The reason is because only the O.S. of the S atoms changes,
while the O.S. of the O atoms remain as -2 (this is because O is
more electronegative than S; oxidation state = formal charge +
electronegativity consideration, meaning O.S. is the charge of an
atom after taking electronegativity into account).
To reiterate, drawing out the structure of the thiosulfate
S2O3 2- ion, you'll see that the two S atoms in thiosulfate S2O3 2-
each has a different O.S. : +0 and +4.
In terms of O.S., 1 of the S atoms (with initial O.S. of
+4) loses 2 electrons, while the other S atom (with initial
O.S. of +0) loses 6 electrons.
This makes a total of 8 electrons, used for reducing 4Cl2 to
8Cl-.
The thiosulfate S2O3 2- ion then rearranges
itself (upon losing the 8 e-) into 2 sulfate(VI) SO4 2-
ions.
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1. When potassium iodide is added to acidified H2O2, a brown
solution of iodine appears. No oxygen is obtained.
[reduction] H2O2 --> H2O
[oxidation] I- --> I2
Balance the half equations above, then combine them to obtain
the balanced redox equation.
(Note that the brown solution is the uninegative triiodide ion
I3-(aq), but unless otherwise specified by the question, it's ok to
write it as I2(aq). Iodine being non-polar, does not dissolve well
in hydrogen-bonded water.)
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'A' Level Qn
Draw the uninegative iodine dichloride ion. State all formal
charges and oxidation states within the ion. State its electron and
ionic geometry.
Balance the following equation :
IO3(-) + I- + H+
+ Cl- ---> ICl2(-) + H2O
Solution :
In the iodine dichloride ion, [Cl-I-Cl]-, the 2 Cl atoms have no
formal charge, and an O.S. of -1. The I atom has a 1- formal
charge, and an O.S. of +1. The sum of formal charges, as well
as the sum of oxidation states, adds up to the uninegative
ionic charge.
Electron geometry = trigonal bipyramidal.
Ionic geometry = linear.
What is happening during the redox reaction : the iodate(V) ion
is reduced, and the iodide ion is oxidized, both becoming the
iodine atom in the iodine dichloride ion (O.S. of +1). Notice that
the O.S. of Cl does not change (-1 before redox, -1 after
redox).
[reduction] 4e- + 6H+ + 2Cl- + IO3(-) ---> [Cl-I-Cl]- +
3H2O
[oxidation] 2Cl- + I- ---> [Cl-I-Cl]- + 2e-
[balanced redox] 6Cl- + 6H+ + 2I- + IO3(-) --->
3[Cl-I-Cl]- + 3H2O