Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, what is the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K)?

We first write the anode reaction that actually takes place. Succinate ion is a doubly-charged anion, $$C_4H_4O_4^{2-}$$. In the Kolbe-type electro-oxidation a dicarboxylate loses both its carboxyl groups as carbon dioxide and the remainder of the skeleton couples intramolecularly to give an alkene. Hence, for one succinate ion we get ethylene and two carbon dioxide molecules:

$$C_4H_4O_4^{2-} \;\;\longrightarrow\;\; C_2H_4 + 2\,CO_2 + 2\,e^-$$

State the electron count: this oxidation liberates $$2$$ electrons per succinate ion.

Now we use the definition of a Faraday. One Faraday corresponds to one mole of electrons. The charge passed is

$$0.2\;\text{F} = 0.2\;\text{mol e}^-$$

Number of succinate ions oxidised:

$$n_{\text{succinate}} = \frac{0.2\;\text{mol e}^-}{2\;\text{mol e}^- \,/\,\text{mol succinate}} = 0.1\;\text{mol}$$

From the stoichiometry of the anode reaction:

• one ethylene per succinate  ⇒  $$n_{C_2H_4} = 0.1\;\text{mol}$$

• two carbon dioxides per succinate  ⇒  $$n_{CO_2} = 2 \times 0.1 = 0.2\;\text{mol}$$

So total moles of gas at the anode are

$$n_{\text{anode\,gases}} = 0.1 + 0.2 = 0.3\;\text{mol}$$

We next examine the cathode. In aqueous solution water is reduced:

$$2\,H_2O + 2\,e^- \;\;\longrightarrow\;\; H_2 + 2\,OH^-$$

This reaction consumes $$2$$ electrons per mole of hydrogen formed. The same $$0.2\;\text{mol e}^-$$ that left the anode must arrive at the cathode, so

$$n_{H_2} = \frac{0.2\;\text{mol e}^-}{2\;\text{mol e}^- \,/\,\text{mol }H_2} = 0.1\;\text{mol}$$

Total moles of gases produced (both electrodes):

$$n_{\text{total gas}} = n_{\text{anode\,gases}} + n_{H_2} = 0.3 + 0.1 = 0.4\;\text{mol}$$

At STP, one mole of any ideal gas occupies $$22.4\;\text{L}$$. Therefore the total volume is

$$V = 0.4\;\text{mol} \times 22.4\;\text{L mol}^{-1} = 8.96\;\text{L}$$

Hence, the correct answer is Option A.