Chemistry Reduction Reaction Help Needed

[VinnyCee]

Chemistry Reduction Reaction Help Needed: MnO4- + S2- --> MnO2 + S8

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

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

$$2MnO_4^-+3S^{2-}+4H_2O\xrightarrow{}2MnO_2+3S_8+8OH^-$$

Does this look correct? Can anyone check? Thanks.

 

[VinnyCee]

Here is a more detailed explanation of the answer I have

Ok, starting with the original equation:

$$MnO_4^-+S^{2-}\xrightarrow{}MnO_2+S_8$$

I then separate into two "half-reactions":

$$MnO_4^-\xrightarrow{}MnO_2$$

$$S^{2-}\xrightarrow{}S_8$$

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

$$X-8+1=0\xrightarrow{}X=7$$

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

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

$$X-4=0\xrightarrow{}X=4$$

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?:

$$MnO_4^-+3e^-\xrightarrow{}MnO_2$$

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:

$$MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2$$

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

$$MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O$$

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

$$S^{2-}\xrightarrow{}S_8$$

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:

$$S^{2-}\xrightarrow{}S_8+2e^-$$

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

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

$$(S^{2-}\xrightarrow{}S_8+2e^-)*3=3S^{2-}\xrightarrow{}3S_8+6e^-$$

Adding them together, I get:

$$2MnO_4^-+3S^{2-}+8H^+\xrightarrow{}2MnO_2+3S_8+4H_2O$$

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

$$2MnO_4^-+3S^{2-}+8H^++8OH^-\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-$$

$$2MnO_4^-+3S^{2-}+8H_2O\xrightarrow{}2MnO_2+3S_8+4H_2O$$

$$2MnO_4^-+3S^{2-}+4H_2O^+\xrightarrow{}2MnO_2+3S_8+8OH^-$$

Does this seem reasonable?

 

[andrewchang]

your final line: $... 4H_2O^{+}$ ?

and the sulfur is not balanced.

 

[VinnyCee]

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.

 

[VinnyCee]

Redoing the problem

I did it again by starting with an actual balanced equation

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

$$MnO_4^-+8S^{2-}\xrightarrow{}MnO_2+S_8$$

I then separate into two "half-reactions":

$$MnO_4^-\xrightarrow{}MnO_2$$

$$8S^{2-}\xrightarrow{}S_8$$

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

$$X-8+1=0\xrightarrow{}X=7$$

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

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

$$X-4=0\xrightarrow{}X=4$$

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?:

$$MnO_4^-+3e^-\xrightarrow{}MnO_2$$

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:

$$MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2$$

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

$$MnO_4^-+3e^-+4H^+\xrightarrow{}MnO_2+2H_2O$$

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

$$8S^{2-}\xrightarrow{}S_8$$

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:

$$8S^{2-}\xrightarrow{}S_8+2e^-$$

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

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

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

Adding them together, I get:

$$2MnO_4^-+24S^{2-}+8H^+\xrightarrow{}2MnO_2+3S_8+4H_2O$$

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

$$2MnO_4^-+24S^{2-}+8H^++8OH^-\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-$$

$$2MnO_4^-+24S^{2-}+8H_2O\xrightarrow{}2MnO_2+3S_8+4H_2O+8OH^-$$

$$2MnO_4^-+24S^{2-}+4H_2O^+\xrightarrow{}2MnO_2+3S_8+8OH^-$$

Does this seem reasonable?

 

[VinnyCee]

Color-coded version of the problem above

There is http://www.chemicalforums.com/index.php?board=7;action=display;threadid=7176;start=0#msg32393", I used their color coding to highlight the important parts of this problem. Check it out and see if I finally did it right. Thanks alot.

 

[VinnyCee]

It's all solved at www.ChemicalForums.com

We figured it out over there, even thought the syntax for typing equations takes some practice to learn, it has a lot of people posting about chemistry!

http://www.chemicalforums.com/index.php?board=7;action=display;threadid=7176;start=0#msg32417".