How Do You Balance Complex Chemical Equations in Ionic and Neutral Forms?

[1question]

Homework Statement

Write the balanced chemical reactions in ionic and neutral forms for:
i) Enargite leaching by sodium bisulfide in basic solution to form chalcocite and thioarsenate.
ii) Oxygen pressure oxidation of As₂S₅ in acid solution to form scorodite. Assume all sulfide sulfur forms sulfate. (Assume that ferric ion is present in solution).

Homework Equations

N/A

The Attempt at a Solution

i) Cu₃AsS₄ + HS^- = Cu₂S + AsS₄^3-
Balancing:
Cu₃AsS₄ + HS^- = 3/2Cu₂S + AsS₄^3-
Cu₃AsS₄ + HS^- = 3/2Cu₂S + AsS₄^3- +H⁺
Cu₃AsS₄ + HS^- +e^-= 3/2Cu₂S + AsS₄^3- +H⁺

? I cannot seem to balance the last equation properly. I do know that the e^- present in the equation means that there should be 2 half reactions, not one as above, but cannot figure those out either.

ii) As₂S₅+O₂ = AsO₄^3-+SO₄^2-

Left side:
As₂S₅ = AsO₄^3-+SO₄^2-
As₂S₅ = 2AsO₄^3-+5SO₄^2-
As₂S₅+28H₂O= 2AsO₄^3-+5SO₄^2-
As₂S₅+28H₂O= 2AsO₄^3-+5SO₄^2-+56H⁺
As₂S₅+28H₂O= 2AsO₄^3-+5SO₄^2-+56H⁺+40e^-

Right side:
O₂ = ?
O₂ = 2H₂O
O₂ + 4H⁺= 2H₂O
O₂ + 4H⁺+4e^-= 2H₂O

Add together (and cancel out e^-):
As₂S₅+28H₂O= 2AsO₄^3-+5SO₄^2-+56H⁺+40e^-
10*(O₂ + 4H⁺+4e^-= 2H₂O)

Ionic Form ===> As₂S₅+8H₂O+10O₂ = 2AsO₄^3-+5SO₄^2-+16H⁺

It is balanced (I checked).

But, I get into trouble when I try to get the neutral form. Nothing on the right requires ferric (the cation I use to put all aq terms into neutral form), thus there is an imbalance of iron. Also, how do I deal with the 2H2O that appears as part of scorodite? Do I just assume that it has become part of the solution?

EDIT: Figured out 1)ii) Use H+ (and NOT ferric, Fe³⁺) as the cation to turn products into neutral compounds (and thus the entire reaction into neutral form).
i.e. As₂S₅+8H₂O+10O₂ = 2H₃AsO₄+5H₂SO₄

 

[Borek]

i) Enargite leaching by sodium bisulfide in basic solution to form chalcocite and thioarsenate.

In basic solution means you should use water and OH^- to balance hydrogen and oxygen. I don't see them in your equation.

Have you tried to decide what is getting oxidized and what is getting reduced?

 

[1question]

I solved it:

Cu3AsS4 + 3/2HS- = 3/2Cu2S + AsS43-

I can't believe I didn't see that earlier. Thanks for the reply!

(Why can't I edit my post after a given amount of time? It would be nice to be able to grab the formatted stuff from there and change it to suit my current needs...)

 

[Borek]

Cu3AsS4 + 3/2HS- = 3/2Cu2S + AsS43-

And where is the hydrogen on the right?

(Why can't I edit my post after a given amount of time? It would be nice to be able to grab the formatted stuff from there and change it to suit my current needs...)

I am afraid we had to do so because some people abuse the editing system. Locking the post means you can't edit the post once caught on posting nonsense.

 

[1question]

@Borek Yes, the 3/2H⁺ on the right exists, I just didn't put it on here, sorry.

Ah, I see. Thanks.

 

[Borek]

What about balancing charge? -3/2 on the left, -3 on the right.