Identifying the Reducing Agent in a Spontaneous Reaction

In summary, in an acidic solution, the hydronium would not be part of any mass or charge balancing and the hypochlorite salt would be created.
  • #1
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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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  • #2
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.
 
  • #3
so the I2 is the reducing agent?
 
  • #4
Yes.
 
  • #5
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?
 
  • #6
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.
 
  • #7
what would we do in an acidic solution when we need to blance charges?
 
  • #8
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.
 

Related to Identifying the Reducing Agent in a Spontaneous Reaction

1. What is a reducing agent?

A reducing agent is a substance that donates electrons in a chemical reaction, causing another substance to be reduced and gain electrons.

2. How do you identify the reducing agent in a spontaneous reaction?

The reducing agent can be identified by looking at the oxidation numbers of each element involved in the reaction. The element that decreases its oxidation number is the reducing agent.

3. Can the reducing agent change during a reaction?

Yes, the reducing agent can change during a reaction. It starts off with a higher oxidation number and donates electrons, causing its oxidation number to decrease and making it the reducing agent.

4. What role does the reducing agent play in a spontaneous reaction?

The reducing agent is essential in a spontaneous reaction as it provides the necessary electrons to drive the reaction forward. Without a reducing agent, the reaction would not occur spontaneously.

5. Are all spontaneous reactions driven by reducing agents?

No, not all spontaneous reactions are driven by reducing agents. Some reactions may be spontaneous due to the thermodynamic properties of the reactants, without the need for a reducing agent. However, many spontaneous reactions do involve reducing agents.

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