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.
Done.
Of course! I'd be happy to help you with balancing the given reaction using the half-reaction method. Here are the steps and an explanation for each step:
Step 1: Divide the reaction into oxidation and reduction half-reactions.
In this given reaction, Fe is being oxidized from Fe+2 to Fe+3, and MnO4- is being reduced from MnO4- to Mn+2. So, we will have one half-reaction for the oxidation of Fe and another half-reaction for the reduction of MnO4-.
Step 2: Balance the elements other than oxygen and hydrogen in each half-reaction.
Start with the oxidation half-reaction:
Fe+2 → Fe+3
There is only one element involved here, which is Fe. In order to balance it, the number of Fe atoms should be the same on both sides of the half-reaction. Since there is one Fe atom on each side, it is already balanced.
Step 3: Balance the oxygen atoms by adding water (H2O) molecules to the side that lacks oxygen.
For the reduction half-reaction:
MnO4- → Mn+2
Count the number of oxygen atoms on each side. There are four oxygen atoms on the left side (in MnO4-) and none on the right side. To balance the oxygen, we need to add four water molecules (H2O) to the right side:
MnO4- → Mn+2 + 4H2O
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen.
For the reduction half-reaction:
MnO4- → Mn+2 + 4H2O
Now, count the number of hydrogen atoms on each side. There are eight hydrogen atoms on the right side (in 4H2O) and none on the left side. To balance the hydrogen, we need to add eight hydrogen ions (H+) to the left side:
MnO4- + 8H+ → Mn+2 + 4H2O
Step 5: Balance the charges by adding electrons (e-) to the more positive side of the equation.
For the oxidation half-reaction:
Fe+2 → Fe+3
In this case, the charge on the Fe ion is changing from +2 to +3, meaning it is losing one electron. To balance the charges, we need to add one electron (e-) to the left side:
Fe+2 + e- → Fe+3
Note that the number of electrons should be the same in both half-reactions, so we have to multiply the oxidation half-reaction by two (2):
2Fe+2 + 2e- → 2Fe+3
Now we can combine both half-reactions into the balanced overall reaction:
2Fe+2 + 2e- → 2Fe+3
MnO4- + 8H+ + 5e- → Mn+2 + 4H2O
Multiplying both half-reactions by appropriate factors such that the number of electrons is the same (in this case, 10), we get:
10Fe+2 + 10e- → 10Fe+3
2MnO4- + 16H+ + 10e- → 2Mn+2 + 8H2O
Finally, add both half-reactions together, and cancel out any species that appear on both sides of the equation (spectator ions):
10Fe+2 + 10e- + 2MnO4- + 16H+ → 10Fe+3 + 2Mn+2 + 8H2O
And there you have it! The given reaction, Fe+2 + MnO4- → Fe+3 + Mn+2, balanced using the half-reaction method.