$$

(d) $\mathrm{Na}_2 \mathrm{SO}_4$ Let the oxidation number of S be $x$.

$$2(+1)+x+4(-2)=0 \text { or } x=+6$$

Oxidation number method

(a)

Balance increase and decrease in oxidation number.

$$6 \mathrm{Fe}^{2+}+\mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}$$

Balance charge by multiplying $\mathrm{H}^{+}$by 14.

$$6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}$$

Balance H and O -atoms by multiplying $\mathrm{H}_2 \mathrm{O}$ by 7 .

$$6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_2 \mathrm{O}$$

This represents a balanced redox reaction.

(b)

Balance increase and decrease in oxidation number

$$\mathrm{I}_2+10 \mathrm{NO}_3^{-} \longrightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}$$

Balance charge by writing $8 \mathrm{H}^{+}$in LHS of the equation.

$$\mathrm{I}_2+10 \mathrm{NO}_3^{-}+8 \mathrm{H}^{+} \longrightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}$$

Balance H -atoms by writing $4 \mathrm{H}_2 \mathrm{O}$ in RHS of the equation.

$$\mathrm{I}_2+10 \mathrm{NO}_3^{-}+8 \mathrm{H}^{+} \longrightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}+4 \mathrm{H}_2 \mathrm{O}$$

Oxygen atoms are automatically balanced. This represents a balanced redox reaction.

(c)

(Multiply $\mathrm{S}_2 \mathrm{O}_3^{2-}$ by 2 because there are 4 S -atoms in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ ion.)

Increase and decrease in oxidation number is already balanced. Charge and oxygen atoms are also balanced.

This represents a balanced redox reaction.

(d)

Increase and decrease in oxidation number is already balanced.

Add $4 \mathrm{H}^{+}$ towards LHS of the equation to balance charge.

$$\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{CO}_2$$

Add $2 \mathrm{H}_2 \mathrm{O}$ towards RHS of the equation to balance H -atoms

$$\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} $$

This represents a balanced redox reaction.

(a) Writing the O.N. on each atom above its symbol, then

Here, the $\mathrm{O} . \mathrm{N}$. of Cl increases from -1 in HCl to O in $\mathrm{Cl}_2$, therefore, $\mathrm{Cl}^{-}$is oxidised and hence HCl acts as the reducing agent.

The O.N. of N decreases from +5 in $\mathrm{HNO}_3$ to +3 in NOCl , therefore, $\mathrm{HNO}_3$ acts as the oxidising agent.

Thus, this reaction is a redox reaction.

(b) Writing the O.N. of each atom above its symbol, we have,

Here, the O.N. of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.

(c)

Here, O.N. of Fe decreases from +3 in $\mathrm{Fe}_2 \mathrm{O}_3$ to 0 in Fe , therefore, $\mathrm{Fe}_2 \mathrm{O}_3$ acts as an oxidising agent. Further, O.N. of C increases from +2 in CO to +4 in $\mathrm{CO}_2$, therefore, CO acts as a reducing agent.

Thus, this reaction is an example of redox reaction.

(d) Writing the O.N. of each atom above its symbol, then

Here, O.N. of none of the atoms undergo a change, therefore, this reaction is not a redox reaction.

(e) Writing the O.N. of each atom above its symbol, then

Here, O.N. of N increases from -3 to 0 in $\mathrm{N}_2$, therefore, $\mathrm{NH}_3$ acts as a reducing agent. Further, $\mathrm{O} . \mathrm{N}$. of O decreases from O in $\mathrm{O}_2$ to -2 in $\mathrm{H}_2 \mathrm{O}$, therefore, $\mathrm{O}_2$ acts as a oxidising agent. Thus, this reaction is a redox reaction.