How to balance equations using the half cell method?

Click For Summary

Discussion Overview

The discussion focuses on the method of balancing redox equations using the half-cell method, specifically addressing the balancing of the reaction between sulfite and permanganate ions. Participants explore the steps involved in this process, including the division into half-reactions and balancing of atoms and charges.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the skeleton equation for the reaction involving sulfite and permanganate ions.
  • Another participant outlines a step-by-step method for balancing redox equations, emphasizing the importance of balancing atoms, oxygen, hydrogen, and charge.
  • A third participant provides a link to an external resource that offers a similar explanation of the half-reaction balancing method.
  • One participant notes the complexity of identifying which species is oxidized or reduced based solely on superscripts, providing the half-reactions for sulfite and permanganate and suggesting a method to equalize the number of electrons before combining them.

Areas of Agreement / Disagreement

There is no explicit consensus on the best approach to balancing the equation, as participants provide different perspectives and methods without resolving the discussion.

Contextual Notes

Some steps in the balancing process may depend on specific assumptions about the reaction conditions or the definitions of oxidation and reduction, which are not fully detailed in the discussion.

nursing
Messages
1
Reaction score
0
SO32- +MnO4-1+H+1 <-> Mn2+ +SO42- +H20(l)
 
Chemistry news on Phys.org
Step 1 Divide the skeleton equation into half-reactions.

Step 2 Balance atoms other than H and O.

Step 3 Balance oxygen atoms by adding H2O to the side that needs O.

Step 4 Balance hydrogen by adding H+ to the side that needs H.

Step 5 Balance the charge by adding electrons.

Step 6 Make the electrons gained equal to the electrons lost and then add the two half-reactions.

Step 7 Cancel anything that is the same on both sides.
 
this problem may seem a bit more tricky than the redox problems that you are used to since it may not be obvious which is getting reduced/oxidized based upon the superscripts alone.

The half-reactions are:

[tex]SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-[/tex]
[tex]MnO_4^{-1} + 8H^+ + 5e^- \rightarrow Mn^{+2} + 4H_2O[/tex]

Make your electrons equal by multiplying the half rxns to make the number of electrons 10 for both, then add the half rxns up and you should get the correct answer.
 

Similar threads

Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
4K
Replies
5
Views
9K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
9
Views
3K
Replies
7
Views
3K