Net charge

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m0286
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Hello Im 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 wouldnt 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 cant 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??
 

Answers and Replies

  • #2
m0286
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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. :smile:
 

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