Identifying the Reducing Agent in a Spontaneous Reaction

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SUMMARY

The reducing agent in the spontaneous reaction involving OCl-, I2, and OH- is I2, as it is oxidized while reducing another species. The hydroxide ion (OH-) does not undergo oxidation or reduction; it serves to stabilize the hypochlorite ion (OCl-) in the reaction. The reaction can be balanced in alkaline conditions, where the presence of hydroxide enhances the stability of hypochlorite. Understanding the roles of reducing and oxidizing agents is crucial in redox chemistry.

PREREQUISITES
  • Understanding of redox reactions and electron transfer.
  • Familiarity with oxidation states and balancing chemical equations.
  • Knowledge of the stability of hypochlorite and hypochlorous acid.
  • Basic concepts of alkaline and acidic solutions in chemical reactions.
NEXT STEPS
  • Study the principles of redox reactions in detail.
  • Learn about the stability and reactions of hypochlorite and hypochlorous acid.
  • Explore balancing chemical equations in both alkaline and acidic conditions.
  • Investigate the role of hydroxide ions in various chemical reactions.
USEFUL FOR

Chemistry students, educators, and professionals involved in redox chemistry and reaction mechanisms will benefit from this discussion.

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Homework Statement



What is the reducing agent in the following spontaneous reaction?

OCl-(aq) + I2(s) + 2 OH-(aq) --->
Cl-(aq) + 2 OI-(aq) + H2O(l)



Homework Equations



The reducing agent is oxidized in the process so it loses electrons.

The Attempt at a Solution



Both I2 and OH- lose electrons so I am not sure which one it is, any help?
 
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Not as much as that. The hydroxide does not change any of its reduction status.

The reducing agent reduces some part of another item; the oxidizing agent oxidizes some part of an item. Essentially, the reducing agent becomes oxidized, and the oxidizing agent becomes reduced. Most of the time, you want to check reduction status of some atom other than oxygens.
 
so the I2 is the reducing agent?
 
Yes.
 
Thanks..so just so i get why:
the I2 is the reduceing agent because it reduces another atom and at the same time is oxidized, but the OH- is only oxidized but didnt cause another atom to be reduced?
 
You made an interesting interpretation, but the hydroxide is not reduced and is not oxidized in the reaction. The hydroxide is used in the balancing of the equation but does not go through oxidation or reduction. One reason why the hydroxide is needed is that it makes the hypochlorite more stable and available for the redox reaction. The acid form, hypochlorous acid, HOCl, is not very stable and can decompose. The neutralized anion, OCl-, is more stable but needs to be the salt form for stability.

When you pump chlorine gas into alkaline solution (hydroxide present), you get hypochlorite salt. This is why your question in post #1 starts with hypochlorite in alkaline solution.
 
what would we do in an acidic solution when we need to blance charges?
 
My best guess for that reaction in acidic solution is that the hydronium would not be part of any mass or charge balancing; I'm really not so certain in these conditions. Maybe someone else can give better understanding. I could only best imagine showing two hydrogen ions on the left and maybe just two HIO on the right, or just two hydrogen ions (therefore redundant) on the rightside. Like I say, my guess for this condition is not secure and someone else should help respond. It's been a long time.
 

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