Thinking Process

Redox reactions represent those reactions which involve change in oxidation number of the interacting species. (i.e., oxidation and reduction)

Answer

(d) Following are the examples of redox reaction

(a) $CuO +H_{2} \longrightarrow Cu +H_{2} O$

(b) $Fe_{2} O_{3}+3 CO \longrightarrow 2 Fe+3 CO_{2}$

(c) $2 ~K+F_{2} \longrightarrow 2 KF$

Option (d) is not an example of redox reaction.

Answer

(d) Given that, $E^{\circ}$ values of

$$ \begin{aligned} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+} & =+0.77 \mathrm{~V} \\ I_{2}(\mathrm{~s}) / I^{-} & =+0.54 \mathrm{~V} \\ \mathrm{Cu}^{2+} / \mathrm{Cu} & =+0.34 \mathrm{~V} \\ \mathrm{Ag}^{+} / \mathrm{Ag} & =+0.80 \mathrm{~V} \end{aligned} $$

Since, $E^{\circ}$ of the redox couple $\mathrm{Ag}^{+} / \mathrm{Ag}$ is the most positive, i.e., $0.80 \mathrm{~V}$, therefore, $\mathrm{Ag}^{+}$is the stronnect nxidicinc anent

Answer

(d) Given that $E^{\circ}$ values of

$$ \begin{aligned} Br_{2} / Br^{-} & =+1.90 ~V \\ Ag / Ag^{+} & =-0.80 ~V \\ Cu^{2+} / Cu(s) & =+0.34 ~V \\ I^{-} / I_{2}(~s) & =-0.54 ~V \\ Br^{-} / Br_{2} & =-1.90 ~V \end{aligned} $$

The $E^{\circ}$ values show that copper will reduce $\mathrm{Br}_{2}$, if the $E^{\circ}$ of the following redox reaction is positive.

Now,

$$ \begin{gathered} 2 Cu+Br_2 \rightarrow CuBr_2 \\ Cu \rightarrow Cu^{2+}+2 e^- ; E^{\circ}=-0.34 ~V \\ \frac{Br_2 +2 e^- \rightarrow 2 Br^- ; E=+1.09 ~V}{Cu+Br_2 \rightarrow CuBr_2 ; E^{\circ}=+0.75 ~V} \end{gathered} $$

Since, $E^{\circ}$ of this reaction is positive, therefore, Cu can reduce $\mathrm{Br}_{2}$. While other reaction will give negative value.

Thinking Process

Calculate the $E^{\circ} _{\text {cell }}$ of the four redox reactions. If $E^{\circ} _{\text {cell }}$ of a reaction is negative, that reaction will not occur.

Answer

(d)

$$ \begin{aligned} & \text { (a) } 2 Fe^{3+}+2 e^- \rightarrow 2 Fe^{2+} ; E^{\circ}=+0.77 ~V \\ & 2 I^- \rightarrow I_{2}+2 e^- ; E^{\circ}=-0.54 ~V \quad \text { (sign of } E^{\circ} \text { is reversed) } \\ & 2 Fe^{3+}+2 I^- \rightarrow 2 Fe^{2+}+I_{2} ; E_{\text {cell }}^{\circ}=+0.23 ~V \end{aligned} $$

This reaction is feasible since $E^{\circ}{ }_{\text {cell }}$ is positive.

$$\text { (b) } \quad Cu \rightarrow Cu^{2+} + 2e^-; E^{\circ} = +0.34 ~V $$

$$ \text { (sign of } \quad E^{\circ} \text{has been reversed )} $$

$$ \begin{aligned} & 2 Ag^{+}+2 e^- \rightarrow 2 Ag ; E^{\circ}=+0.80 ~V \\ & Cu+2 Ag^+ \rightarrow 2 Cu^{2+}+2 Ag ; E^{\circ}=+0.46 ~V \end{aligned} $$

This reaction is feasible since $E^{\circ}{ }_{\text {cell }}$ is positive.

(c)

$$ 2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Fe}^{2+} ; E^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=-0.34 \mathrm{~V} \quad \text{(sign of } E^{\circ} \text{is reversed)}$$

$$ 2 Fe^{3+}+Cu \rightarrow 2 Fe^{2+}+Cu^{2+} ; E^{\circ}=+0.43 ~V $$

This reaction is feasible since $E^{\circ}{ }_{\text {cell }}$ is positive.

(d)

$$ \mathrm{Ag} \rightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; E^{\circ}=-0.80 \mathrm{~V} \quad \quad \text{(sign of } E^{\circ} \text{is reversed)} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} ; E^{\circ}=+0.77 \mathrm{~V} \\ \mathrm{Ag}+\mathrm{Fe}^{3+} \rightarrow \mathrm{Ag}^{+}+\mathrm{Fe}^{2+} ; E^{\circ}=-0.03 \mathrm{~V} \\ \text { This reaction is not feasible since } E_{\text {cell }}^{\circ} \text { is negative. } $$

Answer

(a) $2 S_2^{+2} O_3^{2-}(aq) + I_2^0 s \rightarrow \stackrel{2.5}{~S}_4 O_6^{2-}(aq) + 2 I^-(aq)$

$$ S_{2}^{2-2} O_{3}^{2-}(aq)+2 ~B_{2}(l)+5 H_{2} O(l) \rightarrow 2 \stackrel{+6}{~S}^{2-} O_{4}^{2-}(aq)+4 Br^{-}(aq)+10 H^{+}(aq) $$

Bromine being stronger oxidising agent than $I_{2}$, oxidises $S$ of $S_{2} O_{3}^{2-}$ to $SO_{4}^{2-}$ whereas $I_{2}$ oxidises it only into $S_{4} O_{6}^{2-}$ ion.

Answer

(a) Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen halides, -1 in hydrides and zero in $\mathrm{H}_{2}$ molecule.

