Im confused about this chemistry question, why does it form these products
The question:
Bromine disproportionates when it reacts with potassium hydroxide solution.
Suggest an equation for this reaction.
My answer:
Br2 +KOH -> HBr +KBr +H2O
The actual answer:
Br2 + 2KOH -> KBr + KBrO (why form this and not HBr) +H2O
Bromine disproportionates when it reacts with potassium hydroxide solution.
Suggest an equation for this reaction.
My answer:
Br2 +KOH -> HBr +KBr +H2O
The actual answer:
Br2 + 2KOH -> KBr + KBrO (why form this and not HBr) +H2O
i think your problem is that it asks for a disproportionation ("bromine disproportionates when reacted...") so the Br2 needs to be both oxidised and reduced in one equation.
in your answer the Br is reduced twice: 0 to -1 to KBr and 0 to -1 to HBr; if you look in the actual answer Br is oxidised in KBrO (0 to +1) and reduced in KBr (0 to -1)
in your answer the Br is reduced twice: 0 to -1 to KBr and 0 to -1 to HBr; if you look in the actual answer Br is oxidised in KBrO (0 to +1) and reduced in KBr (0 to -1)
obviously it's really easy to make such mistakes as though your equation is balanced, and would be correct if it was just stating a Br and KOH equation, you need to make sure you do exactly what it's asking for
they really love catching you out on little things like wording especially when it comes to constructing equations haha
they really love catching you out on little things like wording especially when it comes to constructing equations haha
Original post by MP_25
i think your problem is that it asks for a disproportionation ("bromine disproportionates when reacted...") so the Br2 needs to be both oxidised and reduced in one equation.
in your answer the Br is reduced twice: 0 to -1 to KBr and 0 to -1 to HBr; if you look in the actual answer Br is oxidised in KBrO (0 to +1) and reduced in KBr (0 to -1)
in your answer the Br is reduced twice: 0 to -1 to KBr and 0 to -1 to HBr; if you look in the actual answer Br is oxidised in KBrO (0 to +1) and reduced in KBr (0 to -1)
The br2 doesn’t need to be oxidised and reduced, just has to have different oxidation states
Original post by MP_25
i think your problem is that it asks for a disproportionation ("bromine disproportionates when reacted...") so the Br2 needs to be both oxidised and reduced in one equation.
in your answer the Br is reduced twice: 0 to -1 to KBr and 0 to -1 to HBr; if you look in the actual answer Br is oxidised in KBrO (0 to +1) and reduced in KBr (0 to -1)
in your answer the Br is reduced twice: 0 to -1 to KBr and 0 to -1 to HBr; if you look in the actual answer Br is oxidised in KBrO (0 to +1) and reduced in KBr (0 to -1)
Ah alright, Ty!
Original post by fffffffgt
what about HBrO instead of KBrO whch would work too right? and then with no water
You won't make HBrO (which is an acid) when one of the reactants is KOH (which is a base). The acid product would instantly be neutralised, forming KBrO and water.
Original post by Pigster
You won't make HBrO (which is an acid) when one of the reactants is KOH (which is a base). The acid product would instantly be neutralised, forming KBrO and water.
K BrO - very HipHop
Original post by AntonSlayer
The br2 doesn’t need to be oxidised and reduced, just has to have different oxidation states
the definition of something undergoing disproportionation is that it gets reduced (gains some electrons) and also gets oxidised (loses some electrons) in the same equation instead of just having diff oxidation states. i hope that makes sense? ☺️
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