How to balance the following redox reactions using only the oxidation number method? Please illustrate and don't just give the answer!!!!
Zn + HNO3 ------> Zn(NO3)2 + NO2 + H2O
FeS2 + O2 ------> Fe2O3 + SO2
Also please explain how to balance redox reaction when we are given disproportionate reactions, i.e. where we have three different oxidation states of the same element.
Balancing the equations by Oxidation Number Methode.
(a) Zn + HNO3 à Zn(NO3)2 + N2O + H2O
Step 1: Find out the element which undergoes change in Oxidation Number.
Step2: Balance no. of atoms for the corresponding molecules which interchange oxidation no.
Since HNO3 has one N atom and N2O has two Na atom so multiply HNO3 by 2
Thus total increase in O.N= +2 while Total decrease in O.N= 2x4=+8
So equation becomes: Zn + 2HNO3 àZn(NO3)2 + N2O + H2O
Step 3: Since increase in O.N= decrease in O.N in a balanced reaction
Thus Zn should be multiplied by 4 to get change in O.N by +8
So,
4Zn +2 HNO3 àZn(NO3)2 + N2O + H2O
Step4: balance all other atom other than O and H as follows
4Zn + 2HNO3 à4Zn(NO3)2 + N2O + H2O
To balance the no. of N atoms, multiply HNO3 by 5 and balance water molecule by counting total H atom in HNO3
Thus, final balanced equation will be
4Zn + 10HNO3 à4Zn(NO3)2 + N2O + 5H2O
Similarly in case of,
(b) FeS2 + O2 à Fe2O3 + SO2
Step 1: Find change in o.n
Here, O.N of Fe increases by +1 and O.N of S is increased from-1 to +4
Step2: determine total change in o.n
Fe+2 --> Fe+1 increase by +1
S2-1 --> 2S+4 increase by 10
O2 --> 2O-2 decrease by -4
So total increase in O.N = 11 and Decrease= 4
Step 3: Balance total change in O.N. so multiply FeS2 by 4 and O2 by 11 and balance both side .
4FeS2 + 11O2 à 2 Fe2O3 + 8SO2