All the other three statements (b), (c) and (d) are correct.

Answer

(b) $NH_4 NO_3$ is actually $NH_4^+$and $NO_3^-$. It is an ionic compound.

The oxidation number of nitrogen in the two species is different as shown below

Let, oxidation number of $N$ in $\stackrel{+}{N} H_{4}$ is $x$.

$\Rightarrow$ $x+(4 \times 1)=+1$

or $x+4=+1$ or $x=-3$

Let, oxidation number of $N_{\text {in }} NO_{3}^{-}$is $x$

$\Rightarrow \quad x+(3 \times-2)=-1$ or $x-6=-1$ or $x=+5$

Answer

(a) Writing the oxidation number (O.N.) of $\mathrm{Cr}, \mathrm{Cl}$ and $\mathrm{Mn}$ on each species in the four set of ions, then,

(a) $\stackrel{+3}CrO_2^-, \stackrel{+5}ClO_3^-, \stackrel{+6}CrO_4^{2-}, \stackrel{+7}MnO_4^-$

(b) $\stackrel{+5}Cl_3^-, \stackrel{+6}CrO_4^{2-}, \stackrel{+7}Mn O_4^-, \stackrel{+3}C rO_2^-$

(c) $\stackrel{+3}{Cr} O_{2}^{-}, \stackrel{+5}{Cl} O_{3}^{-}, \stackrel{+7}{Mn} O_{4}^{-}, \stackrel{+6}{Cr} O_{4}^{2-}$

(d) $\stackrel{+6}CrO_4^{2-}, \stackrel{+7}Mn O_4^-, \stackrel{+3}CrO_2^-, \stackrel{+5}ClO_3^{3-}$

Only in the arrangement (a), the O.N. of central atom increases from left to right, therefore, option (a) is correct.

Answer

(d) Highest oxidation number of any transition element $=(n-1) d$ electrons $+n s$ electrons. Therefore, large the number of electrons in the $3 d$-orbitals, higher is the maximum oxidation number.

(a) $3 d^{1} 4 s^{2}=3$

(b) $3 d^{3} 4 s^{2}=3+2=5$

(c) $3 d^{5} 4 s^{1}=5+1=6$ and

(d) $3 d^{5} 4 s^{2}=5+2=7$

Thus, option (d) is correct.

Answer

(d) Reactions in which the same substance is oxidised as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions

(a) $ \stackrel{-4}{C}\stackrel{+1}{H_4} + 2\stackrel{0}{O_2} \longrightarrow \stackrel{+4}{C}\stackrel{-2}{O_2} + 2\stackrel{+1}{H_2} \stackrel{-2}{O} $

(b) $ \stackrel{-4}{C}\stackrel{+1}{H_4} + 4\stackrel{0}{Cl_2} \longrightarrow \stackrel{+4}{C}\stackrel{-1}{Cl_4} + 4\stackrel{+1}{H} \stackrel{-1}{Cl} $

(c) $ 2\stackrel{0}{F_2} + 2\stackrel{-2}{O}\stackrel{+1}{H} \longrightarrow 2\stackrel{-1}{F^-} + \stackrel{+2}{O} \stackrel{-1}{F^- _2} + \stackrel{+1}{H_2} \stackrel{-2}{O} $

(d) $ 2\stackrel{+4}{N}\stackrel{-2}{O_2} + 2\stackrel{-2}{O}\stackrel{+1}{H} \longrightarrow \stackrel{+3}{N}\stackrel{-2}{O^- _2} + \stackrel{+5}{N} \stackrel{-2}{O^- _3} + \stackrel{+1}{H_2} \stackrel{-2}{O} $

Thus, in reaction (d), $N$ is both oxidised as well as reduced since the $O . N$. of $N$ increases from +4 in $NO_{2}$ to +5 in $NO_{3}^{-}$and decreases from +4 in $NO_{2}$ to +3 in $NO_{2}^{-}$.

Answer

(c) Being the most electronegative element, $\mathrm{F}$ can only be reduced and hence it always shows an oxidation number of -1 . Further, due to the absence of $d$-orbitals it cannot be oxidised and hence it does not show positive oxidation numbers.

In other words, F cannot be oxidised as well as reduced simultaneously and hence does not show disproportionation reactions.

Answer

$(a, b, c, d)$

Write the oxidation number of each element above its symbol, then

$$ 2 \stackrel{+1+5-2}{~K^{-} ClO_{3}} \longrightarrow 2 \stackrel{+1-1}{~K} Cl^{+} 3 O_{2} $$

(a) The O.N. of $\mathrm{K}$ does not change, $\mathrm{K}$ undergoes neither reduction nor oxidation. Thus, option (a) is not correct.

(b) The O.N. of chlorine decreases from +5 in $\mathrm{KClO}_{3}$ to -1 in $\mathrm{KCl}$, hence $\mathrm{Cl}$ undergoes reduction.

(c) Since, O.N. of oxygen increases from -2 in $KClO_{3}$ to 0 in $O_{2}$, oxygen is oxidised.

(d) This statement is not correct because $\mathrm{Cl}$ is undergoing reduction and $\mathrm{O}$ is undergoing oxidation

Answer

(c, $d)$

Writing the oxidation number of each element above its symbol, so that

$$ \stackrel{0}{Zn} +2 \stackrel{+1-1}{HCl} \longrightarrow \stackrel{+2}{Zn} Cl_{2}^{-1}+\stackrel{0}{H_{2}} $$

(a) The oxidation number of $\mathrm{Zn}$ increases from 0 in $\mathrm{Zn}$ to +2 in $\mathrm{ZnCl}_{2}$, therefore, $\mathrm{Zn}$ acts as a reductant. Thus, option (a) is incorrect.

(b) The oxidation number of chlorine does not change, therefore, it neither acts as a reductant nor an oxidant. Therefore, option (b) is incorrect.

