in-fehlings-test-what-happens-to-the-cupric-ion-when-a-reducing-sugar-is-present

Question

In Fehling's test, what happens to the cupric ion when a reducing sugar is present?

Accepted Answer

It is reduced to form a red cupric oxide precipitate.

Answer

It forms a green solution indicating the presence of reducing sugars.

Answer

It remains unchanged and stays blue.

Answer

It is oxidized to form a blue cupric oxide precipitate.