IChO 1997 · Theory problem 8

Electrodeposition of copper monolayers on Pt(100)

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An electrolyte is prepared from \(\ce{H2SO4}\), \(\ce{CuSO4}\) and distilled water and its volume is \(100.0\ \mathrm{cm^3}\). The concentrations of \(\ce{H+}\) and \(\ce{Cu^2+}\) in the electrolyte are \(c(\ce{H+}) = 1.000\ \mathrm{M}\) \((\mathrm{mol\ dm^{-3}})\) and \(c(\ce{Cu^2+}) = 1.000\times10^{-2}\ \mathrm{M}\) \((\mathrm{mol\ dm^{-3}})\), respectively. Two cubic platinum electrodes are immersed in the electrolyte. Both of the electrodes are single crystals with only one face (100) exposed to the electrolyte (the other five faces are blocked physically by an insulator which is stable in the electrolyte). The exposed surface area of each electrode is equal to \(1.000\ \mathrm{cm^2}\). During an electrolysis a total charge of \(2.0000\ \mathrm{C}\) is passed between the cathode and the anode. At the cathode, two simultaneous processes are occurring: deposition of an epitaxial (layer-by-layer) \(\ce{Cu}\) layer and \(\ce{H2}\) gas generation. At the anode, \(\ce{O2}\) gas is generated. The \(\ce{H2}\) gas is collected in a flask under the following conditions (assume ideal gas behaviour):

\(T = 273.15\ \mathrm{K}\) and \(P(\ce{H2}) = 1.01325\times10^4\ \mathrm{Pa}\); the volume of \(\ce{H2}\) is equal to \(2.0000\ \mathrm{cm^3}\)

8.1 Write equations of the processes taking place at the electrodes.

8.2 Calculate the number of moles of \(\ce{H2}\) gas generated at the cathode and the number of moles of \(\ce{Cu}\) deposited on the electrode.

8.3 Calculate the number of \(\ce{Cu}\) monolayers formed on the \(\ce{Pt}\) (100) cathode.

Note that the lattice constant of \(\ce{Pt}\) is \(a(\ce{Pt}) = 3.9236\times10^{-8}\ \mathrm{cm}\).

Both \(\ce{Pt}\) and \(\ce{Cu}\) have the fcc (face centred cubic) crystallographic structure.

Molar masses and constants:

\(M(\ce{H}) = 1.00795\ \mathrm{g\ mol^{-1}}\)

\(M(\ce{Cu}) = 63.546\ \mathrm{g\ mol^{-1}}\)

\(e = 1.60218\times10^{-19}\ \mathrm{C}\)

\(F = 96485.3\ \mathrm{C\ mol^{-1}}\)

\(R = 8.314510\ \mathrm{J\ K^{-1}\ mol^{-1}} = 0.0820584\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}\)

\(V_m = 22.4141\ \mathrm{dm^3}\)

\(1\ \mathrm{atm} = 1.01325\times10^5\ \mathrm{Pa}\)

\(N_{\mathrm A} = 6.02214\times10^{23}\ \mathrm{mol^{-1}}\)

Official English text of the problem.

Answers

  • 8.1 Cathode: \[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \] \[ \ce{2H+(aq) + 2e- -> H2(g)} \] Anode: \[ \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-} \] The molecular and ionic species are copper(II) ion [Cu+2], hydrogen ion [H+], molecular hydrogen [H][H], water O, and molecular oxygen O=O.

  • 8.2 \[ \boxed{n(\ce{H2})=8.9230\times10^{-6}\ \mathrm{mol}}, \qquad \boxed{n(\ce{Cu})=1.441\times10^{-6}\ \mathrm{mol}}. \]

  • 8.3 \[ \boxed{L=6.681\times10^2\ \text{monolayers}\approx668\ \text{monolayers}}. \]

Solution

8.1

Reduction occurs at the cathode. Copper(II) ions gain two electrons per ion, changing the copper oxidation state from \(+2\) to \(0\) and depositing copper metal:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)}. \]

Simultaneously, hydrogen ions are reduced to molecular hydrogen. Two hydrogen ions require two electrons to form one molecule:

\[ \ce{2H+(aq) + 2e- -> H2(g)}. \]