(c) The oxidation number of hydrogen decreases from +1 in $\mathrm{H}^{+}$to 0 in $\mathrm{H}_{2}$, therefore, $\mathrm{H}^{+}$ acts as an oxidant. Thus, option (c) is correct.

(d) As explained in option (a), Zn acts as reductant, therefore, it cannot act as an oxidant. Thus, option (d) is correct.

Answer

$(b, c, d)$

Elements which have only s-electrons in the valence shell do not show more than one oxidation state. Thus, element with $3 s^{1}$ as outer electronic configuration shows only one oxidation state of +1

Transition element such as elements (b), (c) having incompletely filled d-orbitals in the penultimate shell show variable oxidation states. Thus, element with outer electronic configuration as $3 d^{1} 4 s^{2}$ shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as $3 d^{2} 4 s^{2}$ shows variable oxidation states of $+2,+3$ and +4 .

$p$ - Block elements also show variable oxidation states due to a number of reason such as involvement of $d$-orbitals and inert pair effect e.g., element (d) with $3 s^{2} 3 p^{3}$ as (i.e., P) as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of $d$-orbitals.

Answer

$(c, d)$

Write the O.N. of each element above its symbol, then

In this reaction, $O . N$. of $P$ increases from 0 in $P_{4}$ to +1 in $H_{2} PO_{2}^{-}$and decreases to -3 in $PH_{3}$, therefore, $P$ undergoes both oxidation as well as reduction. Thus, options (a) and (b) are wrong and option (c) is correct.

Further, O.N. of $H$ remains +1 in all the compounds, i.e., $H$ neither undergoes oxidation nor reduction. Thus, option (d) is correct.

Answer

$(a, b)$

All electrodes which have negative electrode potentials are stronger reducing agents than $\mathrm{H}_{2}$ gas and hence acts as anodes when connected to standard hydrogen electrode. Thus, $\mathrm{Al}^{3+} / \mathrm{Al}\left(E^{\ominus}=-1.66 \mathrm{~V}\right)$ and $\mathrm{Fe}^{2+} / \mathrm{Fe}\left(E E^{\ominus}=-0.44 \mathrm{~V}\right)$ act as anode.

Thinking Process

Write the oxidation number of each element above its symbol. and then identify the bleaching reagent by observing the change in oxidation number.

Answer

$\stackrel{0}{C}_{2}(g)+2{\stackrel{-2}{O} H^{-}}^{+1}(aq) \rightarrow \stackrel{+1}{C} IO^{-2}(aq)+\stackrel{-1}{Cl^{-}}(aq)+\stackrel{+1}{H} 2 \stackrel{-2}{O}(l)$

In this reaction, O.N. of $Cl$ increases from 0 (in $Cl_{2}$ ) to 1 (in $ClO^{-}$) as well as decreases from 0 (in $Cl_{2}$ ) to -1 (in $Cl^{-}$). So, it acts both reducing as well as oxidising agent. This is an example of disproportionation reaction. In this reaction, $ClO^{-}$species bleaches the substances due to its oxidising action. [In hypochlorite ion $\left(ClO^{-}\right) Cl$ can decrease its oxidation number from +1 to 0 or -1 .]

Note Disproportionation reactions are a special type of redox reactions. In which an element in one oxidation state is simultaneously oxidised and reduced.

Answer

In $\mathrm{MnO}_{4}^{2-}$, the oxidation number of $\mathrm{Mn}$ is +6 . It can increase its oxidation number (to +7 ) or decrease its oxidation number (to $+4,+3,+2,0$ ).

Hence, it undergoes disproportionation reaction in acidic medium.

In $\mathrm{MnO}_{4}^{-}, \mathrm{Mn}$ is in its highest oxidation state, i.e., +7 . It can only decrease its oxidation number. Hence, it cannot undergo disproportionation reaction.

Answer

Writing the oxidation number of each element above its symbol in the following reactions

(a) $\underset{\text { Basic oxide }}{2 \stackrel{+2}{P} bO^{-2}}+4 \underset{\text { Acid }}{\stackrel{+1-1}{C} Cl} \longrightarrow 2 \stackrel{+2}{PbCl_2^{-1}}+2 \stackrel{+1}{H}_{2} \stackrel{-2}{O}$

In this reaction, oxidation number of each element remains same hence, it is not a redox reaction. In fact, it is an example of acid-base reaction.

(b) $\stackrel{+4}{PbO}_2+4 \stackrel{+1}{H} \stackrel{-1}{Cl} \longrightarrow \stackrel{+2}{~Pb} \stackrel{-1}{Cl}_2+\stackrel{0}{C}_2 +2 \stackrel{+1}{H}_2 O^{-2}$

In $PbO_{2}, ~Pb$ is in +4 oxidation state. Due to inert pair effect $Pb$ in +2 oxidation state is more stable. So, $Pb$ in +4 oxidation state $\left(PbO_{2}\right)$ acts as an oxidising agent.

It oxidises $\mathrm{Cl}^{-}$to $\mathrm{Cl}_{2}$ and itself gets reduced to $\mathrm{Pb}^{2+}$.

Answer

$\mathrm{PbO}$ is a base. It reacts with nitric acid and forms soluble lead nitrate.

$$ PbO +2 HNO_3 \longrightarrow \underset{\text { Soluble }}{Pb \left(NO_{3}\right)_2}+ H_2 O $$

(acid base reaction)

Nitric acid does not react with $PbO_{2}$. Both of them are strong oxidising agents. In $HNO_{3}$, nitrogen is in its maximum oxidation state $(+5)$ and in $PbO_{2}$, lead is in its maximum oxidation state $(+4)$. Therefore, no reaction takes place.

