11th Class Chemistry Redox Reactions

  • 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)

    Answer:

      (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 \[C{{l}_{2}}{{O}_{7}}(g)+4{{H}_{2}}{{O}_{2}}(aq)+2O{{H}^{-}}(aq)\to 2ClO_{2}^{-}+5{{H}_{2}}O\]


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