• # question_answer 61) Write balanced chemical equation for the following reactions :(i) Permanganate ion $(MnO_{4}^{-})$reacts with sulphur dioxide gas in acidic medium to produce $\text{M}{{\text{n}}^{\text{2}+}}$and hydrogen sulphate ion. (Balance by ion electron method)(ii) Reaction of liquid hydrazine $\left( {{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}} \right)$with chlorate ion$(ClO_{3}^{-})$ in basic medium produces nitric oxide gas and chloride ion in gaseous state.(Balance by oxidation number method)(iii) Dichlorine heptaoxide$(C{{l}_{2}}{{O}_{7}})$in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion $(ClO_{2}^{-})$and oxygen gas.(Balance by ion electron method)

(i) The given equation is : $MnO_{4}^{-}+S{{O}_{2}}\to M{{n}^{2+}}+HSO_{4}^{-}$ (Acid medium) Half equations are: $MnO_{4}^{-}\to M{{n}^{2+}}\,\,\,\,\,\,\,\,.....(i)$ $S{{O}_{2}}\to HSO_{4}^{-}\,\,\,\,\,\,\,\,\,\,\,......(ii)$ These equations can be balanced as : $[MnO_{4}^{-}+8{{H}^{+}}5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O]\times 2$ $\frac{\frac{On\,addition:\,[S{{O}_{2}}+2{{H}_{2}}O\to HSO_{4}^{-}+3{{H}^{+}}+2{{e}^{-}}]5}{2MnO_{4}^{-}+5S{{O}_{2}}+{{H}^{+}}\to 2M{{n}^{2+}}+5HSO_{4}^{-}}}{{}}$ (ii) See NCERT exercise 8.19 (b) (iii) $C{{l}_{2}}{{O}_{7}}(g)+{{H}_{2}}{{O}_{2}}(aq)\to ClO_{2}^{-}(aq)+{{O}_{2}}(g)+{{H}^{+}}(aq)$ Step I: Splitting the equation into two half equations: $C{{l}_{2}}{{O}_{7}}(g)\to ClO_{2}^{-};$              ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)$ Step II: Balancing the element other than hydrogen and oxygen:                 $C{{l}_{2}}{{O}_{7}}(g)2ClO_{2}^{-};$                 ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+{{H}^{+}}(aq)$ Step III: Balancing oxygen by adding water molecules:                 $C{{l}_{2}}{{O}_{7}}(g)\to 2ClO_{2}^{-}+3{{H}_{2}}O$                 ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+2{{H}^{+}}$ Step IV: Balancing hydrogen by adding ${{H}^{+}}$ ions. $C{{l}_{2}}{{O}_{7}}(g)+6{{H}^{+}}\to 2ClO_{2}^{-}+3{{H}_{2}}O$ ${{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+2{{H}^{+}}$ Step V: Balancing charge by adding electron, we get $C{{l}_{2}}{{O}_{7}}(g)+6{{H}^{+}}+8{{e}^{-}}\to 2ClO_{2}^{-}+3{{H}_{2}}O$ $[{{H}_{2}}{{O}_{2}}(aq)\to {{O}_{2}}(g)+2{{H}^{+}}+2{{e}^{-}}]\times 4$ Step VI: Adding above equations, we get $C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)\to 2ClO_{2}^{-}+3{{H}_{2}}O+2{{H}^{+}}$This equation can be balanced in basic medium by adding two $O{{H}^{-}}$ ions on both sides: $C{{l}_{2}}{{O}_{7}}+4{{H}_{2}}O(aq)+2O{{H}^{-}}\to 2ClO_{2}^{-}$                                 $+3{{H}_{2}}O+2{{H}^{+}}+2O{{H}^{-}}$                                                 OR $C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)+2O{{H}^{-}}(aq)\to 2ClO_{2}^{-}+5{{H}_{2}}O$