[shanthan]

Hi,
I am really lost in the chemistry class. I am really interested in this subject but i really need some help. How can you balance this particular equation using half reaction method. My textbook does not describe ways to do such a equation. Please *delete me*!!

MnO4- + S2-  --> MnS + S (Basic)

Yours Shanthan

[Borek]

You have to recognize what was oxdidized and what was reduced during reaction.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

[shanthan]

i understand how to do it. I just am not getting how to this particular problem. Please help.

[Borek]

Show what you know and why you think it is wrong.

Same about the other question.

[renge ishyo]

Can you balance this particular equation for an acidic solution? if you can do that (and show your work on here) then I'll show you a clever trick to find the answer for a basic solution from the answer to the balanced acidic equation.

[iwanttopassthisclass]

Ahh... I actually got this assigned to me today too... Can someone help me?
THis is what i did so far:

i made two half reactions:

MnO4- ---> MnS   and S2- ---> S

Then i balanced the charges and elements on both and i got:

4H+ + S + MnO4- + 3e- ---> MnS + 4H2O

and

S2- ---> S + 2 e-

then i balanced the electrons on both the equations so they would cancel out:

2(4H+ + S + MnO4- + 3e- ---> MnS + 4H2O)

and

3(S2- ---> S + 2 e-)

THEN I put the equations together and got:

8H+ + S + 2MnO4- + 3S2- ---> 2MnS + 8H2O + 3S

I'm not sure if i did that last step right...

but after that since there's 8H on the right I added 8OH- to both sides... is that correct?

so the final equation i got was

8OH- + 8H+ + S + 2MnO4- + 3S2- ---> 2MnS + 8H2O + 3S + 8OH-


Did I do any of this right!?
If I'm doing anything wrong... What is it?

Pleeease help 

[Borek]

4H+ + S + MnO4- + 3e- ---> MnS + 4H2O

I would use S^2- on the left, but it would not change the final result. Note however, that initially there is no S present, so this reaction is impossible.

2(4H+ + S + MnO4- + 3e- ---> MnS + 4H2O)

and

3(S2- ---> S + 2 e-)

Both are correctly balanced.

8H+ + S + 2MnO4- + 3S2- ---> 2MnS + 8H2O + 3S

I'm not sure if i did that last step right...

And your doubts are well founded. Two errors - you have not multiplied all molecules/ions by the coefficients and you forgot to cancel out things that are present on both sides.

but after that since there's 8H on the right I added 8OH- to both sides... is that correct?

so the final equation i got was

8OH- + 8H+ + S + 2MnO4- + 3S2- ---> 2MnS + 8H2O + 3S + 8OH-

Equation was wrong, so it can't get better - but obviously it could get worse. You have added OH^-, but you have not done anything with it. It was added for a reason. Besides, the way you did it you have OH^- on both sides, that's not how the corrrectly balanced equation looks like.

[renge ishyo]

4H+ + S + MnO4- + 3e- ---> MnS + 4H2O

The first error took place here (the rest carried on as a result of this), make sure you balance the mass on both sides carefully before moving forward.

Also, on the last step when you add the OH^-, be sure to simplify XH⁺ + XOH^- to XH₂0. Then cancel the water by looking at both sides of the equation. Most of the steps are there, just go back to the step above, make sure the mass is balanced, and work it through again.

[Borek]

My mistake, thanks for checking.

[renge ishyo]

My mistake, thanks for checking.

No worries, it was a pretty subtle mistake on the students part.

[iwanttopassthisclass]

Wait... So I have to balance the equation they give me first.. And THEN split it into two half reactions and balance those?

[Borek]

No, you did almost OK - that is, you have almost correctly determined half-reactions, and then you took a stab at balancing them. What have happened was two simple mistakes at this stage, which made the equation unbalanced at the end.

Then you have added OH^- - but you have not finished the work. It is not just adding, you chould use it to neutralize H⁺ and then cancel out water.