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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