Answer:
Reaction:
\[MnO_{4}^{-}(aq)+{{I}^{-}}(aq)\to
Mn{{O}_{2}}(s)+{{I}_{2}}(s)\]
1st step: Splitting
into two half reactions: (balance the elements other than hydrogen and oxygen)
\[MnO_{4}^{-}\to Mn{{O}_{2}}(s)\] (Reduction)
\[2{{I}^{-}}\to {{I}_{2}}(s)\] (Oxidation)
2nd step: Balance
oxygen atoms by adding water molecule to the oxygen deficient side.
\[MnO_{4}^{-}\to Mn{{O}_{2}}+2{{H}_{2}}O\]
\[2{{I}^{-}}\to
{{I}_{2}}\]
3rd step: Balance
hydrogen by adding \[{{H}^{+}}\] ions to the hydrogen deficient side.
\[MnO_{4}^{-}+4{{H}^{+}}\to
Mn{{O}_{2}}+2{{H}_{2}}O\]
\[2{{I}^{-}}\to {{I}_{2}}\]
4th step: Balance the
charge by adding electrons
\[MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to
Mn{{O}_{2}}+2{{H}_{2}}O\] \[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}\]
5th step: Equations
are added in such a way that electrons are cancelled; we now get the balanced
equation.
\[[MnO_{4}^{-}]+4{{H}^{+}}+3{{e}^{-}}\to
Mn{{O}_{2}}+2{{H}_{2}}O]\times 2\]
\[[2{{I}^{-}}\to
{{I}_{2}}+2{{e}^{-}}]\times 3\]
\[\frac{\frac{{}}{2MnO_{4}^{-}+6{{I}^{-}}+8{{H}^{+}}\to
2Mn{{O}_{2}}+3{{I}_{2}}+4{{H}_{2}}O}}{Balanced\text{ }equation}\]
\[2MnO_{4}^{-}+4{{H}_{2}}O\to
2Mn{{O}_{2}}+3\underset{(Basic\,medium)}{\mathop{{{I}_{2}}+8O{{H}^{-}}}}\,\]
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