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 3 of 3: Calculate KMnO4 needed

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n(\ce{O2}) = \dfrac{0.112}{22.4} = 0.0050\ \text{mol} \Rightarrow n(\ce{KMnO4}) = \dfrac{2}{5} \times 0.0050 = 0.0020\ \text{mol}

Analysis

From the 2 : 5 mole ratio, 0.0050 mol $\ce{O2}$ requires $0.0020$ mol of $\ce{KMnO4}$ . Mass: $m = 0.0020 \times 158.04 = 0.316$ g.

Common pitfall. Equations (1), (2), and (4) may appear atom-balanced on paper, but they violate electron conservation for the known redox couples.