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Question
Balance the following ionic equations.
\[\ce{MnO^{-}4 + SO^{2-}3 + H^{+} -> Mn^{2+} + SO^{2-}4 + H2O}\]
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Solution
Dividing the equation into two half-reactions:
Oxidation half-reaction: \[\ce{SO^{2-}3 -> SO^{2-}4}\]
Reduction half-reaction: \[\ce{MnO^{-}4 -> MN^{2+}}\]
Balancing oxidation and reduction half-reactions separately as:
Oxidation half-reaction:
\[\ce{SO^{2-}3 -> SO^{2-}4}\]
\[\ce{SO^{2-}3 -> SO^{2-}4 + 2e-}\]
Since the reaction occurs in acidic medium,
\[\ce{SO^{2-}3 -> SO^{2-}4 + 2e^{-} + 2H+}\]
\[\ce{SO^{2-}3 + H2O -> SO^{2-}4 + 2H^{+} + 2e-}\] .....(i)
Reduction half-reaction:
\[\ce{MnO^{-}4 -> MN^{2+}}\]
\[\ce{MnO^{-}4 + 5e^{-} -> MN^{2+}}\]
\[\ce{MnO^{-}4 + 5e^{-} -> MN^{2+} + 4H2O}\] .....(ii)
To balance the electrons, multiply equation (i) by 5 and equation (ii) by 2 and add
\[\ce{2MO^{-}4 + 5SO^{2-}3 + 6H+ -> 2Mn^{2+} + 5SO^{2-}4 + 3H2O}\]
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