Question

Write balanced chemical equation for the following reactions:

A) Permanganate ion $\left(Mn{O}_4^{-}\right)$ reacts with sulphur dioxide gas in acidic medium to produce $M{n}^{2+}$ and hydrogen sulphate ion.
(Balance by ion electron method)

B) Reaction of liquid hydrazine $\left(N_2{H}_4\right)$ with chlorate ion $\left(Cl{O}_3^{-}\right)$ in basic medium produces nitric oxide gas and chloride ion in gaseous state.
(Balance by oxidation number method)

C) Dichlorine heptoxide $\left(C{l}_2{O}_7\right)$ in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion $\left(Cl{O}_2^{-}\right)$ and oxygen gas.
(Balance by ion electron method)

Answer 1

Corresponding reaction will be

$Mn{O}_4^{-}+S{O}_2\to M{n}^{2+}+HS{O}_4^{-}$
$Mn=+7S=+4Mn=+2H=+1$
$O=-2O=-2S+6$
$O=-2$
Here $Mn{O}_4^{-}$ is getting reduced to $M{n}^{2+}$ and $S{O}_2$ is getting oxidised to $HS{O}_4^{-}$.

Ionic half-cell reaction (Reduction part)

Here, $Mn$ is changing its oxidation state from $+7$ to $+2$ and thus it has gained $5e^{-}$ to get reduced to $M{n}^2+$.
$Mn{O}_4^{-}\to M{n}^{2+}$

Adding $H_{2}O$ molecules on product side and $H^{+}$ ions on reactant side to balanced Hydrogen and oxygen atoms
$Mn{O}_4^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$

Balancing electrons

$Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$
Balancing number of each atom by multiplying the above equation with $2$:
$2Mn{O}_4^{-}+10e^{-}+16H^{+}\to 2M{n}^{2+}+8H_{2}O$

Ionic half-cell reaction (Oxidation part)
$S{O}_2\to HS{O}_4^{-}$
Balancing the hydrogen and oxygen atom first-
$S{O}_2+2H_{2}O\to HS{O}_4^{-}+3H^{+}$
$+4$ $+6$
$S{O}_2\to HS{O}_4^{-}+2e^{-}$
Balancing number of atoms of each element by multiplying above equation by $5$
$5S{O}_2+10H_{2}O\to 5HS{O}_4^{-}+15H^{+}+10e^{-}$

Thus, reactions are:
$S{O}_2+2H_{2}O\to HS{O}_4^{-}+3H^{+}+2e^{-}---\left(i\right)$
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O---\left(ii\right)$

Addition of Oxidation and Reduction half reactions-

$5S{O}_2+10H_{2}O\to 5HS{O}_4^{-}+15H^{+}+10e^{-}$
$2Mn{O}_4^{-}+16H^{+}+16e^{-}\to 2M{n}^{2+}+8H_{2}O$

$5S{O}_2+2Mn{O}_4^{-}+2H_{2}O+H^{+}\to 5HS{O}_4^{-}+2M{n}^{2+}$

Thus, balanced reaction is:
$5S{O}_2+2Mn{O}_4^{-}+2H_{2}O+H^{+}\to 5HS{O}_4^{-}+2M{n}^{2+}$

B)
Corresponding reaction will be:
$N_2{H}_4+Cl{O}_3^{-}\to NO+C{l}^{-}$
$N=-2Cl=+5N=+2Cl=-1$
$H=+1O=-2O=-2$

Oxidation half reaction
For $N$, oxidation state changes from $-2$ to $+2$ which signifies that $N$ is losing $4$ electrons per $N$ atom
Decrease $=\Delta O.S\times$ no.of. atoms [$\Delta O.S=$ change in oxidation state]
$=\left[\right(+2)-(-2\left)\right]\times$
$2$ [Since $2N$ atoms present in $N_2{H}_2$] $=4\times 2=8$
Corresponding oxidation half-cell reactions:
$N_2{H}_4\to 2NO+8e^{-}---\left(i\right)$