Answer

(a) Ion electron method Write the skeleton equation for the given reaction. $MnO_{4}^{-}(aq)+SO_{2}(~g) \longrightarrow Mn^{2+}(aq)+HSO_{4}^{-}(aq)$

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 : $SO_{2}(~g) \longrightarrow HSO_{4}^{-}(aq)$

To balance reduction half equation

In acidic medium, balance $H$ and $O$-atoms

$$ MnO_{4}^{-}(aq)+8 H^{+}(aq)+5 e^{-} \longrightarrow Mn^{2+}(aq)+H_{2} O(l) $$

To balance the complete reaction

$ 2MnO_{4(aq)} + 16H^+ _{(aq)} + 10e^- \longrightarrow Mn^{2+} _{(aq)} +8H _2O _{(l)} $

$5 \mathrm{SO}_2(g)+10 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 5 \mathrm{HSO}_4^{-}(\mathrm{aq})+15 \mathrm{H}^{+}(\mathrm{aq})+10 e^{-}$

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

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

$$ N_{2} H_{4}(l)+ ClO_{3}^{-}(aq) \longrightarrow NO(g) + Cl^{-}(g) $$

O.N. increases by 4 per $\mathrm{N}$-atom

Multiply $NO$ by 2 because in $N_{2} H_{4}$ there are $2 ~N$ atoms

$$ N_{2} H_{4}(l)+ ClO_{3}^{-}(aq) \longrightarrow 2 NO(g)+Cl^{-}(aq) $$

Total increase in O.N. of N $=2 \times 4=8$ ( $8 e^{-}$lost)

Total decrease in O.N. of $\mathrm{Cl}=1 \times 6=6$ (6 $e^{-}$gain)

Therefore, to balance increase or decrease in O.N. multiply $N_{2} H_{4}$ by $3,2 NO$ by 3 and $ClO_{3}^{-}, Cl^{-}$by 4

$$ 3 ~N_{2} H_{4}(l)+4 ClO_{3}^{-}(aq) \longrightarrow 6 NO(g)+4 Cl^{-}(aq) $$

Balance $\mathrm{O}$ and $\mathrm{H}$-atoms by adding $6 \mathrm{H}_{2} \mathrm{O}$ to RHS

$$ 3 ~N_{2} H_{4}(l)+4 ClO_{3}^{-}(aq) \longrightarrow 6 NO(g)+4 Cl^{-}(aq)+6 H_{2} O(l) $$

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

$$ Cl_{2} O_{7}(~g)+H_{2} O_{2}(aq) \longrightarrow ClO_{2}^{-}(aq)+O_{2}(~g) $$

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

Divide the given skeleton equation into two half equations.

Reduction half equation : $Cl_{2} O_{7} \longrightarrow ClO_{2}^{-}$

Oxidation half equation : $H_{2} O_{2} \longrightarrow O_{2}$

To balance the reduction half equation

$$ Cl_{2} O_{7}(~g)+6 H^{+}(aq)+8 e^{-} \longrightarrow 2 ClO_{2}^{-}(aq)+3 H_{2} O(l) $$

To balance the oxidation half equation

$$ H_{2} O_{2}(aq) \longrightarrow O_{2}(~g)+2 H^{+}+2 e^{-} $$

To balance the complete reaction

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

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

This represents the balanced redox reaction.

Answer

(a) Suppose that the O.N. of $\mathrm{P}$ in $\mathrm{HPO}_{3}^{2-}$ be $x$.

(b) Suppose that the O.N. of $\mathrm{P}$ in $\mathrm{PO}_{4}^{3-}$ be $x$.

Answer

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

(a) $Na_2 ~S_2 O_3$ Let us consider the structure of $Na_2 ~S_2 O_3$.

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 $x$.

$$ \begin{aligned} & 2(+1)+3 \times(-2)+x+1(-2)=0 \\ & \text { For } \mathrm{Na} \quad \text { For } \mathrm{O} \text {-atoms } \quad \text { For coordinate } \mathrm{S} \text {-atom } \\ & x=+6 \end{aligned} $$

Therefore, the two sulphur atoms in $Na_2 ~S_2 O_3$ have -2 and +6 oxidation number.

(b) $Na_2 ~S_4 O_6$ Let us consider the structure of $Na_2 ~S_4 O_6$.

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 $\mathrm{S}$-atoms) S-atom be $x$.

$$ \begin{gathered} 2(+1)+6(-2)+2 x+2(0)=0 \\ \text { For } Na \text { For } O \\ 2-12+2 x=0 \text { or } x=+\frac{10}{2}=+5 \end{gathered} $$

Therefore, the two central $\mathrm{S}$-atoms have zero oxidation state and two terminal $\mathrm{S}$-atoms have +5 oxidation state each.

(c) $Na_{2} SO_{3}$ Let the oxidation number of $S$ in $Na_{2} SO_{3}$ be $x$.

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

(d) $Na_{2} SO_{4}$ Let the oxidation number of $S$ be $x$.

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

Answer

Oxidation number method

(a)

(Multiply $\mathrm{Cr}^{3+}$ by 2 because there are $2 \mathrm{Cr}$ atoms in $Cr_{2} O_{7}^{2-}$ ion.)

Balance increase and decrease in oxidation number.

$$ 6 Fe^{2+}+H^{+}+Cr_{2} O_{7}^{2-} \longrightarrow 2 Cr^{3+}+6 Fe^{3+}+H_{2} O $$

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

$$ 6 Fe^{2+}+14 H^{+}+Cr_{2} O_{7}^{2-} \longrightarrow 2 Cr^{3+}+6 Fe^{3+}+H_{2} O $$

Balance $\mathrm{H}$ and $\mathrm{O}$-atoms by multiplying $\mathrm{H}_{2} \mathrm{O}$ by 7 .

$$ 6 Fe^{2+}+14 H^{+}+Cr_{2} O_{7}^{2-} \longrightarrow 2 Cr^{3+}+6 Fe^{3+}+7 H_{2} O $$

This represents a balanced redox reaction.

