IChO 2023 · Theory problem 2

Electrochemical CO2 Reduction

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Electrochemical \(\ce{CO2}\) reduction

In recent years, the electrochemical conversion of \(\ce{CO2}\) into higher value products has been considered a promising and technologically feasible approach to mitigate the negative climatic effects caused by its increasing levels in the atmosphere. Several technologies have been developed to achieve this goal. Among these, \(\ce{CO2}\) reduction through electrochemical means (\(\ce{CO2}\)RR) warrants particular attention due to its ability to be powered by renewable energy to transform environmentally harmful \(\ce{CO2}\) into platform chemicals. Electrocatalysts are essential not only to accelerate the intrinsically slow \(\ce{CO2}\)RR but also to direct the electrolysis reaction towards the desired reaction products. In this context, it is not only the chemical nature of the catalyst itself which governs the resulting \(\ce{CO2}\)RR product distribution but also its morphological characteristics on various length scales. A new concept of \(\ce{CO2}\)RR catalyst design relies on the electrodeposition of foam-type materials, which offer a large surface area that is accessible to reactants (e.g. \(\ce{H2O}\), \(\ce{H2}\), and \(\ce{CO2}\)). Copper-based materials are the only known metallic \(\ce{CO2}\)RR catalysts that can produce hydrocarbons and alcohols in significant amounts by \(\ce{CO2}\) electrolysis.

Given below are thermodynamic data of selected substances:

\(\Delta_\mathrm{f}H^\ominus\) (\(\mathrm{kJ\,mol^{-1}}\)) \(S^\ominus\) (\(\mathrm{J\,K^{-1}\,mol^{-1}}\))
\(\ce{H2O(l)}\) −285.83 69.95
\(\ce{O2(g)}\) 0 205.15
\(\ce{H2(g)}\) 0 130.68
\(\ce{CO2(g)}\) −393.52 213.79
ethanol(l) −276.00 159.86
n-propanol(l) −302.54 192.80

Table 1. Standard formation enthalpy \(\Delta_\mathrm{f}H^\ominus\) and standard entropy \(S^\ominus\) for some substances under standard conditions (\(T=298.15\ \mathrm{K}\), \(p^\ominus=1\ \mathrm{bar}\))

Cell reaction \(E^\ominus\) vs. SHE (V)
\(\ce{Cu^2+ + 2e- = Cu}\) +0.34
\(\ce{2H+ + 2e- = H2}\) 0.00

Table 2. Selected half-cell reactions and standard potentials under standard conditions

2.1 Write and balance the chemical equation of the half-cell reactions for the following electrochemical reduction processes in acidic environment: (i) \(\ce{CO2}\) to ethanol; (ii) \(\ce{CO2}\) to n-propanol.

2.2 Combine the half-cell of the reduction process with an \(\ce{H2}/\ce{2H+}\) half-cell where the latter acts as anode. Calculate the value of the standard cell potential of the \(\ce{CO2}\) to ethanol reduction.

2.3 Write all the reduction and oxidation half-cell reactions taking place at the cathode and the anode, respectively.

Figure 1 shows the principle of dynamic hydrogen bubble-templated metal deposition.

Figure 2 displays top-down scanning electron microscopy images of five different \(\ce{Cu}\) foams obtained upon interruption of the metal deposition at different times: 5 s, 20 s, 80 s.

\(\ce{Cu}\) foam electrodeposition processes (see Figure 1) were carried out in an aqueous 1.5 M sulfuric acid solution containing 0.2 M copper sulfate (\(\ce{CuSO4 * 5H2O}\)) as the copper source. A \(\ce{Cu}\) disk (\(1\ \mathrm{cm^2}\)) and a \(\ce{Pt}\) foil served as the cathode and the anode, respectively.

