Answer:-

The preparation of a solution of cerium(IV) [Ce(IV)] in an acidic medium is essential to prevent the reduction of Ce(IV) to cerium(III) [Ce(III)]. This reduction occurs due to the presence of reducing agents, and by maintaining an acidic environment, the reduction is minimized. Let’s illustrate this with a balanced chemical equation:

Ce(IV) is commonly prepared as cerium(IV) sulfate, Ce(SO4)2. When Ce(IV) is exposed to a reducing agent in an alkaline (basic) solution, it undergoes reduction to form cerium(III) ions. The reaction can be represented as follows:

Ce(IV) + e⁻ → Ce(III)

Now, to prevent this reduction, an acidic medium is used. The balanced chemical equation for the reduction of Ce(IV) in an acidic medium involves the reduction of hydrogen ions (H⁺) instead of Ce(IV):

2Ce(IV) + 3H₂O + 6e⁻ → 2Ce(III) + 6OH⁻

In this reaction, cerium(IV) ions (Ce⁴⁺) are reduced to cerium(III) ions (Ce³⁺), while hydrogen ions (H⁺) from the acidic medium are simultaneously reduced to form hydroxide ions (OH⁻). By maintaining an acidic environment, you have an abundance of H⁺ ions, which are preferentially reduced instead of Ce(IV), thus preserving the cerium(IV) oxidation state.

So, to prepare a solution of Ce(IV) and ensure that it remains in the +4 oxidation state, it is essential to use an acidic medium to minimize the reduction of Ce(IV) to Ce(III) by reducing agents.