(b)

Balance increase and decrease in oxidation number

$$ I_2+10 NO_3^- \longrightarrow 10 NO_2+2 IO_3^- $$

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

$$ I_2+10 NO_3^-+8 H^+ \longrightarrow 10 NO_2+2 IO_3^- $$

Balance $\mathrm{H}$-atoms by writing $4 \mathrm{H}_{2} \mathrm{O}$ in $\mathrm{RHS}$ of the equation.

$$ I_2+10 NO_3^-+8 H^+ \longrightarrow 10 NO_2+2 IO_3^-+4 H_2 O $$

Oxygen atoms are automatically balanced.

This represents a balanced redox reaction.

(c)

(Multiply $S_2 O_3^2-$ by 2 because there are $4 ~S$-atoms in $S_4 O_6^2-$ 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.

$$ MnO_{2}+C_{2} O_{4}^{2-}+4 H^{+} \longrightarrow Mn^{2+}+2 CO_{2} $$

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

$$ MnO_{2}+C_{2} O_{4}^{2-}+4 H^{+} \longrightarrow Mn^{2+}+2 CO_{2}+2 H_{2} O $$

This represents a balanced redox reaction.

Answer

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

$$ 3 \stackrel{+1}{H}-1 I(aq)+\stackrel{+1+5-2}{H} NO_{3}(aq) \longrightarrow \stackrel{0}{C} l_{2}(g)+\stackrel{+3-2}{~N} O^{-1} Cl(g)+2 \stackrel{+1}{H}_{2}-2(l) $$

Here, the O.N. of $\mathrm{Cl}$ increases from -1 in $\mathrm{HCl}$ to $\mathrm{O}$ in $\mathrm{Cl}_{2}$, therefore, $\mathrm{Cl}^{-}$is oxidised and hence $\mathrm{HCl}$ acts as the reducing agent.

The O.N. of $N$ decreases from +5 in $HNO_{3}$ to +3 in $NOCl$, therefore, $HNO_{3}$ 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,

$$ \stackrel{+2}{\mathrm{HgCl}}{ }^{-1}(\mathrm{aq})+2 \mathrm{~K}^{+-1} \mathrm{I}(\mathrm{aq}) \longrightarrow \stackrel{+2-1}{\longrightarrow} \mathrm{Hg} \mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{~K}^{+1} \mathrm{Cl}^{-1}(\mathrm{aq}) $$

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

$\stackrel{+3}{\mathrm{Fe}_2} \stackrel{-2}{\mathrm{O}}_3(\mathrm{~s})+3 \stackrel{+2-2}{\mathrm{C}} \mathrm{O}^{\Delta}(\mathrm{g}) \xrightarrow{\Delta} 2 \mathrm{~F}^{\mathrm{Fe}}(\mathrm{s})+3 \stackrel{+4}{\mathrm{C}}^{-2}{ }_2(\mathrm{~g})$

Here, $O . N$. of $Fe$ decreases from +3 in $Fe_2 O_3$ to 0 in $Fe$, therefore, $Fe_2 O_3$ acts as an oxidising agent. Further, $O . N$. of $C$ increases from +2 in $CO$ to +4 in $CO_2$, 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

${\stackrel{+3}{\mathrm{P}} \stackrel{-1}{\mathrm{Cl} _3}}(l)+3 \stackrel{+1}{\mathrm{H}} _2 \stackrel{-2}{\mathrm{O}}(l) \rightarrow 3 \stackrel{+1}{\mathrm{H}}\stackrel{-1}{\mathrm{Cl}}(\mathrm{aq})+\stackrel{+1}{\mathrm{H}} _3 \stackrel{+3}{\mathrm{P}}\stackrel{-2}{\mathrm{O_3}}(\mathrm{aq})$

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

$4 \stackrel{-3}{\mathrm{N}}\stackrel{+1}{\mathrm{H_3}}(\mathrm{aq})+3 \stackrel{0}{\mathrm{O}_2}(g) \longrightarrow 2 \stackrel{0}{\mathrm{~N}_2}(g)+6 \stackrel{+1}{\mathrm{H}}_2 \stackrel{-2}{\mathrm{O}}(l)$

Here, O.N. of $N$ increases from -3 to 0 in $N_2$, therefore, $NH_3$ acts as a reducing agent. Further, O.N. of $O$ decreases from 0 in $O_2$ to -2 in $H_2 O$, therefore, $O_2$ acts as a oxidising agent. Thus, this reaction is a redox reaction.

Answer

(a) Write the O. N. of all atoms above their respective symbols.

$\mathrm{O}$. N. decreases by, 3 per $\mathrm{Cr}$-atom

Divide the given equation into two half reactions

Reduction half reaction : $Cr_{2} O_{7} \rightarrow Cr^{3+}$

Oxidation half reaction : $\mathrm{I}^{-} \rightarrow \mathrm{I}_{2}$

To balance reduction half reaction.

$$ Cr_{2} O_{7}^{2-}+14 H^{+}+6 e^{-} \longrightarrow 2 Cr^{3+}+7 H_{2} O $$

To balance oxidation half reaction

$$ 2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-} $$

To balance the reaction by electrons gained and lost

$$ \begin{gathered} Cr_{2} O_{7}^{2-}+14 H^{+}+6 e^{-} \longrightarrow 2 Cr^{3+}+7 H_{2} O \\ 6 I^{-} \longrightarrow 3 I_{2}+6 e^{-} \\ Cr_{2} O_{7}^{2-}+14 H^{+}+6 I^{-} \longrightarrow 2 Cr^{3+}+3 I_{2}+7 H_{2} O \end{gathered} $$

This gives the final balanced ionic equations.

