[vanderwaals]

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–  2 H2O(l) E° = 1.23 V
Br2(l) + 2e–  2 Br–(aq) E° = 1.07 V
2 H2O(l) + 2 e–  H2(g) + 2 OH– E° = –0.80 V
K+(aq) + e–  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?

[AWK]

Br2 is a liquid slightly soluble in water. It also reacts with water and phenophtalein

[vanderwaals]

Thanks for the response! So just to confirm, Br2(l) and not Br2(g) forms at the anode, right?

[AWK]

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).

[Borek]

Thanks for the response! So just to confirm, Br2(l) and not Br2(g) forms at the anode, right?

I would say Br₂(aq).