Question
Hydrogen peroxide oxidises $${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}$$ to $${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$$ in acidic medium but reduces $${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$$ to $${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}\,$$ in alkaline medium. The other products formed are respectively:
A.
$$\left( {{H_2}O + {O_2}} \right)\,{\text{and}}\,{H_2}O$$
B.
$$\left( {{H_2}O + {O_2}} \right)\,{\text{and}}\,\left( {{H_2}O + O{H^ - }} \right)$$
C.
$${H_2}O\,\,{\text{and}}\,\left( {{H_2}O + {O_2}} \right)$$
D.
$${H_2}O\,\,{\text{and}}\,\left( {{H_2}O + O{H^ - }} \right)$$
Answer :
$${H_2}O\,\,{\text{and}}\,\left( {{H_2}O + {O_2}} \right)$$
Solution :
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }}\, + \frac{1}{2}{H_2}{O_2} + {H^ + } \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}\, + {H_2}O$$
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + \frac{1}{2}{H_2}{O_2} + {H^ - } \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} + {H_2}O + \frac{1}{2}{O_2}$$