IChO 2021 · Theory problem 4

The Redox Chemistry of Zinc

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Figure 1

Zinc has long been used as alloys for brass and steel materials. The zinc contained in industrial wastewater is separated by precipitation to detoxify the water, and the obtained precipitate is reduced to recover and reuse it as metallic zinc.

PART A

The dissolution equilibrium of zinc hydroxide \(\ce{Zn(OH)2(s)}\) at 25 °C and the relevant equilibrium constants are given in eq. 1–4.

Figure 2

The solubility, \(S\), of zinc (concentration of zinc in a saturated aqueous solution) is given in eq. 5.

\[ S=[\ce{Zn^2+(aq)}]+[\ce{Zn(OH)2(aq)}]+[\ce{Zn(OH)4^2-(aq)}] \tag{5} \]

4.1 When the equilibria in eq. 1–4 are established, calculate the pH range in which \([\ce{Zn(OH)2(aq)}]\) is the greatest among \([\ce{Zn^2+(aq)}]\), \([\ce{Zn(OH)2(aq)}]\) and \([\ce{Zn(OH)4^2-(aq)}]\).

4.2 A saturated aqueous solution of \(\ce{Zn(OH)2(s)}\) with pH = 7.00 was prepared and filtered. NaOH was added to this filtrate to increase its pH to 12.00. Calculate the molar percentage of zinc that precipitates when increasing the pH from 7.00 to 12.00. Ignore the volume and temperature changes.

PART B

Next, the recovered zinc hydroxide is heated to obtain zinc oxide according to the reaction below:

Figure 3

The zinc oxide is then reduced to metallic zinc by reaction with hydrogen:

Figure 4

4.3 In order for reaction (7) to proceed at a hydrogen pressure kept at 1 bar, it is necessary to reduce the partial pressure of the generated water vapor. Calculate the upper limit for the partial pressure of water vapor to allow reaction (7) to proceed at 300 °C. Here, the Gibbs formation energies of zinc oxide and water vapor at 300 °C and 1 bar for all gaseous species are \(\Delta G_{\ce{ZnO}}(300\,^\circ\mathrm{C})=-2.90\times10^2\ \mathrm{kJ\,mol^{-1}}\) and \(\Delta G_{\ce{H2O}}(300\,^\circ\mathrm{C})=-2.20\times10^2\ \mathrm{kJ\,mol^{-1}}\), respectively.

Metallic zinc is used as a negative electrode (anode) material for metal-air batteries. The electrode consists of \(\ce{Zn}\) and \(\ce{ZnO}\). It uses the following redox reaction to generate electricity with the electromotive force (e.m.f.) at 25 °C and pressure of 1 bar, \(E^\circ\).

Figure 5

4.4 A zinc–air battery was discharged at 20 mA for 24 hours. Calculate the change in mass of the negative electrode (anode) of the battery.

Figure 6

Mt. Fuji

4.5 Consider the change of e.m.f. of a zinc–air battery depending on the environment. Calculate the e.m.f. at the summit of Mt. Fuji, where the temperature and altitude are −38 °C (February) and 3776 m, respectively. The atmospheric pressure is represented by

\[ P\,[\mathrm{bar}] =1.013\times \left(1-\frac{0.0065h}{T+0.0065h+273.15}\right)^{5.257} \tag{9} \]

at altitude \(h\) [m] and temperature \(T\) [°C]. The molar ratio of oxygen in the atmosphere is 21%. The Gibbs energy change of reaction (8) is \(\Delta G_{\ce{ZnO}}(-38\,^\circ\mathrm{C})=-3.26\times10^2\ \mathrm{kJ\,mol^{-1}}\) at −38 °C and 1 bar.

4.6 Calculate the Gibbs energy change for reaction (6) at 25 °C. Note that the standard reduction potentials, \(E^\circ(\ce{Zn^2+/Zn})\) and \(E^\circ(\ce{O2/H2O})\) at 25 °C and 1 bar are given as (10) and (11), respectively.