Reduction half reaction
For $Cl$, oxidation state changes from $+5$ to $-1$ which signifies that it is getting reduced by accepting $6$ electrons per $Cl$ atom.
Decrease in oxidation state
$=\Delta O.S\times$ no.of.atoms
$=[5-(-1\left)\right]\times 1$
[Since one $Cl$ atom is present in $Cl{O}_3^{-}$]
$=6\times 1=+6$
Corresponding reduction half-cell reactions: $Cl{O}_3^{-}+6e^{-}\to C{l}^{-}----\left(ii\right)$

Adding the two half redox reactions and Balancing change in Oxidation states-
(So, there are $8$ electrons lost and $6$ electron gain per half ionic reaction. Taking $LCM$ of $6,8$ we get $24$. So, in order to balance change in oxidation number, we multiply the two half reactions with such a number so as to get $24$ electrons in exchange (which upon addition of the two reactions, finally cancels out.)
Multiplying equation $\left(i\right)$ with $3$ and equation $\left(ii\right)$ with $4$ to balance increase and decrease in oxidation number, then adding these two equations:

$3N_2{H}_2\to 6NO+24e^{-}$
$4Cl{O}_3^{-}+24e^{-}\to 4C{l}^{-}$

$3N_2{H}_2+4Cl{O}_3^{-}\to 6NO+4C{l}^{-}$

Balancing oxygen atoms on both sides by adding $H_{2}O$
Adding $6H_{2}O$ on product side to balance reaction
$3N_2{H}_4+4Cl{O}_3^{-}\to 6NO+4C{l}^{-}+6H_{2}O$
Thus, balanced reaction is:
$3N_2{H}_4+4Cl{O}_3^{-}\to 6NO+4C{l}^{-}+6H_{2}O$

C)
Corresponding reaction will be:
$C{l}_2{O}_7+H_2{O}_2\to Cl{O}_2^{-}+O_2$
$Cl=+7H=+1Cl=+3O_2=0$
$O=-2O=-1O=-2$
Here $C{l}_2{O}_7$ is getting reduced to $Cl{O}_2^{-}$ and $H_2{O}_2$ is getting oxidised to $O_2$

Ionic Half cell reaction (Oxidation )
$H_2{O}_2\to O_2+2e^{-}$
Number of electrons lost $=$
$\Delta OS\times$ number of atoms
$=[0-(-1\left)\right]\times 2$
[ $2$ oxygen atoms present in $H_2{O}_2$]
$=1\times 2=2$

Balancing Oxygen and Hydrogen atoms
Oxygen atoms are already balanced, thus adding $2H+$ ions on product side to balance $H$ atoms.
$H_2{O}_2\to O_2+2e^{-}+2H^{+}$

Ionic Half cell reaction (Reduction)
$C{l}_2{O}_7+8e^{-}\to 2Cl{O}_2^{-}$
No. of electrons gained
$=\Delta OS\times$ number of atoms
$=(7-3)\times 2$
($2$ chlorine atoms present in $C{l}_2{O}_7)$
$=4\times 2=8$

Balancing Oxygen and Hydrogen atoms
Adding $3H_{2}O$ on product side and $6H^{+}$ on reactant side to balance hydrogen and oxygen atoms
$C{l}_2{O}_7+6H^{+}+8e^{-}\to 2Cl{O}_2^{-}+3H_{2}O$

Addition of the two Ionic half reactions
Thus, reactions are:
$H_2{O}_2\to O_2+2e^{-}+2H^{+}---\left(i\right)$
$C{l}_2{O}_7+6H^{+}+8e^{-}\to 2Cl{O}_2^{-}+3H_{2}O---\left(ii\right)$
Multiplying equation $\left(i\right)$ with $4$ and adding both equation
$4H_2{O}_2\to 4O_2+8e^{-}+8H^{+}$
$C{l}_2{O}_7+6H^{+}+8e^{-}\to 2Cl{O}_2^{-}+3H_{2}O$

$C{l}_2{O}_7+4H_2{O}_2\to 2Cl{O}_2^{-}+3H_{2}O+4O_2+2H^{+}$
Thus, balanced equation is:
$C{l}_2{O}_7+4H_2{O}_2\to 2Cl{O}_2^{-}+3H_{2}O+4O_2+2H^{+}$