Figure 1

Figure 1. Depiction of the electrodeposition of foam-type materials. In the course of electrochemical metal deposition (black arrows) at high current densities in aqueous media, the Hydrogen Evolution Reaction (HER) takes place on the metallic (\(\ce{Cu}\)) support (1). The surface becomes thus partially covered by \(\ce{H2}\) bubbles (2). The \(\ce{H2}\) bubbles act as a template for metal deposition (3). As a result a highly porous metal foam emerges (4)

Figure 2

Figure 2. \(\ce{Cu}\) foams obtained by galvanostatic \(\ce{Cu}\) electrodeposition at a current density of \(j=-3\ \mathrm{A\,cm^{-2}}\). The deposition was, in each case, interrupted at distinct elapsed duration: 5 s, 20 s and 80 s. The scale bar is the same in all panels

2.4 Considering this mechanism, assign the correct deposition time to the \(\ce{Cu}\) foams shown in Figure 2 on the answer sheet (white boxes upper left corner).

Bimetallic \(\ce{CuAg}\) systems are excellent catalysts for the electrosynthesis of alcohols from \(\ce{CO2}\). 5.4 mg of a bimetallic \(\ce{Cu}\)-\(\ce{Ag}\) foam catalyst (90 wt. % \(\ce{Cu}\), \(M_{\ce{Cu}}=63.546\ \mathrm{g\,mol^{-1}}\); 10 wt. % \(\ce{Ag}\), \(M_{\ce{Ag}}=107.868\ \mathrm{g\,mol^{-1}}\)) was galvanostatically deposited onto a \(\ce{Cu}\) foil (\(1\ \mathrm{cm^2}\)) at a current density of \(j=-3\ \mathrm{A\,cm^{-2}}\) applied for 20 s (the minus sign accounts for a reductive/cathodic process).

2.5 Calculate the Faradaic efficiency (FE in %) of this metal deposition process. FE is defined as \((Q_\mathrm{product}/Q_\mathrm{total})\times100\%\), where \(Q\) denotes the charge.

We consider a \(\ce{CO2}\) electrolysis experiment carried out in 35 mL \(\ce{CO2}\)-saturated 0.5 M \(\ce{KHCO3}\) electrolyte solution over the bimetallic \(\ce{Cu}\)-\(\ce{Ag}\) foam catalyst (90 wt. % \(\ce{Cu}\); 10 wt. % \(\ce{Ag}\)). The \(\ce{CO2}\) electrolysis was carried out at a constant (total) current density of \(j(\mathrm{tot})=-30\ \mathrm{mA\,cm^{-2}}\) for 3600 s (note the current density is normalized to the geometric surface area of \(1\ \mathrm{cm^2}\); the minus sign accounts for a reductive/cathodic process). A product analysis, carried out after the electrolysis, revealed mass concentrations of \(41.3\ \mathrm{mg\,L^{-1}}\) and \(7.4\ \mathrm{mg\,L^{-1}}\) for ethanol and n-propanol, respectively. Both alcohols are liquid reaction products and accumulate in the electrolyte in the course of the electrolysis reaction. We assume that gaseous hydrogen (\(\ce{H2}\)) is formed as the only by-product of the process.

2.6 Calculate the current densities required for the formation of (a) ethanol (\(M_\mathrm{EtOH}=46.08\ \mathrm{g\,mol^{-1}}\)) and (b) n-propanol (\(M_\mathrm{PrOH}=60.10\ \mathrm{g\,mol^{-1}}\)) assuming that the current densities do not change with electrolysis time.

2.7 Calculate the volume of the formed hydrogen on the \(1\ \mathrm{cm^2}\) catalyst area at 298.15 K and 1 bar, assuming ideal behavior of the formed hydrogen, and its complete release into the gas phase. If you did not get a result in Task 2.6, continue with FE(EtOH) = 45,1 % and FE(PrOH) = 4.8 %.

Official English text of the problem.

