Question

overall equation :

IO3- (aq) + 6H+(aq) + 6S2O32-(aq) I-(aq) + 3S4O62-(aq) +3H2O(l)

Write the oxidation state of each atom in the overall equation. Show the oxidizing and reducing agents.

Answer 1

The oxidation state of each atom in the overall equation is :

• The overall equation given is :
• IO3- (aq) + 6H+(aq) + 6S2O32-(aq) —> I-(aq) + 3S4O62-(aq) +3H2O(l)

• Oxidation state is known as oxidation number which describes the degree of oxidation of an atom in the chemical compound.

• Now in this the oxidation state of the above atoms are : IO3- (aq) = for I , I+ (- 2)*3 = -1 , I = +5 & for O = -2.

• H+ , in this H have +1 oxidation state.

• S2O32-(aq) , in this S have , S2+ (- 2)*3 = -2 , S = -2+6/2 , S = +2 & O = -2 .

• I-(aq) , it have -1 oxidation state.

• S4O62-(aq), in this S have , S4+ (-2)*6= -2 , S = 10/4 ,. S = +2.5 & O have -2 .

• H2O(l) , in this H = +1 & O = -2.

• Oxidising agent is one which has ability to oxidise the other substance , means it will reduce itself means gain of electrons . So , here the oxidising agent is IO3- as in this I oxidation state goes from +5 to -1.

• Reducing agent has ability to reduce the other substance , means it will oxidise itself means loss of electrons .So. here the reducing agent is S2O32- as S oxidation state goes from +2 to +2.5.