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

Question

A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________

Answer 1

To solve the problem of determining the total volume of gases produced during the oxidation of succinate ion, we need to understand the reactions and apply Faraday's laws of electrolysis.

The reaction for the oxidation of succinate ion can be represented as:

C_4H_4O_4^{2-} \rightarrow C_2H_4 + 2CO_2 + 2e^-\

This indicates that one succinate ion produces one molecule of ethylene gas ($C_2H_4$) and two molecules of carbon dioxide gas ($CO_2$), releasing two electrons in the process.

Since 1 Faraday of electricity corresponds to 1 mole of electrons, passing 0.2 Faraday results in the oxidation of:

0.1 \, \text{mol of} \, C_4H_4O_4^{2-}

This indicates:

0.1 mole of $C_2H_4$
0.2 moles of $CO_2$

Next, we calculate the total volume of gases at STP. At STP, 1 mole of gas occupies 22.4 liters.

Therefore, the total volume is calculated as follows:

Volume of $C_2H_4$: 0.1 \, \text{mol} \times 22.4 \, \text{L/mol} = 2.24 \, \text{L}
Volume of $CO_2$: 0.2 \, \text{mol} \times 22.4 \, \text{L/mol} = 4.48 \, \text{L}

Adding these volumes gives the total gas volume produced at the anode:

2.24 \, \text{L} + 4.48 \, \text{L} = 6.72 \, \text{L}

At the cathode, due to water reduction, we assume hydrogen production:

2H_2O + 2e^- \rightarrow H_2 + 2OH^-\

Amount of $H_2$ formed (0.2 Faraday, 0.1 mole of $H_2$):

0.1 \, \text{mol} \times 22.4 \, \text{L/mol} = 2.24 \, \text{L}

The total volume of gases at both the cathode and anode is:

6.72 \, \text{L} + 2.24 \, \text{L} = 8.96 \, \text{L}

Thus, the correct answer is: 8.96 \, \text{L}.

Answer 2