Elemental iodine from Potassium Iodide - Reaction Equations

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SUMMARY

The discussion centers on the chemical reaction to produce elemental iodine (I2) from an aqueous solution of potassium iodide (KI) using concentrated hydrochloric acid (HCl) and hydrogen peroxide (H2O2). The net reaction is confirmed as 2I- + 2K+ + 2H+ + 2Cl- + H2O2 → I2 + 2KCl + 2H2O. The role of H2O2 as an oxidizing agent is emphasized, particularly in acidic conditions where protons are present. The discussion also touches on the potential for the reaction to occur in alkaline conditions, where the perhydroxyl ion (HO-2) serves as the oxidant.

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  • Understanding of redox reactions and oxidation states
  • Familiarity with chemical equations and stoichiometry
  • Knowledge of acid-base chemistry, particularly the role of protons
  • Basic concepts of electrochemistry, including redox potentials
NEXT STEPS
  • Research the redox potentials of common oxidizing agents, including H2O2 and iodine
  • Explore the effects of pH on redox reactions, particularly in alkaline solutions
  • Study the autoprotolysis of water and its implications in chemical reactions
  • Investigate the mechanisms of oxidation for halides, focusing on chloride and iodide
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Chemistry students, particularly those studying redox reactions, as well as educators and researchers interested in the behavior of iodine and hydrogen peroxide in various chemical environments.

Astrum
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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 H2O2 must be oxidizing something in the K+ + Cl- + H+ + I- + H2O
 
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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.
 
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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 ?
 
Astrum said:
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.
 
DrDu said:
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.
 
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.
 
Chloride is significantly harder to oxidize than iodide. Look up the redox potentials.
 

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