Need help with redox reactions?

Question

I'm trying to work out a chemistry lab report, and I don't get it at all. For example, one of the reactions is I-(aq) + NO3-(aq) --> I2(aq) + NO2(g). Can someone please explain how to balance this using the half-cell method?

Answer 1

Best Answer - Chosen by Asker

half reactions:

I-(aq) --> I2(aq)
NO3-(aq) --> NO2(g)

balance in acid:

2I-(aq) --> I2(aq) + 2e-
e- + 2H+ + NO3-(aq) --> NO2(g) + H2O

equalize electrons and add:

4H+ + 2I-(aq) + 2NO3-(aq) --> I2(aq) + 2NO2(g) + 2H2O

More redox:

http://www.chemteam.info/Redox/Redox.htm…

Asker's Rating:

Asker's Comment:

Thank you so much! I understand it a little better now, and I realize what I did wrong.

Answer 2

i think it should be NO and not NO2 in the products side.
well it doesn't matters,we will balance the given red-ox reaction step wise.

Also the reaction medium is not specified (acidic/basic or neutral)

We will assume acidic medium for our convenience.

Step 1: separate the oxidation and reduction half

I-(aq) ---> I2(aq)

NO3-(aq) ---> NO2(g)

Step 2: Balance that atom whose oxidation no. is changing. (Iodine and Nitrogen in this case)

2I-(aq) ---> I2(aq)

NO3-(aq) ---> NO2(g)

Step 3: Now Balance the oxygen atoms by adding moles of H2O on the reqd side. And then balance the hydrogen atoms by adding H+ ions on reqd side.

2I-(aq) ---> I2(aq)

NO3-(aq) ---> NO2(g) + H2O (due to one less O atom in products side)

2H+ + NO3-(aq) ---> NO2(g) + H2O (due to two less H atoms on reactant side, after adding H2O)

Step 4: Now add the reqd moles of electrons so as to balance the net charge on the reactant and products side.

2I-(aq) ---> I2(aq) + 2e (since there were 2 I- on products side which contributed 2 -ve charges)

this half reaction is overall neutral.

e + 2H+ + NO3-(aq) ---> NO2(g) + H2O
this half reaction is also overall neutral.

Step 5: now add both the half reactions eliminating the electrons.

Final Reaction:

2I-(aq) + 4H+ + 2NO3-(aq) ---> 2NO2(g) + 2H2O + I2(aq)

this should be it. if the medium is basic then simply add same no. of OH- ions as that of H+ on both sides. On the side already containing H+, H+ will combine with OH- to form H2O.

Hope this helps. :)

Source(s):

Chem Class

Answer 3

separate into I and NO3^1- reactions
2I- ---> I2
NO3^1- --->NO2

balance the O's by adding water
2I- ---> I2
NO3^1- --->NO2 + H2O

balance the H's by adding H+
2I- ---> I2
NO3^1- + 2H+--->NO2 + H2O

balance the charges by adding electrons
2I- ---> I2 +2e-
NO3^1- + 2H+ +1e- ---> NO2 + H2O

equalize the number of electrons exchanged by multiplying the second equation by 2
2I- ---> I2 +2e-
2NO3^1- + 4H+ +2e- ---> 2NO2 + 2H2O

add the equations together
2I- + 2NO3^1- + 4H+ + 2e- ---> I2 + 2e- + 2NO2 + 2H2O

add out any common items (usually e-, H+ and H2O)
2I- + 2NO3^1- + 4H+ ---> I2 + 2NO2 + 2H2O

Check mass balance: 2I, 2N, 6O, 4H
Check charge balance: 2- + 2- + 4+ = 0 + 0 + 0, check.

That's it.

Answer 4

2I-(aq) -----------> I2(s) + 2e-
NO3-(aq) + e- + 2 H+(aq) ---------> NO2(g) + H2O

2 NO3-(aq) + 4 H+(aq) + 2I-(aq) ----------> 2 NO2(g) + 2 H2O + I2(s)