IChO 1977 · Theory problem 2

Permanganate and hydrogen peroxide

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The reaction of permanganate ions with hydrogen peroxide in an acidic solution gives Mn(II) salt and at the same time oxygen is released:

Figure 1

Problems:

  1. The possible ratios of the reactants in the above equations express:

a) all equations ☐
b) only some of the equations ☐
c) only one equation ☐
d) none ☐

Indicate the correct assertion by a cross in the corresponding square and explain your decision.

  1. Which of the reactants is an oxidising agent and which is a reducing one?

  2. How much potassium permanganate is needed to release 112 \(\mathrm{cm^3}\) of oxygen at STP conditions from an excess of hydrogen peroxide in acidic solution?

Official English text of the problem.

Answers

  1. c) only one equation: the third equation. Place a cross in the square beside c). The reacting mole ratio is \[ n(\ce{MnO4^-}):n(\ce{H2O2})=2:5. \]

  2. Oxidising agent: permanganate ion, \(\ce{MnO4^-}\), [O-][Mn](=O)(=O)=O.
    Reducing agent: hydrogen peroxide, \(\ce{H2O2}\), OO.

  3. \(0.316\ \mathrm{g}\) of potassium permanganate, \(\ce{KMnO4}\), [K+].[O-][Mn](=O)(=O)=O.

Solution

The other species used below are hydrogen ion, \(\ce{H+}\), [H+]; manganese(II) ion, \(\ce{Mn^2+}\), [Mn+2]; dioxygen, \(\ce{O2}\), O=O; and water, \(\ce{H2O}\), O. Potassium permanganate contains potassium ions, \(\ce{K+}\), [K+], and permanganate ions. Permanganate is tetrahedral; its SMILES gives a formal bond representation rather than the geometry or resonance delocalisation.

1. Selecting the reaction equation

All four proposed equations balance atoms and total charge. To identify the equation describing the stated redox reaction, we must consider the oxidation and reduction processes.

Manganese changes from oxidation state \(+7\) in permanganate to \(+2\), accepting five electrons per manganese atom. Its reduction half-reaction in acidic solution is

\[ \ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O}. \]

Hydrogen peroxide is oxidised to dioxygen. Each oxygen atom changes from oxidation state \(-1\) to \(0\), so each peroxide molecule supplies two electrons:

\[ \ce{H2O2 -> O2 + 2H+ + 2e^-}. \]

Multiplying the reduction half-reaction by two and the oxidation half-reaction by five transfers ten electrons in each direction. Adding and cancelling electrons and excess hydrogen ions gives

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

Thus the third equation expresses the stoichiometry of this redox reaction, and the correct choice is c).

2. Identifying the oxidising and reducing agents

  • Permanganate is the oxidising agent: it accepts electrons and is itself reduced, with manganese changing from \(+7\) to \(+2\).
  • Hydrogen peroxide is the reducing agent: it supplies electrons and is itself oxidised, with oxygen changing from \(-1\) to \(0\).

Hydrogen ions provide the acidic medium; their oxidation state does not change.

3. Calculating the required potassium permanganate

Using the conventional STP molar gas volume,

\[ V_m=22.4\ \mathrm{dm^3\,mol^{-1}}, \]

the amount of oxygen is

\[ n(\ce{O2}) =\frac{112\times10^{-3}\ \mathrm{dm^3}} {22.4\ \mathrm{dm^3\,mol^{-1}}} =5.00\times10^{-3}\ \mathrm{mol}. \]

The balanced equation produces five moles of oxygen for every two moles of permanganate. Each mole of potassium permanganate supplies one mole of permanganate ions, so

\[ n(\ce{KMnO4}) =\frac{2}{5}n(\ce{O2}) =2.00\times10^{-3}\ \mathrm{mol}. \]

Using rounded standard atomic masses,

\[ M(\ce{KMnO4}) =39.1+54.9+4(16.0) =158.0\ \mathrm{g\,mol^{-1}}. \]

Therefore,

\[ m(\ce{KMnO4}) =(2.00\times10^{-3}\ \mathrm{mol}) (158.0\ \mathrm{g\,mol^{-1}}) =\boxed{0.316\ \mathrm{g}}. \]

The solution was checked against the official solution.