Problem 2

The reaction of permanganate ions with hydrogen peroxide in acidic solution yields Mn(II) and releases oxygen gas. Four unbalanced schemes are proposed: (1)

2\ce{MnO4-} + 1\ce{H2O2} + 6\ce{H+} -> 2\ce{Mn^{2+}} + 3\ce{O2} + 4\ce{H2O}

; (2)

2\ce{MnO4-} + 3\ce{H2O2} + 6\ce{H+} -> 2\ce{Mn^{2+}} + 4\ce{O2} + 6\ce{H2O}

; (3)

2\ce{MnO4-} + 5\ce{H2O2} + 6\ce{H+} -> 2\ce{Mn^{2+}} + 5\ce{O2} + 8\ce{H2O}

; (4)

2\ce{MnO4-} + 7\ce{H2O2} + 6\ce{H+} -> 2\ce{Mn^{2+}} + 6\ce{O2} + 10\ce{H2O}

. (a) Which of the above represents the actual reaction? Explain using electron transfer. (b) Identify the oxidising and reducing agents. (c) How much potassium permanganate (in g) is needed to liberate 112 cm

^3

of oxygen at STP from an excess of

\ce{H2O2}

in acidic solution? (

M(\ce{KMnO4}) = 158.04

g mol

^{-1}

)

Step 1 of 3: Electron balance selects equation 3

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\ce{Mn^{VII} + 5e- -> Mn^{II}}, \quad (\ce{O2})^{-II} - 2e^- -> \ce{O2^0} \Rightarrow 2\ce{Mn^{VII}} + 5(\ce{O2})^{-II} -> 2\ce{Mn^{II}} + 5\ce{O2^0}

Analysis

Manganese is reduced from +7 to +2 (accepts 5 electrons); the peroxo oxygen atoms are oxidised from -1 to 0 (donate 2 electrons per molecule of $\ce{H2O2}$ ). The lowest common multiple of 5 and 2 is 10, requiring 2 $\ce{MnO4-}$ per 5 $\ce{H2O2}$ — only equation 3 is chemically valid.