Identify Spectator: Bromide Reacts with Acidified NaClO4 - Test Tomorrow

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The discussion revolves around identifying bromide as a spectator ion in the reaction between potassium bromide (KBr) and acidified sodium perchlorate (NaClO4). Participants clarify that bromide cannot be a spectator if it is oxidized to bromine (Br2) during the reaction, as indicated by the question asking for the mass of bromine produced. The correct approach involves writing the balanced equation, which reveals that the reaction is a redox process rather than a simple double displacement. The conversation emphasizes that the presence of Br2 as a product confirms the reaction's nature and the role of bromide. Understanding the reaction dynamics and the specific question asked is crucial for determining whether an ion is a spectator.
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I am really confused on how to identify the spectator. Like this question
what mass of bromine will be produced if a potassium bromide solution reacts with 50.0mL of a 1.20 mol/L solution of acidified NaClO4?
How is the mass formed, when bromide is a spectator?
i have a test tomorrow, if anyone can help me...
ty!
 
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This should really be in the homework section, but let's start with the obvious:

How can bromide be a spectator when it is oxidized from Br- to Br2?

Try writing the balanced equation for the reaction. (Actually, this should always be the first step to solving a problem)
 
PhaseShifter said:
How can bromide be a spectator when it is oxidized from Br- to Br2?

Try writing the balanced equation for the reaction. (Actually, this should always be the first step to solving a problem)

But how would you know if bromide is oxidized ? is it just because the question asked what mass of bromine is formed?
For this, if sodium phosphate react with calcium bromide,then phosphate and bromide will be the spectator..
But how do determine whether bromide is a spectator or not?
btw,KBr+NaClO4-->KClO4+NaBr
is that right?
ty!
 
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The question states that reactants are KBr (aq) and NaClO4 (aq), and one of the products is Br2.

If the bromine starts as Br- and ends up as Br2, it must have been oxidized by something...and therefore can't be a spectator.

It's also no coincidence that perchlorates are powerful oxidizing agents.
 
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KBr+NaClO4-->KClO4+NaBr
Sorry, it still doesn't make sense to me...
I don't see how is Br2 formed...
unless my equation is wrong..
 
Suy said:
KBr+NaClO4-->KClO4+NaBr
Sorry, it still doesn't make sense to me...
I don't see how is Br2 formed...
unless my equation is wrong..

Your equation is wrong...the problem states that Br2 is a product, so it must be a redox reaction rather than double displacement reaction.
 
PhaseShifter said:
Your equation is wrong...the problem states that Br2 is a product, so it must be a redox reaction rather than double displacement reaction.

Is that mean, if the question didn't state Br2 as a product, then Br will be a spectator?
 
Not always...for example the problem could ask about one of the other products of the same reaction.

But what is relevant in this case, is the problem asks how much Br2 is formed...which means an equation without Br2 among the products is not useful for solving the problem, and another equation must be used.

Also, your equation doesn't work because both products are soluble as well as both reactants...meaning all four compounds would be completely dissociated into solvated ions, which would mean all the ions are spectators, and no reaction occurs according to that equation.

Consider the electronegativities of bromine and chlorine, and look at the oxidation numbers in both reactants.
 
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Thanks, that's make sense to me now. It's all depend how the question ask..
 
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