What is the oxidation state of Fe in the product formed when acidified potassium ferrocyanide is treated with \(H_2O_2\)?
Answer & explanation
Correct answer: option 4
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.