IChO 1992 · Theory problem 8

Pourbaix diagrams of water, nitrogen and manganese

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The Pourbaix diagrams for water, nitrogen and manganese are depicted in Fig. 1.

8.1 Write the formula of the species of nitrogen that is predominant

i) in \(\ce{O2}\)-rich lakes of \(\mathrm{pH}\approx 6\),

ii) in highly \(\ce{O2}\)-depleted lakes that are strongly contaminated with acid rain (\(\mathrm{pH}\approx 3\)),

8.2 Which species of manganese is predominant

i) in \(\ce{O2}\)-rich lakes of \(\mathrm{pH}\approx 6\),

ii) in highly \(\ce{O2}\)-depleted lakes that are strongly contaminated with bases (\(\mathrm{pH}\approx 12\))?

8.3 People often find that clear, slightly acidic (\(\mathrm{pH}\approx 5\)) water drawn from wells deposits a black manganese-containing solid on standing in toilet bowls.

i) Write the chemical formula of the black solid?

ii) Write the formula for the species of manganese found in well water while it is still underground ?

8.4 According to Pourbaix diagrams two species of nitrogen should oxidize \(\ce{Mn(s)}\) to \(\ce{Mn^2+(aq)}\).

i) Write the formulas of these two nitrogen species.

ii) Which of the two species of nitrogen does not oxidize \(\ce{Mn(s)}\) (in practice) at room temperature ?

8.5 According to the Poubaix diagrams, some of the chemical forms of manganese should oxidize \(\ce{NH3(aq)}\) or \(\ce{NH4+(aq)}\) to \(\ce{N2(g)}\).

Choose in the following list the forms of Mn that should do it:

\[ \ce{Mn},\ \ce{Mn(OH)2},\ \ce{Mn^2+},\ \ce{Mn3O4},\ \ce{Mn2O3},\ \ce{MnO2},\ \ce{MnO4^2-},\ \ce{MnO4^-} \]

8.6 Ammonium permanganate, \(\ce{NH4MnO4}\), is a well-known salt, but ammonium manganate, \(\ce{(NH4)2MnO4}\), is a rarely known salt.

i) Is \(\ce{NH4MnO4}\) expected to be thermodynamically stable? (YES or NO)

ii) Is \(\ce{(NH4)2MnO4}\) expected to be thermodynamically stable? (YES or NO)

iii) Write and balance an equation for the decomposition of \(\ce{NH4MnO4}\) to give \(\ce{MnO2}\) a \(\ce{N2}\).

iv) Write and balance an equation for the decomposition of \(\ce{(NH4)2MnO4}\) to give \(\ce{Mn}\) a \(\ce{N2}\).

8.7 According to the Pourbaix diagrams, is it dangerous to grind together in a mortar and pestle

i potassium nitrate and manganese metal,

ii) potassium nitrate and manganese dioxide?

8.8 The standard reduction potential, \(E^0\), for the reduction of \(\ce{MnO4^-}\) to \(\ce{MnO2}\) is \(1.692\ \mathrm{V}\). Applying the Nernst equation calculate the reduction potential, \(E\), for the reduction of \(0.00100\ \mathrm{M}\) \(\ce{MnO4^-}\) solution at a \(\mathrm{pH}=4.0\).

Figure 1

Figure 1

Official English text of the problem.

Answers

Names and SMILES are given at first identification. The manganese oxides, manganese hydroxide, and manganese metal are extended solids, so they have no discrete molecular SMILES representing their solid structures.

  • 8.1(i): Nitrate ion, \(\ce{NO3^-}\), O=[N+]([O-])[O-].
  • 8.1(ii): Ammonium ion, \(\ce{NH4+}\), [NH4+].
  • 8.2(i): Manganese(IV) oxide (manganese dioxide), \(\ce{MnO2(s)}\).
  • 8.2(ii): Manganese(II) hydroxide, \(\ce{Mn(OH)2(s)}\).
  • 8.3(i): Manganese dioxide, \(\ce{MnO2(s)}\).
  • 8.3(ii): Manganese(II) ion, \(\ce{Mn^2+(aq)}\), [Mn+2].
  • 8.4(i): Nitrate ion, \(\ce{NO3^-}\), and dinitrogen, \(\ce{N2}\), N#N.
  • 8.4(ii): Dinitrogen, \(\ce{N2}\).
  • 8.5: Manganese(II,III) oxide, \(\ce{Mn3O4}\); manganese(III) oxide, \(\ce{Mn2O3}\); manganese dioxide, \(\ce{MnO2}\); manganate(VI) ion, \(\ce{MnO4^2-}\), [O-][Mn](=O)(=O)[O-]; and permanganate ion, \(\ce{MnO4^-}\), [O-][Mn](=O)(=O)=O.
  • 8.6(i): NO.
  • 8.6(ii): NO.
  • 8.6(iii): \[ \boxed{\ce{2NH4MnO4 -> 2MnO2 + N2 + 4H2O}} \]
  • 8.6(iv): \[ \boxed{\ce{(NH4)2MnO4 -> Mn + N2 + 4H2O}} \]
  • 8.7(i): Yes.
  • 8.7(ii): No, according to the redox criterion of the diagrams.
  • 8.8: \(\boxed{E\approx 1.32\ \mathrm{V}}\) versus the standard hydrogen electrode, assuming \(25^\circ\mathrm{C}\).

