How does varying the solution ph and sodium ethanoate concentration affect the rate and yield of ethane production from its electrolysis?

In the literature, various approaches for Kolbe electrolysis have been published. To calculate the yield of ethane, Kunugi (1951) employed titration to determine the final concentrations of ethanoate, methanol, and methanal solution. This was hazardous and impractical because it required using significant amounts of chromic acid (potassium dichromate (VI) and sulfuric acid) as a titrant for methanol and took two hours for each. However, a small-scale qualitative test for the presence of methanol following electrolysis can be performed using acidified potassium dichromate (VI) solution in the Jones oxidation (Clark, 2020).

 

3CH₃OH _(aq) + 2Cr₂O₇^2-_(aq) + 16H⁺_(aq) → 3HCOOH _(aq) + 4Cr³⁺_(aq) + 11H₂O _(l).

 

By oxidizing to the equivalent carboxylic acid, ethanoic acid, when heated, methanol transforms orange dichromate (VI) into green chromate (III).

Equipment and Glassware

Here, the kinetics of the reaction is discussed. The Kolbe reaction often occurs in the second order. (Atherton et al. 1966 and Ziogas et al. 2020). We may write the following for Kolbe electrolysis using the primary step from the previous sections.

 

\({𝑑[CH_3COO•] \over dt}\) = 𝑘₁[CH3COO⁻] − 𝑘₂[CH₃COO •]

 

The step is quick. Therefore, we may exclude it from the treatment and state the rate at which Ethane is formed due to the carboxylate radical.

 

\({𝑑[C_2H_6]\over dt}\)= 𝑘₃[CH₃COO •]²

 

Using the steady-state approximation that the concentration of the intermediate radical stays constant through the experiment, \({𝑑[CH_3COO•] \over dt}\)= 0

 

Using and [CH₃COO •] = \({k_1 \over k_2}\) [CH₃COO⁻]. So, the rate of ethane production is

 

\({𝑑[C_2H_6] \over dt}={k_3k_1 \over k_2}\) [CH₃COO⁻]²

 

The rate constant depends upon the voltage applied to drive the electrolytic reaction.

 

Due to the absence of information on the kinetics of the Hofer-Moest reaction, I will make an educated guess at a potential rate equation for the methanol production process.

 

\({𝑑[•OH] \over dt}\)= 𝑘₄[OH⁻] − 𝑘₅[• OH]

 

\({𝑑[CH_3OH] \over dt}\)= 𝑘₆[CH₃COO•][•OH]

 

Using the steady-state approximation, we can write.

 

\({𝑑[CH_3OH] \over dt} ={k_6k_4k_1 \over k_5k_2}\)[CH₃COO⁻][OH⁻]

 

With these results, we can hypothesize what the effect of pH will have on the rate of formation of Ethane and methanol.

 

The 1.0 mol dm^-3 ethanoic acid equation shows the link between pH and ethanoate concentration.

 

pH = 4.76 + log₁₀([CH₃COO⁻])  ∴ [CH₃COO⁻] = 10^(pH−4.76)

 

Combined with rate equation we have \({𝑑[C_2H_6] \over dt}={𝑘_3𝑘_1 \over k_2}\)\(\big[\)10^{(pH − 4.76)}\(\big]\)²

 

When sodium hydroxide is in excess, gives

 

pH = 14 + log₁₀([OH⁻]) ∴ [OH⁻] = 10^(pH−14)

 

Combined with rate equation  we have \({𝑑[CH_3OH] \over dt}\) = \({k_6k_4K_1 \over k_5k_2}\)10^{(pH − 4.76)} × 10^{(pH − 14)}

 

Using the simplifying assumption that the rate constants in and are the same, a sketch of this would result in Figure 7 on the following page. Furthermore, according to the rate equation, altering ethanoate concentration ought to have a quadratic relationship with the rate of ethane production.

Ethanoic acid has a p𝐾_𝑎 = 4.76, 𝐾_𝑎 = 1.74 ∙ 10⁻⁵, so it partially dissociates in water according to. Sodium hydroxide neutralizes ethanoic acid to form sodium ethanoate.

