Question

Balance each redox reaction occurring in acidic aqueous solution.

NO-3 (aq) + Sn2+ (aq) -----> Sn4+ (aq) + NO (g)

Answer 1

Follow the steps:

First, split the above equation in two parts and then balance individually and then combine as:

NO3-(aq) ---------> NO(g) ------(1) and Sn2+(aq) ----------> Sn4+(aq) -------(2)

First balance eq (1) step by step as:

NO3-(aq) ---------> NO(g)

NO3-(aq) ---------> NO(g)   + 2H2O (balance oxygen)

NO3-(aq) + 4H+ ---------> NO(g)   + 2H2O     (balance H)

NO3-(aq) + 4H+ + 3e- ---------> NO(g)   + 2H2O ------(3)    (balance charge)

Now, balance equation (2) as:

Sn2+(aq) ----------> Sn4+(aq)             

Sn2+(aq) ----------> Sn4+(aq) + 2e- --------------(4) (balance charge)

Now, multiply eq (3) by 2:

2NO3-(aq) + 8H+ + 6e- ---------> 2NO(g)   + 4H2O ----------(5)

and multiply eq (4) by 3:

3Sn2+(aq) ----------> 3Sn4+(aq) + 6e- -----------(6)

Finally, add eq (5) and (6):   

2NO3-(aq) + 8H+ + 6e- + 3Sn2+(aq) ------>   2NO(g)   + 4H2O +    3Sn4+(aq) + 6e-

Now, cancell the electrons on either side to get balanced equation:

2NO3-(aq) + 8H+ +  3Sn2+(aq)   ------>   2NO(g)   + 4H2O +    3Sn4+(aq)