By the electrolysis of aqueous sodium succinate, ethene is formed at ………..along with ………..A. Anode, { H }_{ 2 }B. Cathode, { H }_{ 2 }C. Anode, { CO }_{ 2 }D. Cathode, { CO }_{ 2 }

In the electrolysis of aqueous sodium succinate (Na₂C₄H₄O₄), the sodium succinate will undergo electrolysis to produce products at the anode and cathode.

At the anode (the positive electrode), oxidation occurs, which means that electrons are lost. In this case, water molecules in the aqueous solution will be oxidized to produce oxygen gas (O₂) and release protons (H⁺ ions). The overall reaction at the anode can be represented as follows:

2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻

So, at the anode, oxygen gas is formed along with protons (H⁺ ions).

At the cathode (the negative electrode), reduction occurs, which means that electrons are gained. In this case, sodium ions (Na⁺) will be reduced to form sodium metal (Na) and release electrons. The overall reaction at the cathode can be represented as follows:

2Na⁺(aq) + 2e⁻ → 2Na(s)

So, at the cathode, sodium metal (Na) is formed along with the release of electrons.

Therefore, the correct answer to your question is:

C. Anode, { CO }_{ 2 }