What is the oxidation state of Fe in the product formed when acidified potassium ferrocyanide is treated with \(H_2O_2\)?

+3

The correct answer is option 4. +3.

When acidified potassium ferrocyanide is treated with \(H_2O_2\), the oxidation state of Fe changes from +2 to +3. The product formed is potassium ferricyanide, which has the formula \(K_3[Fe(CN)_6]\).

In potassium ferrocyanide, the oxidation state of Fe is +2. This is because the cyanide ion \((CN^-)\) has a charge of -1, and there are six cyanide ions surrounding the Fe atom. Therefore, the Fe atom must have a charge of +2 to balance the charge of the cyanide ions.

In potassium ferricyanide, the oxidation state of Fe is +3. This is because the cyanide ion still has a charge of -1, but there are only three cyanide ions surrounding the Fe atom. Therefore, the Fe atom must have a charge of +3 to balance the charge of the cyanide ions.

Here is the balanced chemical equation for the reaction:

\(2K_4[Fe(CN)_6] + H_2O_2 + K_2SO_4 → 2K_3[Fe(CN)_6] + K_2SO_4 + 2H_2O\)

As you can see, the oxidation state of Fe changes from +2 to +3 in the course of the reaction.