Figure 7

Official English text of the problem.

Answers

  • 4.1: \(\boxed{8.41\leq\mathrm{pH}\leq11.80}\). The dissolved neutral species is strictly most abundant between the boundaries; it ties with another species at each boundary.
  • 4.2: \(\boxed{99.48\%}\) of the zinc precipitates.
  • 4.3: The limiting water-vapor pressure is \(\boxed{4.2\times10^{-7}\ \mathrm{bar}}\). Forward reduction requires a pressure below this limit.
  • 4.4: \(\boxed{\Delta m=+0.143\ \mathrm{g}}\): the anode gains mass.
  • 4.5: \(\boxed{E=1.68\ \mathrm{V}}\).
  • 4.6: \(\boxed{\Delta_{\mathrm r}G^\circ=+3.38\ \mathrm{kJ\,mol^{-1}}}\).

Solution

Use \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\), \(F=96485\ \mathrm{C\,mol^{-1}}\), and \(M(\ce{O})=16.00\ \mathrm{g\,mol^{-1}}\). Pure solids and liquid water have unit activity; gas activities are \(p/p^\circ\), with \(p^\circ=1\ \mathrm{bar}\).

Species names and SMILES used below are zinc(II) ion [Zn+2], hydroxide ion [OH-], hydrogen ion [H+], water O, hydrogen [H][H], oxygen O=O, and sodium hydroxide [Na+].[OH-]. Dissolved zinc hydroxide can be represented by the dative-bond SMILES [OH-]->[Zn+2]<-[OH-], and tetrahydroxozincate(II) by [OH-]->[Zn+2](<-[OH-])(<-[OH-])<-[OH-]. These representations omit solvent coordination and do not encode the tetrahedral coordination geometry of tetrahydroxozincate. The solid phases have extended structures rather than unique molecular SMILES.

4.1

Let \[ x=\frac{[\ce{OH-}]}{c^\circ}, \qquad c^\circ=1\ \mathrm{mol\,L^{-1}}. \]

The three dissolution equilibria give \[ \frac{[\ce{Zn^2+}]}{c^\circ}=\frac{K_{\mathrm{sp}}}{x^2}, \qquad \frac{[\ce{Zn(OH)2}]}{c^\circ}=K_1, \qquad \frac{[\ce{Zn(OH)4^2-}]}{c^\circ}=K_2x^2, \] where \[ K_{\mathrm{sp}}=1.74\times10^{-17},\quad K_1=2.62\times10^{-6},\quad K_2=6.47\times10^{-2}. \]

For the neutral dissolved species to have the greatest concentration, both conditions must hold: \[ K_1\geq\frac{K_{\mathrm{sp}}}{x^2}, \qquad K_1\geq K_2x^2. \]

Therefore, \[ \sqrt{\frac{K_{\mathrm{sp}}}{K_1}} \leq x\leq \sqrt{\frac{K_1}{K_2}}, \] or \[ 2.577\times10^{-6}\leq x\leq6.364\times10^{-3}. \]

Since \(K_{\mathrm w}=1.00\times10^{-14}\), \[ \mathrm{pH}=14.00+\log_{10}x. \]

Thus, \[ \boxed{8.41\leq\mathrm{pH}\leq11.80}. \]

At the lower boundary, \([\ce{Zn(OH)2}]=[\ce{Zn^2+}]\); at the upper boundary, \([\ce{Zn(OH)2}]=[\ce{Zn(OH)4^2-}]\). Strict dominance holds inside the interval.

