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College Chemistry : Balancing Redox Reactions

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Example Question #11 : Redox Reactions

Balance the following redox reaction in acidic conditions:

$\textrm{SO}_{3}\textrm{ }^{2-}(aq)\textrm{ + MnO}_{4}^{-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)\textrm{ + Mn}^{2+}(aq)$

Possible Answers:

$\textrm{2 Mn}^{2+}(aq)\textrm{ + 5 SO}_{4}\textrm{ }^{2-}\textrm{ + 3 H}_{2}\textrm{O}(l)\rightarrow5\textrm{ SO}_{3}\textrm{ }^{2-}(aq)\textrm{ + 2 MnO}_{4}\textrm{ }^{-}(aq)\textrm{ + 6 H}^{+}(aq)$

$\textrm{SO}_{3}\textrm{ }^{2-}(aq)\textrm{ + MnO}_{4}\textrm{ }^{-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)\textrm{ + Mn}^{2+}(aq)\textrm{ + 3 }e^{-}$

$5\textrm{ SO}_{3}\textrm{ }^{2-}(aq)\textrm{ + 2 MnO}_{4}\textrm{ }^{-}(aq)\textrm{ + 6 H}^{+}(aq)\rightarrow\textrm{2 Mn}^{2+}(aq)\textrm{ + 5 SO}_{4}\textrm{ }^{2-}\textrm{ + 3 H}_{2}\textrm{O}(l)$

$\textrm{SO}_{3}\textrm{ }^{2-}(aq)\textrm{ + MnO}_{4}^{-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)\textrm{ + Mn}^{2+}(aq)$

$\textrm{6 H}^{+}(aq)\textrm{ + SO}_{3}\textrm{ }^{2-}(aq)\textrm{ + MnO}_{4}\textrm{ }^{-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)\textrm{ + Mn}^{2+}(aq)\textrm{ + 3 H}_{2}\textrm{O}(l)$

Correct answer:

Explanation:

Start by writing the half reactions for the equation

$\textrm{SO}_{3}\textrm{ }^{2-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)$

$\textrm{MnO}_{4}\textrm{ }^{-}(aq)\rightarrow\textrm{Mn}^{2+}(aq)$

All atoms except for O and H are already balanced, so we can go straight to balancing O and H atoms. First balance O atoms by adding H₂O

$\textrm{H}_{2}\textrm{O}(l)\textrm{ + SO}_{3}\textrm{ }^{2-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)$

$\textrm{MnO}_{4}\textrm{ }^{-}(aq)\rightarrow\textrm{Mn}^{2+}(aq)\textrm{ + 4 H}_{2}\textrm{O}(l)$

Now balance H atoms with H⁺

$\textrm{H}_{2}\textrm{O}(l)\textrm{ + SO}_{3}\textrm{ }^{2-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)\textrm{ + 2 H}^{+}(aq)$

$\textrm{8 H}^{+}(aq)\textrm{ + MnO}_{4}\textrm{ }^{-}(aq)\rightarrow\textrm{Mn}^{2+}(aq)\textrm{ + 4 H}_{2}\textrm{O}(l)$

Balance each half reaction's electrical charge with electrons

$\textrm{H}_{2}\textrm{O}(l)\textrm{ + SO}_{3}\textrm{ }^{2-}(aq)\rightarrow\textrm{SO}_{4}\textrm{ }^{2-}(aq)\textrm{ + 2 H}^{+}(aq)\textrm{ + 2 }e^-$

$\textrm{5 }e^{-}\textrm{ + 8 H}^{+}(aq)\textrm{ + MnO}_{4}\textrm{ }^{-}(aq)\rightarrow\textrm{Mn}^{2+}(aq)\textrm{ + 8 H}_{2}\textrm{O}(l)$

Now combine both half reactions to get the overall reaction. Be sure to cancel out electrons by multiplying the oxidation reaction by 5 and the reduction reaction by 2. This will give 10 e^- on each side of the overall reaction so that they cancel out.

$\textrm{5 H}_{2}\textrm{O}(l)\textrm{ + 5 SO}_{3}\textrm{ }^{2-}(aq)\rightarrow\textrm{5 SO}_{4}\textrm{ }^{2-}(aq)\textrm{ + 10 H}^{+}(aq)\textrm{ + 10 }e^-$

$\textrm{10 }e^{-}\textrm{ + 16 H}^{+}(aq)\textrm{ + 2 MnO}_{4}\textrm{ }^{-}(aq)\rightarrow\textrm{2 Mn}^{2+}(aq)\textrm{ + 8 H}_{2}\textrm{O}(l)$

Overall Reaction:

$\textrm{5 H}_{2}\textrm{O}(l)\textrm{ + 16 H}^{+}(aq)\textrm{ + 5 SO}_{3}\textrm{ }^{2-}(aq)\textrm{ + 2 MnO}_{4}\textrm{ }^{-}(aq)\rightarrow$

$\textrm{5 SO}_{4}\textrm{ }^{2-}(aq)\textrm{Mn}^{2+}(aq)\textrm{ + 8 H}_{2}\textrm{O}(l)\textrm{ + 10 H}^{+}(aq)$

Notice that we have H₂O and H⁺ on both sides of the equation, so we need to simplify. Subtract the 5 H₂O on the reactant side from the 8 H₂O on the product side which leaves 3 H₂O on the reactant side. Then subtract the 10 H⁺ on the product side from the 16 H⁺ on the reactant side which leave 6 H⁺ on the reactant side. Now we have the simplified overall reaction

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College Chemistry : Balancing Redox Reactions

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