Chemistry Reduction Reaction Help Needed

  • Thread starter VinnyCee
  • Start date
  • #1
489
0
Chemistry Reduction Reaction Help Needed: MnO4- + S2- --> MnO2 + S8

I was wondering if I had done this reduction reaction equation correctly:
[tex]MnO_4^-+S^{2-}\xrightarrow{}MnO_2+S_8[/tex]

After going through the reduction reaction problem solving process (I think) I get this equation as the final equation in the reaction:

[tex]2MnO_4^-+3S^{2-}+4H_2O\xrightarrow{}2MnO_2+3S_8+8OH^-[/tex]

Does this look correct? Can anyone check? Thanks.
 
Last edited:

Answers and Replies

  • #2
489
0
Here is a more detailed explanation of the answer I have

Ok, starting with the original equation:

[tex]MnO_4^-+S^{2-}\xrightarrow{}MnO_2+S_8[/tex]

I then seperate into two "half-reactions":

[tex]MnO_4^-\xrightarrow{}MnO_2[/tex]

[tex]S^{2-}\xrightarrow{}S_8[/tex]

Figureing the oxidation charge for the left side of the first "half-reaction":

[tex]X-8+1=0\xrightarrow{}X=7[/tex]

Therefore, the oxidation number of Mn on the left side is 7, right?

Then I do the right side of the first "half-reaction":

[tex]X-4=0\xrightarrow{}X=4[/tex]

This means that the oxidation number of Mn on the right side is 4. Meaning that I have to add 3 electrons to the left side of the first "half-reaction" equation, right?:

[tex]MnO_4^-+3e^-\xrightarrow{}MnO_2[/tex]

Now, I added H+ ions to the left side to bring it's overall charge down to zero, because the right side has an overall charge of zero:

[tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2[/tex]

Finally, I add H2O to even out the oxygens in the first "half-reaction" equation:

[tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O[/tex]

Going on to the second of the two "half-reaction" equations:

[tex]S^{2-}\xrightarrow{}S_8[/tex]

The oxidation number for the left hand side of the second "half-reaction" equation is -2 and for the right side it is 0, right? So then I added 2 electrons to the right side of the second "half-reaction" equation:

[tex]S^{2-}\xrightarrow{}S_8+2e^-[/tex]

I then multiply the two results from the "half-reaction" equations in order to eliminate the electrons:

[tex](MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O)*2=2MnO_4^-+6e^-+8H^+\xrightarrow{}2MnO_2+4H_2O[/tex]

[tex](S^{2-}\xrightarrow{}S_8+2e^-)*3=3S^{2-}\xrightarrow{}3S_8+6e^-[/tex]

Adding them together, I get:

[tex]2MnO_4^-+3S^{2-}+8H^+\xrightarrow{}2MnO_2+3S_8+4H_2O[/tex]

Finally, I add OH- to balance and eliminate the H2O's on the right side:

[tex]2MnO_4^-+3S^{2-}+8H^++8OH^-\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-[/tex]

[tex]2MnO_4^-+3S^{2-}+8H_2O\xrightarrow{}2MnO_2+3S_8+4H_2O[/tex]

[tex]2MnO_4^-+3S^{2-}+4H_2O^+\xrightarrow{}2MnO_2+3S_8+8OH^-[/tex]

Does this seem reasonable?
 
  • #3
217
0
your final line: [itex] ... 4H_2O^{+} [/itex] ?

and the sulfur is not balanced.
 
  • #4
489
0
How do I balance the equation then? Did I do the "half-reaction" equations correctly?

Do you see where I made a mistake?

Thanks again.
 
  • #5
489
0
Redoing the problem

I did it again by starting with an actual balanced equation:eek:

Ok, starting with the original equation (now balanced):

[tex]MnO_4^-+8S^{2-}\xrightarrow{}MnO_2+S_8[/tex]

I then seperate into two "half-reactions":

[tex]MnO_4^-\xrightarrow{}MnO_2[/tex]

[tex]8S^{2-}\xrightarrow{}S_8[/tex]

Figureing the oxidation charge for the left side of the first "half-reaction":

[tex]X-8+1=0\xrightarrow{}X=7[/tex]

Therefore, the oxidation number of Mn on the left side is 7, right?

Then I do the right side of the first "half-reaction":

[tex]X-4=0\xrightarrow{}X=4[/tex]

This means that the oxidation number of Mn on the right side is 4. Meaning that I have to add 3 electrons to the left side of the first "half-reaction" equation, right?:

[tex]MnO_4^-+3e^-\xrightarrow{}MnO_2[/tex]

Now, I added H+ ions to the left side to bring it's overall charge down to zero, because the right side has an overall charge of zero:

[tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2[/tex]

Finally, I add H2O to even out the oxygens in the first "half-reaction" equation:

[tex]MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O[/tex]

Going on to the second of the two "half-reaction" equations:

[tex]8S^{2-}\xrightarrow{}S_8[/tex]

The oxidation number for the left hand side of the second "half-reaction" equation is -2 and for the right side it is 0, right? So then I added 2 electrons to the right side of the second "half-reaction" equation:

[tex]8S^{2-}\xrightarrow{}S_8+2e^-[/tex]

I then multiply the two results from the "half-reaction" equations in order to eliminate the electrons:

[tex](MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O)*2=2MnO_4^-+6e^-+8H^+\xrightarrow{}2MnO_2+4H_2O[/tex]

[tex](8S^{2-}\xrightarrow{}S_8+16e^-)*3=24S^{2-}\xrightarrow{}3S_8+48e^-[/tex]

Adding them together, I get:

[tex]2MnO_4^-+24S^{2-}+8H^+\xrightarrow{}2MnO_2+3S_8+4H_2O[/tex]

Finally, I add OH- to balance and eliminate the H2O's on the right side:

[tex]2MnO_4^-+24S^{2-}+8H^++8OH^-\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-[/tex]

[tex]2MnO_4^-+24S^{2-}+8H_2O\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-[/tex]

[tex]2MnO_4^-+24S^{2-}+4H_2O^+\xrightarrow{}2MnO_2+3S_8+8OH^-[/tex]

Does this seem reasonable?
 
Last edited:

Related Threads on Chemistry Reduction Reaction Help Needed

Replies
2
Views
23K
  • Last Post
Replies
1
Views
2K
Replies
16
Views
2K
  • Last Post
Replies
10
Views
20K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
6
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
1K
Top