Re: Balancing of halfreactions.
- From: "Mathias Rocher" <MXR_no_554_spam@xxxxxxxxxx>
- Date: Sat, 14 Jan 2006 13:13:21 -0000
Sorry Oddveig but it's my turn to ask questions !
Did you read my two last answers ? In the first mail I explained how to
balance a half reaction, taking an exemple with your second halfreaction,
the oxidation of S to S03(2-). In the second I gave you the answer of the
whole problem ! What can I do next ? I am not a teacher, and that's not my
job to explain you what your teacher in chemistry can explain better than me
! In particular, I don't want to solve an exercise on which you should work
with what you learned. If you want to understand how to balance halfreaction
and be able to re-do it yourself, I can show you on other examples, but
don't expect to understand anything if you don't try by yourself. That's why
I will gave a few exercises at the end for you. Come back with your answer !
Let's take the general oxidation of a metal to a cation: it's a very simple
reaction to balance. You need to go from M, the metal, to the cation M(n+),
the same metal minus three electrons. The halfreaction will simply be:
M -> M(n+) + n.e(-)
A little bit more difficult : the formation of an oxide. It's the case of
the oxidation of S to SO3(2-), it can be also applied to the oxidation of
manganese Mn to permanganate MnO(4)(-), of iron Fe to Fe(2)O(3), etc... What
you want to do in a general case is to go from M to M(x)O(y)(n-). It means
you need x M atoms, and y O atoms. let's ignore the electrical charge at
this state. The x M atoms have to be there at the begining, as well as the y
O atoms. For oxygen, we will consider it comes from water. It means that we
will introduce H atoms in the reactions, which atoms will be there at the
end of course. We will get them as H(+) ions. The last step is to
equilibrate the charges with electrons so that we have the same global
charge on each side of the halfreaction, in this case 0. The general
solution is then:
x.M + y.H(2)O -> M(x)O(y)(n-) + 2y.H(+) + (2y - n).e(-)
Don't try to remember this general equation, it's not useful and to
complicated ! Remember the method and its logic, because the same logic can
be applied to more general case. It's the same logic for a reduction. Last
thing you need for your exercise: how to equilibrate a halfreaction in basic
conditions. It's very simple: follow the same logic until you get the
equilibrated reaction as shown above (that means, in neutral or acidic
aqueous conditions). Then add to your equation as much HO(-) as you want
(that means 2y HO(-) in that case), change all the [H(+) + HO(-)] into
H(2)O, and erase all the H(2)O molecules that you don't need on each side of
the reaction. In that case it means you have to remove y HO(-) on each side.
General solution:
x.M + 2y.HO(-) -> M(x)O(y)(n-) + y.H(2)O + (2y - n).e(-)
Now you can answer those halfreaction (I will not say if they are reductions
or oxidations):
from dihydrogen H(2) to H(+)
from chlorine Cl(2) to chloride Cl(-)
from nitrobenzene PhNO(2) to aniline PhNH(2) (Ph is for "phenyl", C(6)H(5))
in acidic conditions
from hypochloride ClO(4)(-) to chlorine Cl(2) in basic conditions (that's
what happens when you mix bleach with caustic soda, it causes many deaths
each year)
.
- References:
- Balancing of halfreactions.
- From: Oddveig
- Re: Balancing of halfreactions.
- From: Mathias Rocher
- Re: Balancing of halfreactions.
- From: Oddveig
- Balancing of halfreactions.
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