 

CH₃COOH _(aq) + OH^- _(aq) ⇌ CH₃COO^- _(aq) + H₂O _(l)

 

As ethanoate is the conjugate base of ethanoic acid, Henderson-Hasselbalch allows us to calculate the pH of the solution when the ethanoic acid is partially neutralized.

 

pH = p𝐾_𝑎 + log₁₀\(\big({[A^-]\over [HA]}\big)\)= 4.76 + log₁₀ \(\big({[CH_3COO^-]\over CH_3COOH}\big)\)

 

When the ethanoic acid is completely neutralized by hydroxide, the solution is weakly basic. Ethanoate reacts with water to release hydroxide ions.

 

CH₃COO^- _(aq) + H₂O _(l) ⇌ CH₃COOH _(aq) + OH^- _(aq)

 

𝐾_𝑏 = \({10^{-14}\over 1.74 ^{.}10^{-5}}\) = 5.71 ∙ 10⁻¹⁰  and

 

𝐾_𝑏 = \({[CH_3COOH][OH^-]\over [CH_3COO^−]}\)

 

But [CH₃COOH] = [OH⁻] and [CH₃COOH] ≪ [CH₃COO−] so 𝐾𝑏 = \({[OH^-]^2 \over [CH_3COO^-]}\)

 

[OH⁻] = \(\sqrt{𝐾_𝑏[CH^3COO^−]}\) ∴ pH = 14 + log₁₀ \((\sqrt{𝐾_𝑏[CH3COO^−]})\)

 

When sodium hydroxide is in excess, the pH derives entirely from the hydroxide ion concentration.

 

𝐾_𝑤 = [H₃O⁺][OH⁻] ∴ [H₃O⁺] = \({k_w \over [OH^-]}\) = \({10^{14} \over [OH^-]}\)

 

pH = − log₁₀([H₃O⁺]) ∴ pH = − log₁₀ \(\big({10^{-14} \over [OH^-]}\big)\) = 14 + log₁₀([OH⁻])

Fairweather (1926) used a direct method to examine the gaseous byproducts when electrolyzing propionate. A Bone and Wheeler apparatus was used to collect anode gas that had developed over mercury (Figure 8). Volume was calculated using the difference after CO₂ and oxygen were removed using potash (potassium hydroxide), pyrogallol, etc. Although better than water since gases are insoluble, this approach is inappropriate because it uses mercury and antiquated equipment.

 

Unfortunately, none of the Kolbe electrolysis procedures described in the literature were appropriate for the lab at the school. I chose to utilize a Hofmann apparatus with a variable power source for this experiment. This enables ongoing volume monitoring over time. Anode gas might be taken out and put through concentrated sodium hydroxide, like in Fairweather, to measure and collect the volume of ethane with more accuracy. As explained below, the gases could also be examined for methane and carbon dioxide.

 

• It is possible to create a hazy suspension of calcium carbonate through a CO₂ reaction with limewater (aqueous calcium hydroxide)

 

CO_{2 (g)} + Ca(OH)_{2 (aq)} → CaCO_{3 (s)} + H₂O _(l)

 

• It is possible to employ the free radical reaction of ethane with bromine in UV light (see reaction mechanism below). However, a control is required since ethene and other unsaturated hydrocarbons also interact electrophilically with bromine. Ethane exposed to UV light should discolour bromine water, while ethane not exposed to UV light should not.

 

• Br₂ −ℎ𝑣 → 2 Br•  } Initiation (UV light)
• Br• + CH₃CH₃ → •CH₂CH₃ + HBr
• •CH₂CH₃ + Br₂ → CH₃CH₂Br + Br• } Propagation
• 2 •CH₂CH₃ → CH₃(CH₂)₂CH₃
• 2 Br• → Br₂ }Termination
• Br• + •CH₂CH₃ → CH₃CH₂Br

 

The ideal method to establish the existence of ethane and any other small gaseous side products would have been gas-liquid chromatography, as described in Woolford, et al. (1966), but this was not available. This test will at least imply the existence of an alkane, even though it cannot prove the presence of methane.