Solution

A Pourbaix diagram shows the thermodynamically predominant species as a function of pH and reduction potential. Higher potentials favor oxidized forms; lower potentials favor reduced forms. For the lake questions, oxygen-rich conditions are read near the upper water-stability boundary, while highly oxygen-depleted conditions are interpreted as strongly reducing, near the lower boundary.

Water is \(\ce{H2O}\), SMILES O, and dioxygen is \(\ce{O2}\), SMILES O=O. The manganese oxyanion SMILES above specify formal connectivity and charges; both oxyanions have tetrahedral arrangements of oxygen atoms, which these strings do not depict spatially.

8.1

i) At \(\mathrm{pH}\approx6\), the oxygen/water boundary lies in the nitrate region of the nitrogen diagram. Thus the predominant nitrogen species under oxygen-rich conditions is

\[ \boxed{\ce{NO3^-(aq)}}. \]

ii) At \(\mathrm{pH}\approx3\), strongly reducing conditions lie below the boundary between nitrogen gas and ammonium. The acidic solution also favors ammonium over ammonia. Therefore,

\[ \boxed{\ce{NH4+(aq)}}. \]

8.2

i) At \(\mathrm{pH}\approx6\), the potential associated with oxygen-rich water falls within the manganese dioxide stability region:

\[ \boxed{\ce{MnO2(s)}}. \]

ii) At \(\mathrm{pH}\approx12\), the strongly reducing part of the water-stability region falls within the manganese(II) hydroxide field:

\[ \boxed{\ce{Mn(OH)2(s)}}. \]

Although manganese metal is stable at still lower potentials, its region lies below the lower water-stability boundary.

8.3

i) When the well water stands exposed to air, dissolved oxygen raises its redox potential. At \(\mathrm{pH}\approx5\), this favors insoluble, black manganese dioxide:

\[ \boxed{\ce{MnO2(s)}}. \]

ii) Underground, the water is oxygen-poor and has a lower redox potential. At this pH, dissolved manganese is predominantly

\[ \boxed{\ce{Mn^2+(aq)}}. \]

The precipitation upon aeration can be represented by

\[ \ce{2Mn^2+(aq) + O2(g) + 2H2O(l) -> 2MnO2(s) + 4H+(aq)}. \]

Here \(\ce{H+}\) is the hydrogen ion, SMILES [H+]; in water it represents a solvated proton.

8.4

i) A species can thermodynamically oxidize manganese metal when its reduction potential is above the \(\ce{Mn^2+/Mn}\) boundary at the same pH. Both nitrate and nitrogen gas satisfy this condition:

\[ \boxed{\ce{NO3^-}\ \text{and}\ \ce{N2}}. \]

Nitrate can accept electrons by reduction to a lower nitrogen oxidation state. Nitrogen gas can accept electrons by reduction to ammonium; for example, the corresponding thermodynamic reaction with manganese is

\[ \ce{3Mn(s) + N2(g) + 8H+(aq) -> 3Mn^2+(aq) + 2NH4+(aq)}. \]

ii) The species that does not appreciably oxidize manganese at room temperature is

\[ \boxed{\ce{N2}}. \]

Reduction of nitrogen requires activation of its strong triple bond and has a large kinetic barrier. The Pourbaix diagram predicts thermodynamic favorability, not reaction rate.

8.5

Nitrogen has oxidation state \(-3\) in both ammonia, \(\ce{NH3}\), SMILES N, and ammonium. Conversion to \(\ce{N2}\), where nitrogen has oxidation state \(0\), releases three electrons per nitrogen atom.