Answers

  • 2.1 (i) Ethanol, CCO: \[ \ce{2CO2(g) + 12H+(aq) + 12e- -> CH3CH2OH(l) + 3H2O(l)} \] (ii) n-Propanol (propan-1-ol), CCCO: \[ \ce{3CO2(g) + 18H+(aq) + 18e- -> CH3CH2CH2OH(l) + 5H2O(l)} \]

  • 2.2 The combined cell reaction is \[ \ce{2CO2(g) + 6H2(g) -> CH3CH2OH(l) + 3H2O(l)}, \qquad \boxed{E^\ominus_\mathrm{cell}=+0.0824\ \mathrm{V}}. \]

  • 2.3 Cathode: \[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \] \[ \ce{2H+(aq) + 2e- -> H2(g)} \] Anode: \[ \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-} \]

  • 2.4 Fill the white boxes from left to right with 80 s, 5 s, 20 s.

  • 2.5 \(\boxed{\mathrm{FE}_\mathrm{metal}=25.4\%}\).

  • 2.6 (a) \(\boxed{j_\mathrm{EtOH}=-10.1\ \mathrm{mA\,cm^{-2}}}\); (b) \(\boxed{j_\mathrm{PrOH}=-2.08\ \mathrm{mA\,cm^{-2}}}\).

  • 2.7 Hydrogen, [H][H]: \(\boxed{V_{\ce{H2}}=8.25\ \mathrm{mL}}\), using the results calculated in 2.6.

Solution

The molecular and ionic species used in the solution have the following names and SMILES. Physical states and electron transfer are specified in the reaction equations.

Species SMILES
Carbon dioxide O=C=O
Water O
Hydrogen [H][H]
Oxygen O=O
Ethanol CCO
n-Propanol (propan-1-ol) CCCO
Hydrogen ion (proton) [H+]
Copper(II) ion [Cu+2]
Silver(I) ion [Ag+]

The attached pages do not list numerical physical constants. I use \[ F=96485\ \mathrm{C\,mol^{-1}}, \qquad R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}. \]

2.1 — Reduction half-reactions

For ethanol, two carbon dioxide molecules supply the two carbon atoms. Balancing oxygen requires three water molecules on the product side. The products then contain twelve hydrogen atoms, requiring twelve protons on the reactant side. Twelve electrons balance the charge:

\[ \boxed{\ce{2CO2(g) + 12H+(aq) + 12e- -> CH3CH2OH(l) + 3H2O(l)}} \]

For n-propanol, three carbon dioxide molecules supply the three carbon atoms. Five water molecules balance oxygen, eighteen protons balance hydrogen, and eighteen electrons balance charge:

\[ \boxed{\ce{3CO2(g) + 18H+(aq) + 18e- -> CH3CH2CH2OH(l) + 5H2O(l)}} \]

Thus the electron requirements are \(12\ \mathrm{mol}\) of electrons per mole of ethanol and \(18\ \mathrm{mol}\) per mole of n-propanol.

2.2 — Standard cell potential

Hydrogen undergoes oxidation at the anode:

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

Multiplying this equation by six and adding it to the ethanol half-reaction gives

\[ \ce{2CO2(g) + 6H2(g) -> CH3CH2OH(l) + 3H2O(l)}. \]

For this reaction, the standard reaction enthalpy is

\[ \begin{aligned} \Delta_\mathrm{r}H^\ominus &=\sum\nu\Delta_\mathrm{f}H^\ominus_\mathrm{products} -\sum\nu\Delta_\mathrm{f}H^\ominus_\mathrm{reactants}\\ &=[-276.00+3(-285.83)]-[2(-393.52)+6(0)]\\ &=-346.45\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

The standard reaction entropy is

\[ \begin{aligned} \Delta_\mathrm{r}S^\ominus &=[159.86+3(69.95)]-[2(213.79)+6(130.68)]\\ &=-841.95\ \mathrm{J\,K^{-1}\,mol^{-1}}\\ &=-0.84195\ \mathrm{kJ\,K^{-1}\,mol^{-1}}. \end{aligned} \]

At \(298.15\ \mathrm{K}\),

\[ \begin{aligned} \Delta_\mathrm{r}G^\ominus &=\Delta_\mathrm{r}H^\ominus-T\Delta_\mathrm{r}S^\ominus\\ &=-346.45-298.15(-0.84195)\\ &=-95.4226\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

The cell reaction transfers twelve electrons, so

\[ E^\ominus_\mathrm{cell} =-\frac{\Delta_\mathrm{r}G^\ominus}{12F} =\frac{95422.6\ \mathrm{J\,mol^{-1}}} {12(96485\ \mathrm{C\,mol^{-1}})} =\boxed{+0.0824\ \mathrm{V}}. \]

Since the standard hydrogen electrode has a reduction potential of \(0.00\ \mathrm{V}\), this is also the standard reduction potential of the carbon dioxide/ethanol half-cell versus SHE.