4.2

The total dissolved zinc concentration at saturation is \[ S=c^\circ\left(\frac{K_{\mathrm{sp}}}{x^2}+K_1+K_2x^2\right). \]

At pH \(7.00\), \(x=10^{-7}\), so \[ \begin{aligned} S_7 &=\left( \frac{1.74\times10^{-17}}{10^{-14}} +2.62\times10^{-6} +(6.47\times10^{-2})(10^{-14}) \right)\mathrm{mol\,L^{-1}}\\ &=1.74262\times10^{-3}\ \mathrm{mol\,L^{-1}}. \end{aligned} \]

At pH \(12.00\), \(x=10^{-2}\), giving \[ \begin{aligned} S_{12} &=\left( \frac{1.74\times10^{-17}}{10^{-4}} +2.62\times10^{-6} +(6.47\times10^{-2})(10^{-4}) \right)\mathrm{mol\,L^{-1}}\\ &=9.09000\times10^{-6}\ \mathrm{mol\,L^{-1}}. \end{aligned} \]

Precipitation occurs principally through \[ \ce{Zn^2+(aq) + 2OH-(aq) -> Zn(OH)2(s)}. \]

Because the volume is unchanged, the fraction precipitated is the decrease in dissolved concentration divided by its initial value: \[ \begin{aligned} \%\text{ precipitated} &=100\left(1-\frac{S_{12}}{S_7}\right)\\ &=100\left(1-\frac{9.09000\times10^{-6}} {1.74262\times10^{-3}}\right)\\ &=\boxed{99.48\%}. \end{aligned} \]

This includes zinc remaining dissolved as tetrahydroxozincate at the final pH.

4.3

For \[ \ce{ZnO(s) + H2(g) -> Zn(s) + H2O(g)}, \] the standard Gibbs energy change at \(300\,^\circ\mathrm C\) is \[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=\Delta_{\mathrm f}G^\circ(\ce{H2O(g)}) -\Delta_{\mathrm f}G^\circ(\ce{ZnO(s)})\\ &=(-220)-(-290)\\ &=+70.0\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

The elemental substances have zero standard Gibbs energies of formation. The reaction quotient is \[ Q=\frac{p_{\ce{H2O}}/p^\circ}{p_{\ce{H2}}/p^\circ} =\frac{p_{\ce{H2O}}}{p_{\ce{H2}}}. \]

At the limiting pressure, \(\Delta_{\mathrm r}G=0\): \[ 0=\Delta_{\mathrm r}G^\circ+RT\ln Q. \]

Using \(T=573.15\ \mathrm K\) and \(p_{\ce{H2}}=1\ \mathrm{bar}\), \[ \begin{aligned} p_{\ce{H2O},\mathrm{lim}} &=(1\ \mathrm{bar}) \exp\left(-\frac{70000}{8.314(573.15)}\right)\\ &=4.17\times10^{-7}\ \mathrm{bar}. \end{aligned} \]

Hence forward reduction requires \[ \boxed{p_{\ce{H2O}}<4.2\times10^{-7}\ \mathrm{bar}}. \]

Equality corresponds to equilibrium.

4.4

The overall discharge reaction is \[ \ce{Zn(s) + 1/2 O2(g) -> ZnO(s)}. \]

Zinc changes oxidation state from \(0\) to \(+2\), so each mole of zinc converted transfers two moles of electrons.

The charge passed is \[ Q=It=(0.020\ \mathrm A)(24\times3600\ \mathrm s) =1728\ \mathrm C. \]

Thus, \[ n_{\ce{Zn},\mathrm{converted}} =\frac{Q}{2F} =\frac{1728}{2(96485)} =8.955\times10^{-3}\ \mathrm{mol}. \]

The electrode contains both zinc and zinc oxide. Zinc remains in the electrode, while one oxygen atom is incorporated per zinc atom oxidized. Therefore, \[ \begin{aligned} \Delta m &=n_{\ce{Zn},\mathrm{converted}} \left[M(\ce{ZnO})-M(\ce{Zn})\right]\\ &=n_{\ce{Zn},\mathrm{converted}}M(\ce{O})\\ &=(8.955\times10^{-3})(16.00)\\ &=\boxed{+0.143\ \mathrm g}. \end{aligned} \]

The anode gains mass.

