Balancing the Equations with Half-Cell Method: Step 4

[m0286]

Hello I am working on questions for balancing the equations using the half-cell method. I have read the text and I think I get it..well up until step 4 (that they gave me) Its balancing the ionic charge the example was:
MnO4^- + SO3^2- -> Mn^2+ + SO4^2-
(7+)...(4+)...(2+)...(6+)(ignore the periods they are there for spaces that wouldn't show up??)
They tell me to write the oxidation numbers and determine which ones change. I have seen whut the numbers are They are written below the equation. I write how much each element increases or decreases by Mn decreases by 5 and S increases by 2. It says find the lowest common multiple of the decrease and increases os the oxidation numbers. which is 10. NEXT IS WHERE IM CONFUSED!
im supposed to determine the net charge for the left and right side it says the left side is 12- and the right side is 6-? I can't even think of how to determine the net charge, and i need to know how to do this for the questions i do have to answer its probably simple but can someone please help??

 

[m0286]

Wow my brain must not be working i figured it out.. it was simple:
since the lowest common multiple was 10.. you put a 2 in front of Mn and a a 5 infront of SO.. so
2MnO4^- + 5SO3^2- -> 2Mn^2+ + 5SO4^2-
and -2 + (5 * 2-=-10)=-12 THEN (2 * 2=4) + (5 * 2-=-10)=-6. Thanks for looking though.

 

[ravenprp]

Hi there,

Step 4 in balancing equations using the half-cell method involves determining the net charge on the left and right side of the equation. This is important because we want the overall charge to be balanced on both sides.

In your example, the left side has a charge of 12- (7- from MnO4^- and 5- from SO3^2-) and the right side has a charge of 6- (2+ from Mn^2+ and 4- from SO4^2-).

To determine the net charge, we simply add up all the individual charges on each side. So for the left side, we have 7- + 5- = 12-. And for the right side, we have 2+ + 4- = 6-.

Since the net charge on the left side is 12- and on the right side is 6-, we can see that there is a difference of 6- between the two sides. This means that there is an overall excess of negative charge on the left side, which needs to be balanced out.

To do this, we can add 6 electrons to the right side of the equation. This will balance out the negative charge and make the overall charge on both sides equal.

I hope this helps clarify the process for you. Keep practicing and you'll get the hang of it! Let me know if you have any other questions.