(b) Write the skeletal equation of the given reaction

$$ Cr_{2} O_{7}^{2-}(aq)+Fe^{2+}(aq) \longrightarrow Cr^{3+}(aq)+Fe^{3+}(aq) $$

Write the $\mathrm{O}$. $\mathrm{N}$. of all the elements above their respective symbols.

Divide the given equation into two half reactions

Oxidation half reaction : $Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)$

reduction half reaction : $Cr_{2} O_{7}^{2-}(aq) \rightarrow Cr^{3+}(aq)$

To balance oxidation half reaction

$$ Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)+e^{-} $$

To balance reduction half reaction

$$ Cr_{2} O_{7}^{2-}(aq)+6 e^{-} \longrightarrow 2 Cr^{3+}(aq) $$

Balance charge by adding $\mathrm{H}^{+}$ions.

$$ Cr_{2} O_{7}^{2-}(aq)+14 H^{+}(aq)+6 e^{-} \longrightarrow 2 Cr^{3+}(aq) $$

Balance $\mathrm{O}$ atoms by adding $\mathrm{H}_{2} \mathrm{O}$ molecules

$$ Cr_{2} O_{7}^{2-}(aq)+14 H^{+}(aq)+6 e^{-} \longrightarrow 2 Cr^{3+}(aq)+7 H_{2} O(l) $$

To balance the reaction

$$ \begin{gathered} 6 Fe^{2+}(aq) \longrightarrow 6 Fe^{3+}(aq)+6 e^{-} \\ Cr_{2} O_{7}^{2-}(aq)+14 H^{+}(aq)+6 e^{-} \longrightarrow 2 Cr^{3+}(aq)+7 H_{2} O(l) \end{gathered}$$

$$ \begin{gathered} Cr_{2} O_{7}^{2-}(aq)+6 Fe^{2+}(aq)+14 H^{+}(aq) \longrightarrow 2 Cr^{3+}(aq)+7 H_{2} O(l)+6 Fe^{3+}(aq) \end{gathered} $$

(c) Write the O. N. of all atoms above their respective symbols.

Divide the skeleton equation into two half-reactions.

Reduction half reaction: $\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}$

Oxidation half reaction : $SO_{3}^{2-} \longrightarrow SO_{4}^{2-}$

To balance reduction half reaction

$$ MnO_{4}^{-}+8 H^{+}+5 e^{-} \longrightarrow Mn^{2+}+4 H_{2} O $$

To balance oxidation half reaction

$$ SO_{3}^{2-} \longrightarrow SO_{4}^{2-}+2 e^{-} $$

Balance charge by adding $H^{+}$ions.

$$ SO_{3}^{2-} \longrightarrow SO_{4}^{2-}+2 H^{+}+2 e^{-} $$

Balance $O$-atoms by adding $H_{2} O$ molecules

$$ SO_{3}^{2-}+H_{2} O \longrightarrow SO_{4}^{2-}+2 H^{+}+2 e^{-} $$

To balance the reaction

$$ \begin{gathered} 2 MnO_{4}^{-}+16 H^{+}+10 e^{-} \longrightarrow 2 Mn^{2+}+8 H_{2} O \\ 5 SO_{3}^{2-}+5 H_{2} O \longrightarrow 5 SO_{4}^{2-}+10 H^{+}+10 e^{-} \end{gathered} $$

$$ \begin{gathered} 2 MnO_{4}^{-}+5 SO_{3}^{2-}+6 H^{+} \longrightarrow 2 Mn^{2+}+5 SO_{4}^{2-}+3 H_{2} O \end{gathered} $$

This represents the correct balanced redox equation.

(d) Write the $\mathrm{O}$. N. of all the atoms above their respective symbols.

Divide skeleton equation into two half reactions

Reduction half reaction $\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}$

Oxidation half reaction $\mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2}$

To balance reduction half reaction

$$ MnO_{4}^{-}+8 H^{+}+5 e^{-} \longrightarrow Mn^{2+}+4 H_{2} O $$

To balance oxidation half reaction

$$ 2 \mathrm{Br}^{-} \longrightarrow \mathrm{Br}_{2}+2 \mathrm{e}^{-} $$

To balance the reaction

$$ 2 MnO_4^- +16 H^+ +10 e^- \longrightarrow 2 Mn^{2+} +8 H_2 O \\ 10 Br^- \longrightarrow 5 Br_2 +10 e^-\\ $$

$$ 2 MnO_4^- +10 Br^- +16 H^+ \longrightarrow 2 Mn^{2+} +5 Br_2 +8 H_2 O $$

This represents the correct balanced ionic equation.

Answer

A. $\rightarrow(4)$

B. $\rightarrow(5)$

C. $\rightarrow(3)$

D. $\rightarrow(1)$

Suppose that $x$ be the oxidation states of central atoms.

A. Oxidation number of $Cr$ in $Cr_{2} O_{7}^{2-}$

$$ \begin{aligned} 2 x+7(-2) & =-2 \\ 2 x-14 & =-2 \\ 2 x & =+12 \\ x & =+6 \end{aligned} $$

B. Oxidation number of $\mathrm{Mn}$ in $\mathrm{MnO}_{4}^{-}$

$$ \begin{aligned} x+4(-2) & =-1 \\ x-8 & =-1 \\ x & =+7 \end{aligned} $$

C. Oxidation number of $\mathrm{V}$ in $\mathrm{VO}_{3}^{-}$

$$ \begin{aligned} x+3(-2) & =-1 \\ x-6 & =-1 \\ x & =+5 \end{aligned} $$

D. Oxidation number of $\mathrm{Fe}$ in $\mathrm{FeF}_{6}^{3-}$

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

Answer

A. $\rightarrow(5)$

B. $\rightarrow(4)$

C. $\rightarrow(3)$

D. $\rightarrow$ (2)

E. $\rightarrow(6)$

A. lons having positive charge - Cation

B. The sum of oxidation number of all atoms in a neutral molecule - Zero

C. Oxidation number of hydrogen ion $\left(\mathrm{H}^{+}\right)-+1$

D. Oxidation number of fluorine in $\mathrm{NaF}– -1$

E. Ions having negative charge - Anion

Assertion and Reason

In the following questions a statement of assertion (A) followed by a statement of reason $(\mathrm{R})$ is given. Choose the correct option out of the choices given below in each question.

