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Topic: Electrolysis of KBr  (Read 4509 times)

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Offline vanderwaals

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Electrolysis of KBr
« on: February 14, 2017, 03:32:10 PM »
Use the standard reduction potentials to determine what is
observed at the cathode during the electrolysis of a 1.0 M
solution of KBr that contains phenolphthalein. What
observation(s) is(are) made?
O2(g) + 4 H+(aq) + 4 e–  :rarrow: 2 H2O(l) E° = 1.23 V
Br2(l) + 2e–  :rarrow: 2 Br–(aq) E° = 1.07 V
2 H2O(l) + 2 e–  :rarrow: H2(g) + 2 OH– E° = –0.80 V
K+(aq) + e–  :rarrow: K(s) E° = –2.92 V
(A) Solid metal forms.
(B) Bubbles form and a pink color appears.
(C) Dark red Br2(aq) forms.
(D) Bubbles form and the solution remains colorless.


The answer is (B). So I know that H2 gas and OH- are formed at the cathode, so that's where the pink color comes from. But what exactly forms at the anode? Is it Br2 gas? Why wouldn't it be Br2 liquid?

Offline AWK

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Re: Electrolysis of KBr
« Reply #1 on: February 14, 2017, 03:46:46 PM »
Br2 is a liquid slightly soluble in water. It also reacts with water and phenophtalein
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Offline vanderwaals

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Re: Electrolysis of KBr
« Reply #2 on: February 14, 2017, 03:53:12 PM »
Thanks for the response! So just to confirm, Br2(l) and not Br2(g) forms at the anode, right?

Offline AWK

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Re: Electrolysis of KBr
« Reply #3 on: February 14, 2017, 04:12:53 PM »
I use this method for preparing small amounts of bromine water (eg for test of double bond in organic compounds). Phenols (phenolphtalein) also easily react with bromine.
Br2 is a volatile liquid (not gas).
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Offline Borek

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Re: Electrolysis of KBr
« Reply #4 on: February 14, 2017, 06:25:10 PM »
Thanks for the response! So just to confirm, Br2(l) and not Br2(g) forms at the anode, right?

I would say Br2(aq).
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