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Question
Balance the following ionic equations.
\[\ce{MnO^{-}4 + H^{+} + Br^{-} -> Mn^{2+} + Br2 + H2O}\]
Solution
Dividing the equation into two half-reactions:
Oxidation half-reaction: \[\ce{Br^{-} -> Br2}\]
Reduction half-reaction: \[\ce{MnO^{-}4 -> Mn^{2+}}\]
Balancing oxidation and reduction half-reactions separately as:
Oxidation half-reaction:
\[\ce{Br^{-} -> Br2}\]
\[\ce{2Br^{-} -> Br2}\]
\[\ce{2Br^{-} -> Br2 + 2e-}\] .....(i)
Reduction half-reaction:
\[\ce{MnO^{-}4 -> Mn^{2+}}\]
\[\ce{MnO^{-}4 + 5e^{-} -> Mn^{2+}}\]
\[\ce{MnO^{-}4 + 8H^{+} + 5e^{-} -> Mn^{2+}}\]
\[\ce{MnO^{-}4 + 8H^{+} + 5e^{-} -> Mn^{2+} + 4H2O}\] .....(ii)
To balance the electrons, multiply equation (i) by 5 and equation (ii) by 2 and add
\[\ce{2MnO^{-}4 + 10Br^{-} + 16H^{+} -> 2Mn^{2+} + 5Br2 + 8H2O}\]
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