- Joined
- Dec 18, 2006
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When Chad is doing the nonspontaneous (electrolytic) reaction
NaBr(l) ---> Na + Br2
he says the half reactions are
Na^+ + e^- ---> Na
2Br^- ---> Br2 + 2e^-
Cool, I get that because Na gets reduced, then Na is the negative cathode
and that because 2Br^- gets oxidized, then Br2 is the positive anode.
I understand the anode is positive because these elements (Na and Br2) cannot exist in elemental form in the natural world, but can be made to exist by a nonspontaneous reaction.
What I don't understand is how the Br2 (the positive anode) is receiving electrons (from electricity) and is forming an anion in solution: 2Br^-
If Br2 as the anode is consuming electrons (from electricity) and passes on those electrons (LOSES electrons) to the cathode...
Shouldn't a 2Br^+ form in solution instead of 2Br^- ???
Because the way the half reaction is written...it looks wrong to me, but idk...
.....helppppp please!
NaBr(l) ---> Na + Br2
he says the half reactions are
Na^+ + e^- ---> Na
2Br^- ---> Br2 + 2e^-
Cool, I get that because Na gets reduced, then Na is the negative cathode
and that because 2Br^- gets oxidized, then Br2 is the positive anode.
I understand the anode is positive because these elements (Na and Br2) cannot exist in elemental form in the natural world, but can be made to exist by a nonspontaneous reaction.
What I don't understand is how the Br2 (the positive anode) is receiving electrons (from electricity) and is forming an anion in solution: 2Br^-
If Br2 as the anode is consuming electrons (from electricity) and passes on those electrons (LOSES electrons) to the cathode...
Shouldn't a 2Br^+ form in solution instead of 2Br^- ???
Because the way the half reaction is written...it looks wrong to me, but idk...
.....helppppp please!
