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?