4.5

First calculate the atmospheric pressure: \[ \begin{aligned} P &=1.013\left( 1-\frac{0.0065(3776)} {-38+0.0065(3776)+273.15} \right)^{5.257}\ \mathrm{bar}\\ &=0.6011\ \mathrm{bar}. \end{aligned} \]

The oxygen partial pressure is therefore \[ p_{\ce{O2}}=0.21P=0.1262\ \mathrm{bar}. \]

At \(-38\,^\circ\mathrm C\), \(T=235.15\ \mathrm K\). Obtain the standard e.m.f. at this temperature from the supplied Gibbs energy: \[ E^\circ(T) =-\frac{\Delta_{\mathrm r}G^\circ(T)}{2F} =\frac{326000}{2(96485)} =1.6894\ \mathrm V. \]

For \[ \ce{Zn(s) + 1/2 O2(g) -> ZnO(s)}, \] the reaction quotient is \[ Q=\left(\frac{p_{\ce{O2}}}{p^\circ}\right)^{-1/2}. \]

The Nernst equation consequently gives \[ \begin{aligned} E &=E^\circ(T)-\frac{RT}{2F}\ln Q\\ &=E^\circ(T)+\frac{RT}{4F} \ln\left(\frac{p_{\ce{O2}}}{p^\circ}\right)\\ &=1.6894+ \frac{8.314(235.15)}{4(96485)}\ln(0.1262)\\ &=1.6789\ \mathrm V. \end{aligned} \]

Thus, \[ \boxed{E=1.68\ \mathrm V}. \]

4.6

The target reaction is \[ \ce{Zn(OH)2(s) -> ZnO(s) + H2O(l)}. \]

Construct a Hess cycle at \(25\,^\circ\mathrm C\).

First, combine zinc oxidation with oxygen reduction: \[ \ce{Zn(s) -> Zn^2+(aq) + 2e-}, \] \[ \ce{1/2 O2(g) + 2H+(aq) + 2e- -> H2O(l)}. \]

Their sum is \[ \ce{Zn(s) + 1/2 O2(g) + 2H+(aq) -> Zn^2+(aq) + H2O(l)}. \qquad\text{(A)} \]

The cell potential and corresponding Gibbs energy are \[ E_A^\circ=1.23-(-0.77)=2.00\ \mathrm V, \] \[ \Delta G_A^\circ=-2FE_A^\circ =-385.94\ \mathrm{kJ\,mol^{-1}}. \]

Next, combine dissolution, \[ \ce{Zn(OH)2(s) <=> Zn^2+(aq) + 2OH-(aq)}, \] with neutralization, \[ \ce{2H+(aq) + 2OH-(aq) -> 2H2O(l)}. \]

The resulting reaction is \[ \ce{Zn(OH)2(s) + 2H+(aq) -> Zn^2+(aq) + 2H2O(l)}. \qquad\text{(B)} \]

Its equilibrium constant is \[ K_B=\frac{K_{\mathrm{sp}}}{K_{\mathrm w}^2} =\frac{1.74\times10^{-17}}{(1.00\times10^{-14})^2} =1.74\times10^{11}. \]

Therefore, \[ \begin{aligned} \Delta G_B^\circ &=-RT\ln K_B\\ &=-8.314(298.15)\ln(1.74\times10^{11})\\ &=-64.16\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Finally, reaction (8) is \[ \ce{Zn(s) + 1/2 O2(g) -> ZnO(s)}, \] with \[ \Delta G_8^\circ=-2F(1.65) =-318.40\ \mathrm{kJ\,mol^{-1}}. \]

Adding reaction (B), the reverse of reaction (A), and reaction (8) cancels the ions, oxygen, and metallic zinc, leaving exactly \[ \ce{Zn(OH)2(s) -> ZnO(s) + H2O(l)}. \]

Hence, \[ \begin{aligned} \Delta G_6^\circ &=\Delta G_B^\circ-\Delta G_A^\circ+\Delta G_8^\circ\\ &=-64.16-(-385.94)-318.40\\ &=\boxed{+3.38\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

The solution was checked against the official solution.