Balancing Redox?

Question

The following reaction occurs in aqueous acid solution:
NO3- + I- = IO3- + NO2
In the balanced equation the coefficient of NO3- is:

A. 2
B. 3
C. 4
D. 5
E. 6

Answer 1

E. 6

The charges and number of moles of each element (except O) need to be balanced:

NO3- + e- => NO2
oxidation # of N in NO3 - is +5
oxidation # of N in NO2 is +4

I- => IO3- + 6e-
oxidation # of I- is -1
oxidation # of I in IO3- is +5

Add the two equations together (canceling the e's):

6NO3- + I- => IO3- + 6NO2

You can finish balancing the equation by adding H2O's and H+'s if you would like, but here you already have the answer - E. 6

(If you would like to finish balancing:
There are 18 O's on the LHS but 15 on the RHS, so add 3 molecule H2O on the RHS and 6 molecules H+ on the left to balance out the H's:

6NO3 - + I- + 6H+ => IO3 - + 6NO2 + 3H2O

Check at the end to see that the atoms and charges are balanced.)

Answer 2

Original
NO3 + I = IO3 + NO2
Balanced Equation is
3 NO3 + I = IO3 + 3 NO2
Answer is B. 3

Answer 3

NO3-+I- =IO3-+NO2
on L.H.S N=1,O=3,I=1 => on R.H.S I=1,O=5,N=1
so,balance by multiplying by 3 with NO3 on LHS and 3 with NO2on RHS

Answer 4

6

N 5+ + 1 e- --> N 4+
I 1- --> I 5+ + 6 e-

mult top equation by 6 to balance e-

then balance remaining equation by using "2" I - and "2" IO3 1-

Answer 5

B.3

3No3- + I- = IO3- + 3NO2