posted by Nessy on .
Okay, so i've got 8 reactions i am supposed to balance "according to the half-reaction method". Only problem is, i was gone for the lesson, and my borrowed notes look like an entirely different language!
So if you could please show me the steps and explain how to do the first one, that would be MOST appreciated!
Fe+2 + MnO4- ====> Fe+3 + Mn+2
Thank you SO very much!
The half reaction method I use is not standard but I use it because it uses the oxidation states and it was a way for me to make sure students knew oxidation states and how to determine them.
1. Divide the reaction into half reactions
Fe^+2 ==> Fe^+3
MnO4^- ==> Mn^+2
2. Determine the oxidation state. I will do Mn first.
Mn is +7 on the left, +2 on the right.
3. Add electrons to the appropriate side to balance the change of oxidation state.
MnO4^- + 5e ==> Mn^+2
(Note:+7 +5e = +2--not the charge of +2 but the oxidation state of +2).
4. Now count the charge on the left and right. I see -6 on the left; +2 on the right. and
a. If acid solution, add H^+ to balance the charge.
b. If basic solution, add OH^- to balance the charge.
This is acid so we add H^+ to balance the charge which means add 8H^+ to the left.
MnO4^- + 5e + 8H^+ ==> Mn^+2
(I always check it to make sure; I see +2 on the left and right).
5. Now add water to balance the H^+ which means 4H2O.
MnO4^- + 5e + 8H^+ ==> Mn^+2 + 4H2O
6. Balance the other half using the same method. In this case, it is just one step.
Fe^+2 ==> Fe^+3 + e
7.Multiply each equation by a coefficient to make the electron change the same; i.e., multiply the Mn equation by 1 and the Fe equation by 5 and add them.
5Fe^+2 + MnO4^- + 8H^+ + 5e ==> Mn^+2 + 4H2O + 5Fe^+3 + 5e.
8. Cancel anything common to both sides; in this case, cancel the 5e, then check to make sure it balances three ways.
a. the atoms must balance.
b. the charge must balance.
c. the changer of electrons must balance.