Oxidation occurs at the anode. Water supplies the oxygen evolved; oxygen changes from oxidation state \(-2\) in water to \(0\) in molecular oxygen. Balancing atoms and charge in the acidic electrolyte gives:

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

8.2

The stated hydrogen pressure is \(1.01325\times10^4\ \mathrm{Pa}\), corresponding to \(0.100000\ \mathrm{atm}\). Convert the gas volume to SI units:

\[ V=2.0000\ \mathrm{cm^3}=2.0000\times10^{-6}\ \mathrm{m^3}. \]

Using the ideal-gas equation,

\[ \begin{aligned} n(\ce{H2}) &=\frac{PV}{RT}\\ &=\frac{(1.01325\times10^4\ \mathrm{Pa}) (2.0000\times10^{-6}\ \mathrm{m^3})} {(8.314510\ \mathrm{J\ K^{-1}\ mol^{-1}}) (273.15\ \mathrm{K})}\\ &=8.9229558\times10^{-6}\ \mathrm{mol}. \end{aligned} \]

Here \(\mathrm{Pa\,m^3}=\mathrm{J}\), so the resulting unit is moles. Therefore,

\[ \boxed{n(\ce{H2})=8.9230\times10^{-6}\ \mathrm{mol}}. \]

Each mole of hydrogen and each mole of deposited copper consumes two moles of electrons. Consequently, the cathodic charge balance is

\[ Q=2F\,n(\ce{H2})+2F\,n(\ce{Cu}). \]

The anode carries the same total charge; its charge is not an additional contribution to this cathodic balance. Thus,

\[ \begin{aligned} n(\ce{Cu}) &=\frac{Q}{2F}-n(\ce{H2})\\ &=\frac{2.0000\ \mathrm{C}} {2(96485.3\ \mathrm{C\ mol^{-1}})} -8.9229558\times10^{-6}\ \mathrm{mol}\\ &=1.4413173\times10^{-6}\ \mathrm{mol}. \end{aligned} \]

Rounding after the subtraction gives

\[ \boxed{n(\ce{Cu})=1.441\times10^{-6}\ \mathrm{mol}}. \]

As a consistency check, the initial amount of copper(II) ions is

\[ (1.000\times10^{-2}\ \mathrm{mol\ dm^{-3}}) (0.1000\ \mathrm{dm^3}) =1.000\times10^{-3}\ \mathrm{mol}, \]

which is more than sufficient for this deposit.

8.3

In the epitaxial layer model of the problem, a copper monolayer occupies the surface sites of the \(\ce{Pt}\)(100) lattice.

Consider a square of side \(a\) on an fcc (100) face. It contains four corner atoms, each shared among four such squares, and one face-centre atom lying entirely within the square. The number of atoms per square is therefore

\[ 4\left(\frac14\right)+1=2. \]

Hence the surface atomic density is

\[ \sigma_{(100)} =\frac{2}{a^2} =\frac{2}{(3.9236\times10^{-8}\ \mathrm{cm})^2} =1.2991537\times10^{15}\ \mathrm{atoms\ cm^{-2}}. \]

Only the exposed cathode face receives the copper deposit. For its area \(A=1.000\ \mathrm{cm^2}\), one complete monolayer contains

\[ N_1=\sigma_{(100)}A =1.2991537\times10^{15}\ \text{atoms}. \]

The total number of deposited copper atoms, retaining the unrounded mole calculation, is

\[ \begin{aligned} N_{\mathrm{dep}} &=n(\ce{Cu})N_{\mathrm A}\\ &=(1.4413173\times10^{-6}\ \mathrm{mol}) (6.02214\times10^{23}\ \mathrm{mol^{-1}})\\ &=8.6798144\times10^{17}\ \text{atoms}. \end{aligned} \]

Thus the deposit expressed in monolayers is

\[ L=\frac{N_{\mathrm{dep}}}{N_1} =\frac{n(\ce{Cu})N_{\mathrm A}a^2}{2A} =668.11296. \]

Therefore,

\[ \boxed{L=6.681\times10^2\ \text{monolayers}\approx668\ \text{monolayers}}. \]

The fcc lattice and its (100) orientation describe an extended metallic crystal; molecular SMILES do not encode this surface geometry.

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.