[Glitch000]

hi, can someone help me balance this redox reaction please?

ClO₃^- + As₂S₃ > Cl^- + H₂AsO₄^- + SO₄^2-

Thank you

[Borek]

What have you tried so far?

Do you want to use oxidation number method, or half reaction method?

[Glitch000]

first of all i found the oxidation number of the single atoms:

reagents:
Cl = +V
As₂= +III
S₃= -II

products:
Cl^-= -I
As= +V
S= +VI

Then i wrote the half-reaction of oxidation and reduction:

ox: As₂³⁺ > 2AsO₄^3- + 4e^-

is this right? i don't know how to continue the balance of this half-reaction... now i have to balance the charge right?
the reaction is in an acid solution

[Glitch000]

I've also done the other half-reaction btw...

Ox: 14H₂O + S₃^2- > 3SO₄^2- + 24e^- + 28H⁺

Red: 6H⁺ + 6e^- + ClO₃^- > Cl^- + 3H₂O

Is this correct?

[mjc123]

Don't write As₂³⁺. That looks like something with 2 As atoms and a charge of 3+, i.e. an average oxidation number of 1.5. That's not what you mean (I hope). It will lead to confusion in balancing the charges. Write 2As³⁺. Likewise S₃^2- (this is a species that actually exists, so it is important not to get confused with it). You mean 3S^2-. Your equation is unbalanced as regards charge, H and O.

[Glitch000]

Oh you're right, my bad! i meant 2As³⁺ and 3S^2-...

May i ask you a question... since there is the 2 as a stoichiometric coefficient in
 2As³⁺ > 2AsO₄^3- + 4e^-
i have to multiply by 2 also H⁺ and H₂O?
And also, to balance the charge, i have to consider the stoichiometric coefficient? i mean, the total charge of the products is -10 ( since 2As^3- has a total charge of -6 ( i multiplied -3 x 2) and then 4e^-) or -6?
and the same for the reagent... (2As³⁺) is it +3 or +6?
sorry for my bad explanation, i'm not very good in english haha, hope you understand

[mjc123]

You need the right number of H⁺ and H₂O to balance the equation. You also have to balance the charges, and yes you have to consider the stoichiometric coefficient. (What do you think it means?) So your equation as written has +6 on the left and -10 on the right.