2.3 — Electrode reactions during copper-foam deposition

This part concerns the copper-foam electrodeposition described immediately after the question.

At the cathode, copper(II) ions accept electrons and deposit as metallic copper:

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

Protons are also reduced, producing the hydrogen bubbles that template the pores:

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

At the platinum anode, water is oxidized to oxygen:

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

The platinum foil serves as the conducting anode without being consumed in this reaction.

2.4 — Assignment of deposition times

Copper deposits around hydrogen bubbles, leaving pores when the bubbles escape. As deposition proceeds, bubble growth and coalescence produce larger pores.

Because all three panels have the same scale bar, their pore sizes can be compared directly:

  • The left panel has the largest pores: 80 s.
  • The middle panel has the smallest pores: 5 s.
  • The right panel has intermediate pore sizes: 20 s.

Therefore, the white boxes should read

\[ \boxed{\text{Left: }80\ \mathrm{s}\qquad \text{Middle: }5\ \mathrm{s}\qquad \text{Right: }20\ \mathrm{s}}. \]

2.5 — Faradaic efficiency of metal deposition

Use positive charge magnitudes when calculating Faradaic efficiency. The total charge passed is

\[ Q_\mathrm{total}=|j|At =(3\ \mathrm{A\,cm^{-2}})(1\ \mathrm{cm^2})(20\ \mathrm{s}) =60\ \mathrm{C}. \]

The deposited metal masses are

\[ m_{\ce{Cu}}=0.90(5.4\ \mathrm{mg})=4.86\ \mathrm{mg}, \] \[ m_{\ce{Ag}}=0.10(5.4\ \mathrm{mg})=0.54\ \mathrm{mg}. \]

Their amounts are

\[ n_{\ce{Cu}} =\frac{4.86\times10^{-3}\ \mathrm{g}} {63.546\ \mathrm{g\,mol^{-1}}} =7.6480\times10^{-5}\ \mathrm{mol}, \]

\[ n_{\ce{Ag}} =\frac{0.54\times10^{-3}\ \mathrm{g}} {107.868\ \mathrm{g\,mol^{-1}}} =5.0061\times10^{-6}\ \mathrm{mol}. \]

Copper deposition requires two electrons per atom. Taking silver to be deposited from silver(I), its deposition requires one:

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

Hence

\[ \begin{aligned} Q_\mathrm{metal} &=F(2n_{\ce{Cu}}+n_{\ce{Ag}})\\ &=(96485)\left[2(7.6480\times10^{-5})+5.0061\times10^{-6}\right]\\ &=15.2414\ \mathrm{C}. \end{aligned} \]

Therefore,

\[ \boxed{ \mathrm{FE}_\mathrm{metal} =\frac{15.2414}{60}\times100\% =25.4\% }. \]

The competing hydrogen-evolution reaction accounts for the low efficiency of metal deposition in the bubble-templating process.

2.6 — Partial current densities for alcohol formation

The electrolyte volume is \(0.035\ \mathrm{L}\). For a product requiring \(z\) electrons per molecule,

\[ n=\frac{c_mV_\mathrm{el}}{M}, \qquad Q=zFn, \qquad j=-\frac{Q}{At}. \]

The negative sign denotes a cathodic current. The relevant area is the stated geometric area, \(A=1\ \mathrm{cm^2}\).

