Balancing the Equations with Half-Cell Method: Step 4

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SUMMARY

The discussion focuses on balancing redox equations using the half-cell method, specifically addressing the example of MnO4^- and SO3^2-. The participant successfully identifies the changes in oxidation numbers, noting that manganese decreases by 5 and sulfur increases by 2. The lowest common multiple of these changes is determined to be 10, leading to the balanced equation 2MnO4^- + 5SO3^2- -> 2Mn^2+ + 5SO4^2-. The participant clarifies the net charges for both sides of the equation, confirming the left side as 12- and the right side as 6-.

PREREQUISITES
  • Understanding of oxidation states and their changes in redox reactions
  • Familiarity with the half-cell method for balancing equations
  • Basic knowledge of ionic charges and their calculations
  • Ability to determine the lowest common multiple (LCM) in numerical contexts
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  • Study the half-cell method in detail for balancing complex redox reactions
  • Learn how to calculate oxidation states for various elements in compounds
  • Explore examples of balancing redox reactions in acidic and basic solutions
  • Practice determining net charges for various chemical equations
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Chemistry students, educators, and anyone involved in learning or teaching redox reactions and the half-cell method for balancing equations.

m0286
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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??
 
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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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