Elemental iodine from Potassium Iodide - Reaction Equations

[Astrum]

In the reaction to get I₂ from an aqueous solution of KI.

The process. Take the aqeuous KI, and pour concentrated hydrochloric acid, followed by hydrogen peroxide.

K⁺+I^-+H⁺+Cl^-+H₂O$\rightarrow$ KCl + HI + H₂O

The reaction with H₂O₂ is perplexing.

I'm a physics student trying to do chemistry, I'm sure I made some mistake in writing out that chemical equation (I forgot if ions have to be separated in the equation, when in a solute).

What are the specifics of this chain of reactions? The elemental iodine will precipitate out of solution. The H_2O2 must be oxidizing something in the K⁺ + Cl^- + H⁺ + I^- + H₂O

 

[DrDu]

The Iodine gets oxidized:
$\rm 2 I^-\rightarrow I_2+2e^-$
Hydrogen peroxide gets reduced:
$\rm H_2O_2 +2e^-+2H^+\rightarrow 2 H_2O$
The protons on the LHS stem from the hydrochloric acid, that's why you add acid.

 

[Astrum]

I see, hydrogen peroxide is a oxidizing agent, I seemed to have missed that.

Would $H_{2}O_{2}$ react with KI in the absence of any hydrogen ions?

The net equation would be $2I^{-}+2K^{+}+2H^{+}+2Cl^{-}+H_{2}O_{2}\rightarrow I_{2} + 2KCl + 2H_{2}O$ ?

 

[DrDu]

Would $H_{2}O_{2}$ react with KI in the absence of any hydrogen ions?

The net equation is correct. As soon as you use water as a solvent there will always be hydrogen ions around due to the autoprotolysis of water. I am not sure whether the reaction would also take place in an alkaline medium.

 

[hilbert2]

I am not sure whether the reaction would also take place in an alkaline medium.

In alkaline solution of $H_{2}O_{2}$ the actual oxidant is the perhydroxyl ion $HO^{-}_{2}$, for which we have the redox half-reaction

$HO^{-}_{2}+H_{2}O+2e^{-} \rightarrow 3OH^{-}$ $E^{0}$ = +0.87 V

The redox potential is higher than that of iodine (+0.59 V), so the reaction should also happen in alkaline medium.

 

[Astrum]

I'm wondering why the $H_2O_2$ doesn't oxidize the chlorine anion as an equal amount to the oxidation of iodide.

This process probably produces some amount of $Cl_2$.

 

[Corribus]

Chloride is significantly harder to oxidize than iodide. Look up the redox potentials.