(a) Ethanol

The amount produced is

\[ \begin{aligned} n_\mathrm{EtOH} &=\frac{(41.3\times10^{-3}\ \mathrm{g\,L^{-1}}) (0.035\ \mathrm{L})} {46.08\ \mathrm{g\,mol^{-1}}}\\ &=3.13694\times10^{-5}\ \mathrm{mol}. \end{aligned} \]

Using \(z=12\),

\[ Q_\mathrm{EtOH} =12F n_\mathrm{EtOH} =36.3201\ \mathrm{C}. \]

Thus

\[ j_\mathrm{EtOH} =-\frac{36.3201\ \mathrm{C}} {(1\ \mathrm{cm^2})(3600\ \mathrm{s})} =-0.0100889\ \mathrm{A\,cm^{-2}}, \]

\[ \boxed{j_\mathrm{EtOH}=-10.1\ \mathrm{mA\,cm^{-2}}}. \]

(b) \(\boldsymbol{n}\)-Propanol

The amount produced is

\[ \begin{aligned} n_\mathrm{PrOH} &=\frac{(7.4\times10^{-3}\ \mathrm{g\,L^{-1}}) (0.035\ \mathrm{L})} {60.10\ \mathrm{g\,mol^{-1}}}\\ &=4.30948\times10^{-6}\ \mathrm{mol}. \end{aligned} \]

Using \(z=18\),

\[ Q_\mathrm{PrOH} =18F n_\mathrm{PrOH} =7.48441\ \mathrm{C}. \]

Therefore,

\[ j_\mathrm{PrOH} =-\frac{7.48441\ \mathrm{C}} {(1\ \mathrm{cm^2})(3600\ \mathrm{s})} =-0.00207900\ \mathrm{A\,cm^{-2}}, \]

\[ \boxed{j_\mathrm{PrOH}=-2.08\ \mathrm{mA\,cm^{-2}}} \]

(or \(-2.1\ \mathrm{mA\,cm^{-2}}\) to two significant figures).

The corresponding Faradaic efficiencies, retaining unrounded current values, are

\[ \mathrm{FE}_\mathrm{EtOH} =\frac{10.0889}{30}\times100\% =33.63\%, \]

\[ \mathrm{FE}_\mathrm{PrOH} =\frac{2.07900}{30}\times100\% =6.930\%. \]

2.7 — Hydrogen volume

The total electrolysis charge is

\[ Q_\mathrm{total} =(0.030\ \mathrm{A\,cm^{-2}}) (1\ \mathrm{cm^2})(3600\ \mathrm{s}) =108\ \mathrm{C}. \]

Since hydrogen is the only by-product, its formation accounts for all charge remaining after alcohol formation:

\[ \begin{aligned} Q_{\ce{H2}} &=Q_\mathrm{total}-Q_\mathrm{EtOH}-Q_\mathrm{PrOH}\\ &=108-36.3201-7.48441\\ &=64.1955\ \mathrm{C}. \end{aligned} \]

Formation of one hydrogen molecule requires two electrons. Consequently,

\[ n_{\ce{H2}} =\frac{Q_{\ce{H2}}}{2F} =\frac{64.1955}{2(96485)} =3.32671\times10^{-4}\ \mathrm{mol}. \]

Using the ideal-gas law at \(p=1\ \mathrm{bar}=10^5\ \mathrm{Pa}\),

\[ \begin{aligned} V_{\ce{H2}} &=\frac{n_{\ce{H2}}RT}{p}\\ &=\frac{(3.32671\times10^{-4}\ \mathrm{mol}) (8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) (298.15\ \mathrm{K})} {10^5\ \mathrm{Pa}}\\ &=8.24631\times10^{-6}\ \mathrm{m^3}. \end{aligned} \]

Therefore,

\[ \boxed{V_{\ce{H2}}=8.25\ \mathrm{mL}}. \]

Using the question’s optional fallback efficiencies would instead give \[ \mathrm{FE}_{\ce{H2}}=100\%-45.1\%-4.8\%=50.1\% \] and \(V_{\ce{H2}}=6.95\ \mathrm{mL}\). The main answer uses the efficiencies calculated directly from the stated product concentrations.

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.