Here is how you do one of the half reactions.
Let's take NO3^- --> NO
1. Oxidation state N on left is +5; on right is +2
2. Add electrons on appropriate side to balance change of oxidation state.
NO3^- + 3e ==> NO
3. Count charge on left and right. I see -4 on left and zero on right so add H^+ to balance the charge.
NO3^- + 3e + 4H^+ ==> NO
4. Now add H2O to balance the the H+.
NO3^- + 3e + 4H^+ ==> NO + 2H2O
5. Check it.
a. atoms.
N--1 left and right
O--3 on left and 3 right
H -- 4 left and right
b. charge
zero left and right
c. change in oxidation state.
+5 + 3e = +2
Balanced.
If you have follow up questions, show what you've done and explain what you don't understand about the next step.
Will you help me use the half reaction method to balance these equations. I need to add water molecules and hydrogen ions in acid solutions. I also need to keep the balanced equations in net ionic form. thanks
1) Cl-(aq) + NO3-(aq) --> ClO-(aq) +NO(g)
(in acid solution)
2) IO3-(aq) + Br-(aq) --> Br2(l) + IBr(s)
(in acid solution)
3) I2(s) + Na2SO3(aq) --> Na2S2O4(aq) + NaI (aq)
(in acid solution)
Thanks Again clo
1 answer