Chemistry Labs

International Chemistry Olympiad · 1977

Problems

  1. Problem 2The reaction of permanganate ions with hydrogen peroxide in acidic solution yields Mn(II) and releases oxygen gas. Four unbalanced schemes are proposed: (1) 2MnOX4X−+1HX2OX2+6HX+−>2MnX2++3OX2+4HX2O2\ce{MnO4-} + 1\ce{H2O2} + 6\ce{H+} -> 2\ce{Mn^{2+}} + 3\ce{O2} + 4\ce{H2O}; (2) 2MnOX4X−+3HX2OX2+6HX+−>2MnX2++4OX2+6HX2O2\ce{MnO4-} + 3\ce{H2O2} + 6\ce{H+} -> 2\ce{Mn^{2+}} + 4\ce{O2} + 6\ce{H2O}; (3) 2MnOX4X−+5HX2OX2+6HX+−>2MnX2++5OX2+8HX2O2\ce{MnO4-} + 5\ce{H2O2} + 6\ce{H+} -> 2\ce{Mn^{2+}} + 5\ce{O2} + 8\ce{H2O}; (4) 2MnOX4X−+7HX2OX2+6HX+−>2MnX2++6OX2+10HX2O2\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 cm3^3 of oxygen at STP from an excess of HX2OX2\ce{H2O2} in acidic solution? (M(KMnOX4)=158.04M(\ce{KMnO4}) = 158.04 g mol−1^{-1})Solutions: 1