[Like_A_Whisper]

Greetings, At the moment we are studying redox reactions at school. Our teacher showed us a way to solve some redox equations. Some of them are easy, and some are with moderate difficulty. I don't know if this way is international but I would like to make it clearer since there isn't a similar way in my chemistry book(Unfortunately  ).

Here is what I've learned:

• First we need to make an electrochemical scheme(I'm not sure about the name of this term) - This contains the half reaction of oxidation and the other half reaction of reduction
• We need to focus on balancing the atoms of elements which changed their oxidation number during the process

• Balance the number of atoms of all remaining reagents and compounds(Exempt for atoms of hydrogen/oxygen)
• Balance the number of Hydrogen atoms

• Balance the number of Oxygen atoms
• Calculate the total atoms of every single substance which is a part of the reaction and make sure their number is similar on the left and right side(individually of course)

Example:
FeSO₄ + KMnO₄ + H₂SO₄   Fe₂(SO₄)₃ + MnSO₄ + K₂SO₄ + H₂O
We usually don't write the "(aq, l, s or g)" signs here

Step 1:
Fe⁺²S⁺⁶O₄^-2 + K⁺¹Mn⁺⁷O₄^-2 + H₂⁺¹S⁺⁶O₄^-2   Fe₂⁺³(S⁺⁶O₄^-2)₃ + Mn⁺²S⁺⁶O₄^-2 + K₂⁺¹S⁺⁶O₄^-2 + H₂⁺¹O^-2

--We see that the Fe(iron) has changed its oxidation number from +2 to +3 and the Mn(Manganese) has changed its oxidation number from +7 to +2. So we present that with half reactions

2Fe⁺² -1 electron Fe₂⁺³        |   5  - Half reaction of oxidation
                                            |5
Mn⁺⁷ +5 electrons Mn⁺²        |    1 - Half reaction of reduction
                                            

Then we use those coefficients to balance the reaction. Next come Step 2,3,4,5,6
Final Result:
10FeSO₄ + 2KMnO₄ + 8H₂SO₄  = 5Fe₂(SO₄)₃ + 2MnSO₄ + K₂SO₄ + 8H₂O

If there is an easier way, please I need to know how.
Kind regards from a fellow student from Macedonia

[Borek]

This is basically balancing by oxidation numbers:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method

with alternative being balancing by half reactions:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

Problem with ON is that they don't exist in reality (ie there is no measurable property of atoms that can be attributed to ON) and in some molecules they are hard to assign (try carbon in acetone or sulfur in thiosulfate).

[Like_A_Whisper]

Never seen that way before... But I see they have the same equation as I presented. They got KMnO₄ so only the Mn changes its oxidation number. Why do they use MnO₄ on the left side and only Mn on the right side in the half reactions. I haven't experienced that kind of method. It would be nice if it is more secure than my way... I'll try to figure it out. Thank you for your time

[Borek]

Why do they use MnO₄ on the left side and only Mn on the right side in the half reactions.

As far as I can tell (guessing at what you are aiming) "they" don't - just MnO₄^- & Mn²⁺ are listed in one place not as the reaction, but just as a general scheme of what is going on, scheme that is later beefed up with balancing of hydrogen and oxygen.

On the other hand - that's what is really happening, permanganate gets reduced to Mn²⁺...

[salleebrowne]

I've een teaching high school chemistry for quite a ew years and I have always taught the method that you describe. Think of it as a puzzle, and when you've filled out a page and a half of chemical equations - show it to your parents - it'll make them proud! 

[Like_A_Whisper]

@ salleebrowne: Mind explaining it again to me?