themuffinman11

2+ Year Member
Feb 13, 2017
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Hey everyone, I'm having trouble with balancing redox reactions that involve many ions, I'm having difficulty determining the half reactions, for example:

H2SO4(aq)+NaBr(s)⟶Br2(l)+SO2(g)+Na2SO4(aq)+H2O(l)

I can see that Br is getting oxidized (-1 > 0), and is the sulfur being reduced? (+6 in H2SO4 and +4 in SO2)

But what would the half reactions be? I think it's the polyatomic ions that confuse me. Thanks for any help!
 

princecoup

2+ Year Member
Jul 18, 2015
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Status
Pre-Dental
Yes, you're right in that Br is getting oxidized and sulfur is being reduced based on their oxidation states.
I don't know if your question was worded incorrectly, but NaBr is very soluble in water because it is an ionic compound and sodium is a group IA alkali metal, so it should be written as NaBr (aq), which may be throwing you off when you were trying to write the half-reactions (it threw me off at first too lol).

So when you write the 2 half-reactions:
SO42- --> SO2
Br- --> Br2

The balanced redox reaction (I did it in acidic conditions) should look like:
4H+ + SO42- + 2Br- --> Br2 + SO2 + 2H2O

Also, the sodium ion was a spectator ion in the overall reaction, that's why it was canceled out.
Hope this helps!
 
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themuffinman11

themuffinman11

2+ Year Member
Feb 13, 2017
166
119
Yes, you're right in that Br is getting oxidized and sulfur is being reduced based on their oxidation states.
I don't know if your question was worded incorrectly, but NaBr is very soluble in water because it is an ionic compound and sodium is a group IA alkali metal, so it should be written as NaBr (aq), which may be throwing you off when you were trying to write the half-reactions (it threw me off at first too lol).

So when you write the 2 half-reactions:
SO42- --> SO2
Br- --> Br2

The balanced redox reaction (I did it in acidic conditions) should look like:
4H+ + SO42- + 2Br- --> Br2 + SO2 + 2H2O

Also, the sodium ion was a spectator ion in the overall reaction, that's why it was canceled out.
Hope this helps!
What does the "2" mean in front of SO4 in the half reaction? Also what about the electrons? Where would you put them, and they have to be balanced in both equations right (multiply by common integer?)
Thank you for the help!
 

princecoup

2+ Year Member
Jul 18, 2015
37
53
Status
Pre-Dental
What does the "2" mean in front of SO4 in the half reaction? Also what about the electrons? Where would you put them, and they have to be balanced in both equations right (multiply by common integer?)
Thank you for the help!
The balanced redox chemical reaction that I wrote is the overall balanced equation with the electrons already having been canceled out and the charge is equal on both sides. (Also the "2-" in front of the SO4 is the -2 charge of the sulfate ion).

Here are the two half-reactions right before adding them together to get the overall balanced redox rxn:
2e- + 4H+ + SO42- --> SO2 + 2H2O
+ 2Br- --> Br2 + 2e-
------------------------------------------------
4H+ + SO42- + 2Br- --> Br2 + SO2 + 2H2O