Answer:
Skeletal
equation is
\[\overset{+7}{\mathop{MnO_{4}^{-}}}\,+\overset{-1}{\mathop{Br}}\,\to
\overset{+4}{\mathop{Mn{{O}_{2}}}}\,+\overset{+5}{\mathop{BrO_{3}^{-}}}\,\]
Perusal of oxidation number confirms that \[MnO_{4}^{-}\]
is oxidant while \[B{{r}^{-}}\] ion is reductant.
The two half equations can be balanced as:
\[[MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to
Mn{{O}_{2}}+2{{H}_{2}}O]2\,\,\,\,\,\,\,\,\,\,\,......(i)\]\[B{{r}^{-}}+3{{H}_{2}}O\to
BrO_{3}^{-}+6{{H}^{+}}+6{{e}^{-}}\,\,\,\,\,\,\,\,\,\,\,.......(ii)\]
\[\frac{On\text{
}addition}{\frac{2MnO_{4}^{-}(aq)+B{{r}^{-}}+2{{H}^{+}}\to
2Mn{{O}_{2}}+BrO_{3}^{-}+{{H}_{2}}O}{{}}}\]
On adding two \[O{{H}^{-}}\] ions on both sides we get the balanced
equation in basic medium:
\[2MnO_{4}^{-}(aq)+B{{r}^{-}}(aq)+\underbrace{2{{H}^{+}}(aq)+2O{{H}^{-}}(aq)}_{{}}\to
\]
\[2Mn{{O}_{2}}+BrO_{3}^{-}+2O{{H}^{-}}+{{H}_{2}}O\]
\[2MnO_{4}^{-}(aq)+B{{r}^{-}}(aq)+{{H}_{2}}O\to \]
\[2Mn{{O}_{2}}+BrO_{3}^{-}(aq)+2O{{H}^{-}}(aq)\]
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