 

The presence of a combustible gas can be further established by igniting the leftover gases that remained after absorption with carbon dioxide.

• Distilled water
• Sodium ethanoate trihydrate (NaCH₃COO • 3H₂O, Molar Mass = 136.08 g mol^-1)
• Ethanoic Acid (CH₃COOH, Molar Mass = 60.051 g mol^-1), Conc. = 2 mol dm^-3
• Sodium hydroxide solution (NaOH, Molar Mass = 40.00 g mol^-1), Conc. = 2 moldm-3
• Acidified Potassium dichromate (VI) solution, 0.2 mol dm^-3
• Bromine water, 3%
• Limewater, saturated
• Dilute Sodium thiosulfate solution (for neutralizing bromine water)

However, the Hofer-Moest reaction, also known as the non-Kolbe reaction, competes with the Kolbe process and produces side products like methane (HCHO) both methanol (CH₃OH). The Hofer-Moest reaction is favoured under alkaline conditions, while the Kolbe reaction is favoured under acidic conditions, according to Kunugi (1951) and Schafer (1979). Kluh et al. (2020) contend that the addition of an alkaline catalyst, such as sodium hydroxide, increases the rate of reaction because there is a higher concentration of the deprotonated carboxylate anion (CH₃COO^-) available for anodic oxidation to form ethane than there would be if ethanoic acid were to self-ionize, which would result in a much lower concentration. The reduced pH of the ethanoate solution will be explored as an independent variable in this experiment.

 

However, like with other electrolytic cells, altering the concentration of the dissolved ion will significantly impact the rate of electrolysis due to its impact on the solution's conductivity. Due to the dissolved ion's role as a reactant in the Kolbe reaction, its concentration will also affect the rate of electrolysis. The trials must also determine the minimum voltage needed for electrolysis. This serves as the basis for investigating how the speed and yield of ethane generation from its electrolysis are affected by modifying the solution pH and sodium ethanoate concentration.

As mentioned in the introduction, ethanoic acid can be electrolyzed either through the Hofer-Moest side reaction at the anode or the dominant Kolbe reaction.

 

• CH₃COOH _(aq) + H₂O _(l) ⇌ CH₃COO- (aq) + H₃O⁺
• CH₃COO^- _(aq) → CH3COO• _(adsorbed) + e^-
• CH3COO• _(adsorbed) → •CH3 _(adsorbed) + CO_{2 (g)}
• 2 •CH_{3 (adsorbed)} → C₂H₆ (g)

The Kolbe reaction has attracted attention recently because of its possible use in the industrial synthesis of hydrocarbons from more environmentally friendly sources, such as biomass. According to a theoretical process design study by Klüh et al. (2020), biomass-derived organic acids like pentanoic acid (CH₃(CH₂)₃COOH) and the saturated fatty acid dodecanoic acid (CH₃(CH₂)₁₀COOH) could be transformed into the alkane products n-octane (C₈H₁₈), a component of gasoline, and n-docosane (C₂₂H₄₆) via Kolbe To maximize faradic efficiency (the proportion of product yield to charge transfer in an electrochemical cell) and the potential outcome of the Kolbe product, Ziogass et al. (2020) describes the construction of a practical electrochemical reactor using the Kolbe method and investigate various variables, such as electrode material, electrode surface area, and theoretical energy requirements.

 

This study focuses on the Kolbe electrolysis of ethanoic acid, one of the most basic carboxylic acids (CH₃COOH). The cathodic and anodic reactions that take place are shown in Figure 2. As a result, there has recently been interest in ethanolic acid's potential production from biomass as a valuable feedstock (Deng et al., 2016). Ethanoic acid is an important industrial chemical and a natural product from the bacterial fermentation of carbohydrates (for example, vinegar) (Brown, 2020). Ethane (C₂H₆), the equivalent Kolbe product, is a significant component of natural gas and serves as a feedstock for the steam-cracking process used to produce ethene gas, which is used to make polyethene polymers and ethanol, among other things (Rogers, 2013). This reaction has excellent potential to replace the conventional use of nonrenewable resources.

