Formaldehyde can be distinguished from acetaldehyde by the use of :

Formaldehyde (HCHO) and acetaldehyde (CH₃CHO) are both aldehydes, but they can be distinguished based on their structural differences. Formaldehyde has no alkyl group attached to the carbonyl carbon, while acetaldehyde has a methyl group (CH₃-). This difference allows acetaldehyde to undergo specific reactions that formaldehyde cannot.

Let's evaluate each option:

Option A: Schiff's reagent - Both formaldehyde and acetaldehyde give a pink color with Schiff's reagent because both are aldehydes. So, this reagent cannot distinguish between them.

Option B: Tollen's reagent - Both aldehydes reduce Tollen's reagent to produce a silver mirror. Thus, both give a positive test and cannot be distinguished.

Option D: Fehling's solution - Both formaldehyde and acetaldehyde reduce Fehling's solution to give a red precipitate of cuprous oxide (Cu₂O). Therefore, this test is positive for both and fails to distinguish them.

Option C: I₂/Alkali - This is the iodoform test. Acetaldehyde (CH₃CHO) has a methyl group attached to the carbonyl carbon and undergoes the iodoform reaction in the presence of iodine and alkali (NaOH), producing a yellow precipitate of iodoform (CHI₃). The reaction is:

$$ CH_{3}CHO + 3I_{2} + 4NaOH -> CHI_{3} + HCOONa + 3NaI + 3H_{2}O $$

Formaldehyde (HCHO) lacks a methyl group and does not undergo this reaction. Instead, it undergoes the Cannizzaro reaction in strong alkali, but no iodoform is formed. Thus, formaldehyde gives a negative iodoform test.

Therefore, I₂/Alkali (option C) can distinguish formaldehyde from acetaldehyde: acetaldehyde gives a yellow precipitate (positive test), while formaldehyde does not (negative test).

Hence, the correct answer is Option C.