When an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is electrolysed, the product at anode is : (a) \(\mathrm{H}^{-}\) (b) \(\mathrm{OH}^{-}\) (c) \(\mathrm{SO}_{4}^{2-}\) (d) \(\mathrm{O}_{2}\)

In the context of electrolysis, the anode reaction is a fundamental process that involves oxidation, which means losing electrons. In our specific example involving an aqueous solution of \texorpdfstring{\(\mathrm{H}_{2} \mathrm{SO}_{4}\)}, the anode reaction does not produce \texorpdfstring{\(\mathrm{H}^{-}\)} or \texorpdfstring{\(\mathrm{SO}_{4}^{2-}\)} ions; instead, the reaction generates oxygen gas \texorpdfstring{\(\mathrm{O}_{2}\)}.

During electrolysis, water molecules near the anode give up electrons and transform into oxygen gas and hydrogen ions. This reaction can be represented by the equation:
\[2 \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{O}_{2}(g) + 4 \mathrm{H}^{+}(aq) + 4 e^{-}\]
This shows that water is oxidized, releasing oxygen gas, protons \texorpdfstring{\(\mathrm{H}^{+}\)}, and electrons. These electrons travel through the circuit and are used at the cathode, completing the electrical circuit.

Oxidation processes are chemical reactions in which an element or compound loses electrons. They are crucial in the field of electrochemistry and especially pertinent in electrolysis. Understanding oxidation can help clarify why certain products, such as oxygen gas in our exercise, are formed at the anode of an electrolytic cell.

An underlying principle to remember is that oxidation always occurs at the anode of an electrolytic cell. This is because the anode attracts negatively charged ions, or anions, and causes them to lose electrons. In aqueous solutions, though substances like \texorpdfstring{\(\mathrm{SO}_{4}^{2-}\)} ions might be present, water often undergoes oxidation instead since its oxidation process is energetically more favorable.

Oxidation can be represented by a half-reaction, showing only the loss of electrons. Here's a simplified example for water oxidation:
\[\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \frac{1}{2}\mathrm{O}_{2}(g) + 2 \mathrm{H}^{+}(aq) + 2 e^{-}\]
When students approach problems involving electrolysis, identifying which species will undergo oxidation is essential. Factors like electrode potential and concentrations can influence the outcome, which is why understanding the basics of oxidation is so important.

Chemical decomposition is a type of chemical reaction where a compound is broken down into simpler substances. Electrolysis is a classic example of this process, as it uses electrical energy to cause the decomposition. In the electrolysis of aqueous \texorpdfstring{\(\mathrm{H}_{2} \mathrm{SO}_{4}\)}, both the water molecules and the \texorpdfstring{\(\mathrm{H}_{2} \mathrm{SO}_{4}\)} are decomposed.

At the anode, we've seen that water is decomposed to release oxygen. Similarly, at the cathode, \texorpdfstring{\(\mathrm{H}^{+}\)} ions are reduced to \texorpdfstring{\(\mathrm{H}_{2}\)} gas. This breakdown is integral to the process of electrolysis, turning liquid water and solute ions into gaseous products and other ionic forms.

Anode vs. Cathode Decomposition

In electrolytic decomposition, the anode and cathode facilitate different types of reactions. As discussed, oxidation occurs at the anode (loss of electrons), whereas reduction (gain of electrons) occurs at the cathode. Therefore, the decomposition reaction can be seen as two half-reactions occurring simultaneously but separately at each electrode, ultimately leading to the formation of new substances.