Answer

(b) Both assertion and reason are true but reason is not the correct explanation of assertion. Among halogen $\mathrm{F}_{2}$ is the best oxidant because it has the highest $E^{\circ}$ value.

Answer

(c) Assertion is true but reason is false.

$10 KI+2 KMnO_{4}+8 H_{2} SO_{4} \longrightarrow 2 \stackrel{+2}{MnSO}_{4}+6 K_2 SO_4 +8 H_2 O+5 I_2$

Oxidation state of $\mathrm{Mn}$ decreases from +7 to +2 .

Answer

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Thus, the above reaction is an example of disproportionation reaction.

Answer

(a) Both assertion and reason are true reason is the correct explanation of assertion.

Redox couple is the combination of oxidised and reduced form of substance. In the representation $E_{\mathrm{Fe}^{3+}}^{\ominus} / \mathrm{Fe}^{2+}$ and $E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{\prime}}^{\ominus}, \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ and $\mathrm{Cu}^{2+} / \mathrm{Cu}$ are redox couples.

Answer

As we know that, the reactions

$$ \begin{aligned} & 2 Na(s)+Cl_{2}(g) \longrightarrow 2 NaCl(s) \\ & 4 Na(s)+O_{2}(g) \longrightarrow 2 Na_{2} O(s) \end{aligned} $$

are redox reactions because in each of these reactions sodium is oxidised due to the addition of either oxygen or more electronegative element to sodium. Simultaneously, chlorine and oxygen are reduced because of each of these, the electropositive element sodium has been added.

From our knowledge of chemical bonding we also know that, sodium chloride and sodium oxide are ionic compounds and perhaps better written as $\mathrm{Na}^{+} \mathrm{Cl}^{-}(s)$ and $\left(\mathrm{Na}^{+}\right)_{2} \mathrm{O}^{2-}(s)$. Development of charges on the species produced suggests us to rewrite the above reaction in the following manner

For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and other the gain of electrons. As an illustration, we may further elaborate one of these, say, the formation of sodium chloride.

$$ \begin{aligned} 2 \mathrm{Na}(s) & \longrightarrow 2 \mathrm{Na}^{+}(g)+2 \mathrm{e}^{-} \\ \mathrm{Cl}_{2}(g)+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Cl}^{-}(g) \end{aligned} $$

Each of the above steps is called a half reaction, which explicitly shows involvement of electrons. Sum of the half reactions gives the overall reaction:

$$ 2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Na}^{+} \mathrm{Cl}^{-}(s) \text { or } 2 \mathrm{NaCl}(s) $$

The given reactions suggest that half reactions that involved loss of electrons are oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions.

It may not be out of context to mention here that the new way of defining oxidation and reduction has been achieved only by establishing a correlation between the behaviour of species as per the classical idea and their interplay in electron-transfer change.

In the given reactions, sodium, which is oxidised, acts as a reducing agent because it donates electron to each of the elements interacting with it and thus helps in reducing them. Chlorine and oxygen are reduced and act as oxidising agents because these accept electrons from sodium.

To summarise, we may mention that

Oxidation Loss of electron(s) by any species.

Reduction Gain of electron(s) by any species.

Oxidising agent Acceptor of electron(s).

Reducing agent Donor of electron(s).

Answer

As we know that,

$$ E_{Cu^{2+} / Cu}^{\circ} =0.34 V, E_{Zn^{2+} / Zn}^{\circ}=-0.76 V $$

$$ E_{Mg^{2+} / Mg}^{\circ} =-2.37 V, E_{Fe^{2+} / Fe}^{\circ}=-0.74 V $$

$$ E_{Br_{2} / Br^{-}}^{\circ} =+1.08 V, E_{Cl_{2} / Cl^{-}}^{\circ}=+1.36 V $$

$$ E_{Cd^{2+} / Cd}^{\circ} =-0.44 V $$

(a) $E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}$ and $E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}$

$$ \mathrm{Cu}+\mathrm{Zn}^{2+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Zn} $$

In the given cell reaction, Cu is oxidised to $\mathrm{Cu}^{2+}$, therefore, $\mathrm{Cu}^{2+} / \mathrm{Cu}$ couple acts as anode and $\mathrm{Zn}^{2+}$ is reduced to $\mathrm{Zn}$, therefore, $\mathrm{Zn}^{2+} / \mathrm{Zn}$ couple acts as cathode.

$$ \begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=-0.76-(+0.34)=-1.10 \mathrm{~V} \end{aligned} $$

Negative value of $E_{\text {cell }}^{\circ}$ indicates that the reaction will not occur.

(b) $\mathrm{Mg}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$

$$ E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.37 \mathrm{~V} \text { and } E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.74 \mathrm{~V} $$

In the given cell reaction, $\mathrm{Mg}$ is oxidised to $\mathrm{Mg}^{2+}$ hence, $\mathrm{Mg}^{2+} / \mathrm{Mg}$ couple acts as anode and $\mathrm{Fe}^{2+}$ is reduced to $\mathrm{Fe}$ hence, $\mathrm{Fe}^{2+} / \mathrm{Fe}$ couple acts as cathode.

$$ \begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=-0.74-(-2.37)=+1.63 \mathrm{~V} \end{aligned} $$

Positive value of $E^{\circ}{ }_{\text {cell }} 5$ indicates that the reaction will occur.