The manganese species must therefore accept electrons at a potential above the nitrogen/ammonia or nitrogen/ammonium boundary. Comparison of the diagrams selects

\[ \boxed{\ce{Mn3O4},\ \ce{Mn2O3},\ \ce{MnO2},\ \ce{MnO4^2-},\ \ce{MnO4^-}}. \]

These species contain manganese above oxidation state \(+2\), at least in part:

Species Manganese oxidation state
\(\ce{Mn3O4}\) One Mn(II) and two Mn(III) per formula unit
\(\ce{Mn2O3}\) \(+3\)
\(\ce{MnO2}\) \(+4\)
\(\ce{MnO4^2-}\) \(+6\)
\(\ce{MnO4^-}\) \(+7\)

Their relevant reduction boundaries lie above the nitrogen/\(\mathrm{N}(-III)\) boundary.

The remaining candidates do not qualify. Manganese metal is the reduced form, while reduction of \(\ce{Mn^2+}\) or \(\ce{Mn(OH)2}\) to metal requires potentials below the nitrogen/ammonia–ammonium boundary.

8.6

Ammonium permanganate has the constituent-ion SMILES [NH4+].[O-][Mn](=O)(=O)=O. Ammonium manganate has the constituent-ion SMILES [NH4+].[NH4+].[O-][Mn](=O)(=O)[O-]. These disconnected SMILES describe ionic composition, not crystal packing.

i) NO. The ammonium and permanganate stability regions do not overlap. Ammonium can be oxidized while permanganate is reduced, so internal redox decomposition is thermodynamically favored.

ii) NO. Ammonium and manganate likewise do not have a common stability region and are thermodynamically incompatible.

The existence of an isolable salt does not establish thermodynamic stability: an activation barrier can delay decomposition.

iii) In ammonium permanganate, nitrogen changes from \(-3\) to \(0\), losing three electrons per nitrogen atom. Manganese changes from \(+7\) to \(+4\), gaining three electrons per manganese atom. Two formula units supply the two nitrogen atoms needed for one nitrogen molecule:

\[ \boxed{\ce{2NH4MnO4 -> 2MnO2 + N2 + 4H2O}}. \]

Both sides contain two nitrogen atoms, eight hydrogen atoms, two manganese atoms, and eight oxygen atoms.

iv) In ammonium manganate, two nitrogen atoms each lose three electrons. The manganese atom gains all six electrons, changing from \(+6\) to \(0\). With the products specified in the question,

\[ \boxed{\ce{(NH4)2MnO4 -> Mn + N2 + 4H2O}}. \]

Both sides contain two nitrogen atoms, eight hydrogen atoms, one manganese atom, and four oxygen atoms.

8.7

Potassium nitrate, \(\ce{KNO3}\), has the constituent-ion SMILES [K+].[O-][N+](=O)[O-].

i) Yes. Nitrate is an oxidizing agent and manganese metal is a reducing agent. Their widely separated redox boundaries indicate a favorable redox reaction. Grinding brings them into intimate contact and can help initiate the reaction.

ii) No, according to the diagrams. Oxidizing manganese dioxide to manganate or permanganate requires a higher potential than nitrate reduction provides at the same pH. Thus the diagrams do not predict a favorable redox reaction between potassium nitrate and manganese dioxide.

8.8

First balance the reduction half-reaction:

\[ \ce{MnO4^-(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H2O(l)}. \]

Thus \(n=3\). Taking the activities of solid manganese dioxide and liquid water as unity,

\[ Q= \frac{1} {a_{\ce{MnO4^-}}\,a_{\ce{H+}}^{\,4}}. \]

Using the concentration approximation for permanganate and the definition of pH,

\[ a_{\ce{MnO4^-}}\approx 1.00\times10^{-3}, \qquad a_{\ce{H+}}=10^{-4.0}. \]

Therefore,

\[ Q=\frac{1}{(10^{-3})(10^{-4})^4}=10^{19}. \]

The Nernst equation at \(25^\circ\mathrm{C}\) is

\[ E=E^\circ-\frac{RT}{nF}\ln Q =E^\circ-\frac{0.05916\ \mathrm{V}}{n}\log_{10}Q. \]

Substitution gives

\[ \begin{aligned} E &=1.692\ \mathrm{V} -\frac{0.05916\ \mathrm{V}}{3}(19)\\ &=1.31732\ \mathrm{V}. \end{aligned} \]

Allowing for the precision of the supplied pH,

\[ \boxed{E\approx1.32\ \mathrm{V}}. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.