[Sis290025]

What is the chemical formula for a potassium iodide-iodate solution? I know that potassium iodide is KI and potassium iodate is KIO3.

Thank you for any help.

[Alberto_Kravina]

Iodide is oxidized to elementar iodine with iodate.

[Sis290025]

From that, is this setup even remotely correct?

KI + KIO3 --> I2

(I'm trying to understand the oxidation in this situation. Thanks.)

[Alberto_Kravina]

From that, is this setup even remotely correct?

KI + KIO3 --> I2

(I'm trying to understand the oxidation in this situation. Thanks.)

Yeah! Remotely correct! Try to write two half-reactions!

[Sis290025]

I am assuming that this solution is in a basic solution and not acidic:

IO3- + I- --> I2

a. Oxidized: IO3- --> I2
b. Reduced: I- --> I2

a. IO3- --> I2
 12OH- + 12H+ + 2IO3- --> I2  + 6H2O
10e- + 12H2O + + 2IO3- --> I2 + 6H2O + 12OH-
10e- + 6H2O + 2IO3- --> I2 + 12OH-

b. I- --> I2
2I- --> I2
2I- --> I2 + 2e-
5(2I- --> I2 + 2e-)= 10I- --> 5I2 +10e-


10e- + 6H2O + 2IO3- --> I2 + 12OH-
10I- --> 5I2 +10e-
__________________________________________
6H2O + 2IO3- + 10I- --> I2 + 12OH- + 5I2




Thanks.
 

[Albert]

I am assuming that this solution is in a basic solution and not acidic:

WHY?!

IO3- --> I2

This reaction takes place in acidic conditions.

[Sis290025]

Why does it take place under acidic conditions? I don't understand that part.



Thanks again.

Now that this is the method for a potassium iodide-iodate solution, how do I write the equation for the reaction between this iodide-iodate solution with sulfuric acid, H2SO4?

[Albert]

I read your previous post again. The equation WAS balanced also under basic conditions. However, when I performed this very titration, I carried it out under acidic condition. Moreover, it's easier to balance reactions this way.  

[Albert]

Now that this is the method for a potassium iodide-iodate solution, how do I write the equation for the reaction between this iodide-iodate solution with sulfuric acid, H2SO4?

I think it's just potassium iodide which reacts with sulfuric acid.

[Borek]

Oxyacids oxidation potential is higher in low pH, that's whay they are used in solutions of strong acids. Cl^- gets easily oxidized to Cl₂, so sulfuric acid is preferred over hydrochloric for solution acidification.

[Sis290025]

Basically, I'm trying to find the molarity of a titrant sodium thiosulfate solution (containing NaOH and Na2S2O3) through standardization in which a mixture of H2SO4 and the potassium iodide-iodate solution, along with starch indicator, are involved.

The setup of equations is giving some difficulty. Can anyone direct me, or give me hints, on how to write out the equations involved, so I can eventually find the molarity?

Thanks for any help.

[Borek]

IO₃^- + I^- + H⁺ -> I₂ + H₂O

I₂ + S₂O₃^2- -> I^- + S₄O₆^2-

Note these are not balanced.

[Sis290025]

To balance them, how can I tell if it is a basic or an acidic condition?

To find the molarity, I would use the ratio of mol of IO3- to mol S2O3(2+)?

 

[Borek]

Both go in low pH, but it doesn't matter, just balance them.

Amount od iodine produced is defined by mass of iodate used, rest is simple stoichiometry.

[Sis290025]

For second part,

I2 --> 2I-
2e- + I2 --> 2I-

2S2O3(2-) --> S4O6(2-) + 2e-
2e- + I2 --> 2I-
________________________________
2S2O3(2-) + I2 --> 2I- + S4O6(2-)

First part:
using change-in-oxidation method-

IO3- + 5I- +6H+ --> 3I2 + 3H2O