(c)

$$ \begin{array}{r} Br_{2}+2 Cl^{-} \longrightarrow Cl_{2}+2 Br^{-} \\ E_{Br}^{-} / Br_{2}^{\circ}=+1.08 \vee \text { and } E_{Cl}^{-} / Cl_{2}^{\circ}=+1.36 V \end{array} $$

In the given cell reaction, $Cl^{-}$is oxidised to $Cl_{2}$ hence, $Cl^{-} / Cl_{2}$ couple acts as anode and $Br_{2}$ is reduced to $Br^{-}$hence; $Br^{-} / Br_{2}$ couple acts as cathode.

$$ \begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=+1.08-(+1.36)=-0.28 \mathrm{~V} \end{aligned} $$

Negative value of $E^{\circ}{ }_{\text {cell }}$ indicates that the reaction will occur.

(d)

$$ \begin{array}{r} \mathrm{Fe}+\mathrm{Cd}^{2+} \longrightarrow \mathrm{Cd}+\mathrm{Fe}^{2+} \\ E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.74 \mathrm{~V} \text {. and } E_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\circ}=-0.44 \mathrm{~V} \end{array} $$

In the given cell reaction, $\mathrm{Fe}$ is oxidised to $\mathrm{Fe}^{2+}$ hence, $\mathrm{Fe}^{2+} / \mathrm{Fe}$ couple acts as anode and $\mathrm{Cd}^{2+}$ is reduced to $\mathrm{Cd}$ hence, $\mathrm{Cd}^{2+} / \mathrm{Cd}$ couple acts as cathode.

$$ \begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=-0.44-(-0.74)=+0.30 \mathrm{~V} \end{aligned} $$

Positive value $E^{\circ}{ }_{\text {cell }}$ indicates that the reaction will occur.

Answer

In a disproportionation reaction, the same species is simultaneously oxidised as well as reduced. Therefore, for such a redox reaction to occur, the reacting species must contain an element which has atleast three oxidation states.

The element, in reacting species, is present in an intermediate state while lower and higher oxidation states are available for reduction and oxidation to occur (respectively).

Fluorine is the strongest oxidising agent. It does not show positive oxidation state. That’s why fluorine does not show disproportionation reaction.

Thinking Process

A redox couple represents the oxidised and reduced forms of a substance together taking part in an oxidation or reduction half reaction.

Answer

Given that,

$$ \begin{gathered} Cu+Zn^{2+} \longrightarrow Cu^{2+}+Zn \\ Mg+Fe^{2+} \longrightarrow Mg^{2+}+Fe \\ Br_{2}+2 Cl^{-} \longrightarrow Cl_{2}+2 Br \\ Fe+Cd^{2+} \longrightarrow Cd^{2+} Fe^{2+} \end{gathered} $$

(a) $Cu^{2+} / Cu$ and $Zn^{2+} / Zn$

(b) $Mg^{2+} / Mg$ and $Fe^{2+} / Fe$

(c) $Br_{2} / Br^{-}$and $Cl_{2} / Cl^{-}$

(d) $Fe^{2+} / Fe$ and $Cd^{2+} / Cd$

Answer

Suppose that the oxidation number of chlorine in these compounds be $x$.

O.N. of $\mathrm{Cl}$ in $\mathrm{NaClO}_{4} \therefore \quad+1+x+4(-2)=0$ or, $x=+7$

O.N. of $\mathrm{Cl}$ in $\mathrm{NaClO}_{3} \therefore+1+x+3(-2)=0$ or, $x=+5$

O.N. of $\mathrm{Cl}$ in $\mathrm{NaClO} \therefore+1+x+1(-2)=0$ or, $x=+1$

O.N. of $\mathrm{Cl}$ in $\mathrm{KClO}_{2} \quad \therefore \quad+1+x+2(-2)=0$ or, $x=+3$

O.N. of $Cl_{\text {in }} Cl_{2} O_{7} \quad \therefore \quad+2 x+7(-2)=0$ or, $x=+7$

O.N. of $\mathrm{Cl}^{\text {in } \mathrm{ClO}_{3}} \quad \therefore \quad x+3(-2)=0$ or, $x=+6$

O.N. of $Cl_{\text {in }} Cl_{2} O \quad \therefore \quad 2 x+1(-2)=0$ or, $x=+1$

O. N. of $\mathrm{Cl}$ in $\mathrm{NaCl} \quad \therefore \quad+1+x=0$ or, $x=-1$

O. N. of $Cl_{\text {in }} Cl_{2} \quad \therefore \quad 2 x=0$ or, $x=0$

O. N. of $\mathrm{Cl}$ in $\mathrm{ClO}_{2} \quad \therefore \quad x+2(-2)=0$ or, $x=+4$

None of these compounds have an oxidation number of +2 .

Increasing order of oxidation number of chlorine is : $-1,0,+1,+3,+4,+5,+6,+7$

Therefore, the increasing order of oxidation number of $\mathrm{Cl}$ in compounds is

$$ NaCl<Cl_{2}<NaClO<KClO_{2}<ClO_{2}<NaClO_{3}<ClO_{3}<Cl_{2} O_{7} $$

Answer

Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it is negative, it acts as an oxidant.

Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as an reductant or oxidant. e.g., measurement of standard electrode potential of $\mathrm{Zn}^{2+} / \mathrm{Zn}$ electrode using SHE as a reference electrode.

The EMF of the cell comes out to be $0.76 \mathrm{~V}$. (reading of voltmeter is $0.76 \mathrm{~V}$ ). $\mathrm{Zn}^{2+} / \mathrm{Zn}$ couple acts as anode and SHE acts as cathode.

$$ \therefore \quad \begin{aligned} E_{\text {cell }}^{\circ} & =0.76=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ 0.76 & =0-E_{\text {anode }}^{\circ} \\ E_{\text {anode }}^{\circ} & =-0.76 \mathrm{~V} \\ E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ} & =-0.76 \mathrm{~V} . \end{aligned} $$