 

Cathode (-) 2H⁺(aq) + 2e^- → H₂(g)

 

Anode (+) 2CH₃COO^-(aq) → C₂H₆(g) + 2CO₂(g) + 2e^-

 

Overall Reaction

Ethanoic acid is first deprotonated by water in a solution to produce the ethanoate anion and hydronium. The ethanoate radical (CH₃COO•) is created at the anode when an electron is taken from the ethanoate (CH3COO-). This radical quickly disproportionates to create a methyl radical (H₃C•) and carbon dioxide gas (CO₂). Ethane (C₂H₆) is produced due to biradical methyl coupling, which ends the reaction (4a). As well as (Sidney 1975) and (Klüh et al., 2020), This mechanism is displayed on the left in Figure 3.

 

The free radical species (CH₃COO•, H₃C•) are stabilized and adsorbed on the surface of the anode, such as platinum, according to Atherton et al. (1996). The subscript adsorbed in the reaction serves as a representation of this. The authors also suggest an alternative step (4b) in the ethane synthesis, shown in Figure 4, where the methyl radical attacks an ethanoate anion after dissolving from the anode to produce ethane, carbon dioxide, and a free electron.

 

 

•CH3 _(aq) + CH₃COO^-_(aq) → C₂H_{6 (g)} + CO₂ (g) + e^-

German chemist Hermann Kolbe initially described the Kolbe electrolysis, also known as the Kolbe reaction, as an organic electrochemical reaction in 1849. This discovery launched the science of organic electrosynthesis. This reaction occurs when two molecules of carboxylic acid (or carboxylate anions) decarboxylate dimerize to create a linked hydrocarbon product and carbon dioxide (Sidney et al., 1975). The reaction pathway for the Kolbe electrolysis of a carboxylic acid is shown in Figure 1 below. Since then, because of its capacity to "produce C-C bonds under exceptionally gentle conditions," the Kolbe reaction and its modifications have been studied as tools for organic synthesis. 2014 (Marko and Chelle). The authors define mild conditions as operating at room temperature and using harmless solvents like water and alcohol.

The overall new reaction remains the same as in the equation.

 

2CH₃COO^–_(aq) → C₂H_{6 (g)} + 2CO₂ (g) + 2e^– _(aq)

 

As was previously stated, the Hofer-Moest side reactions that produce byproducts like methanol challenge the dominant Kolbe mechanism. Overall, the response is as follows.

 

CH3COO^– _(aq) + OH^- _(aq) → CH₃OH _(g) + CO_{2 (g)} + 2e^– _(aq)

 

Two distinct methods for methanol production have been documented in the literature on this subject. The Hofer-Moest mechanism is more advantageous when the electrolysis occurs in austere conditions. Atherton et al. (1966) proposed the following stages based on their experimental data on the impact of hydroxide concentration on the rate of methanol formation.

 

• OH- _(aq) → •OH _(adsorbed) + e^-
• H₃C• _(adsorbed) + •OH _(adsorbed) → CH3OH _(l)
• H3C• _(aq) + OH^- _(aq) → CH₃OH _(l) + e^-

 

Steps (a) and (b) in the Kolbe reaction are identical for creating the methyl radical required to synthesize methanol. The hydroxyl radical is made when the hydroxide in the solution loses an electron at the anode. It then reacts with the methyl radical to produce methanol (a). Alternately, the methyl radical may directly interact with hydroxide after desorption into solution, producing methanol and liberating a free electron (8b).

We can create a pH curve from and that relates the amount of sodium hydroxide added to a solution of 1.0 mol dm^-3 ethanoic acid (Figure 5).

 

Using this information, with the mechanistic information on the

 

We can conclude that the yields of ethane (estimated by comparing it with ideal current efficiency) will be high up until a pH of around, at which point the concentration of hydroxide will rise noticeably, favouring the Hofer-Moest reaction more and more.