Just one simple question... so simple that it bugs me when I
can't determine which is the preferred answer.
Alright, so we were asked how we can distinguish between I2 and
Br2.
There were 2 options I find quite Ok
First is reaction with sodium thoiosulfate. This seems
reasonable because we know that I- ions will become I2 and Br-
becomes NaBr. If I am not wrong, NaBr should be white (as is all
group I ionic compounds?) and I2 gives a yellow solution
Second option is using aq Fe2+ then add aq NaOH.
we know I2 does not react witih Fe2+... so Br2 WILL, giving Fe3+
and Br -. NaOH and react with Fe3+ to give Fe(OH)3 ppt?
I chose the first answer anyway... was the answer that just
"came" to me at that time. Any other take? Thanks
Method 1 :
Assuming standard molarities, thiosulfate will indeed reduce both
iodine and bromine, to iodide and bromide respectively, and is
itself oxidized to tetrathionate and sulfate(VI)
respectively. All of which are (equally)
colourless.
Method 2 :
Assuming standard molarities, calculating cell potential (ie.
reduction potential at cathode + oxidation potential at anode) will
show that only one halogen is a sufficiently strong oxidizing agent
to oxidize Fe2+ to Fe3+. Thereafter, adding OH-
will precipitate Fe(OH)2(s) and Fe(OH)3(s) in the two separate
test solutions, which have distinct colours of dark
green and reddish brown respectively.