***@hotmail.com wrote:
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Post by f***@hotmail.comPost by p***@woelen.nlSCN(-) --> S(6+) + CN(-) + 6e
Six electron transfer means that sulfur here is 0 oxidation state
(absolute oxidation number) which is possible eg in S4O6(2-), two
central sulfurs have zero oxidation number if its Lewis structure is
[O3S-S-S-SO3](2-). If you wish I can write the details of calculating
ON from Lewis structures.
I have a textbook which explains Lewis structures. The problem with
this is that results sometimes are ambiguous (probably I missed
something in the description).
If I were to determine the ox.num. of the atoms I would come up with
the following structure (please read this in Courier typeface):
O O
|| ||
O = S - S - S - S = O
| |
O(-) O(-)
O: -2
side S have 5 bonds to O's and one bond to S. This makes +5 and 0 for
the S atoms or +6 and -1 for the S atoms. Which one is best, I cannot
decide.
I indeed would like to see your method for determining ox.num.
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Post by f***@hotmail.comI think I don't undertsnad your point. You are implicitly assuming that
S in SCN was in 0 oxidation state.
S(0)CN- -----> S(VI)O4 (2-) + 6e
No, what I do is looking at the whole of the ion. In your reasoning
only the S changes oxidation number, but what if the C and/or the N
also do change their oxidation number at the same time?
I think the following are equivalent: In the first line the sulphur has
oxidation state 0 and goes to oxidation state +6. In the second line
the sulphur has oxidation state -2 and goes to oxidation state +6, but
at the same time the C goes from +4 to +2, which is accounted for by
taking 2 electrons on the left.
[S(0)C(+2)N(-3)] ---> S(+6) + [C(+2)N(-3)] + 6e
[S(-2)C(+4)N(-3)] + 2e ---> S(+6) + [C(+2)N(-3)] + 8e
This is what I mean by looking at the total. Although ox.nums are quite
different in the two equations, the net effect is the same. Used in
this way, it is nothing more than a bookkeeping method.
A nice other example is reduction of the dark blue CrO5 (better
CrO(O2)2) to Cr(3+) on addition of a reductor, such as sulfite. Here we
have oxidation numbers +6 for the central chromium atom, -1 for the
oxygen atoms of the two peroxo groups and -2 for the remaining oxygen
atom . Finally we end up with oxidation numbers +3 and -2.
[Cr(+6)O(-2)O(-1)O(-1)O(-1)O(-1)] + 3e + 4e --> Cr(3+) + 5[O(2-)]
Here we see a formal equation with 4 electrons taken at the left for
the four oxygens in the peroxo groups and 3 electrons taken at the left
for the central chromium atom. In one molecule there are two species
changing their oxidation state.
I hope that this explanation makes my reasoning clear to you.
I'll post a message on determination of oxidation numbers of SCN(-) and
CN(-). See what answers we get now.
Post by f***@hotmail.comCan we try solving the equation algebraically, with lot of
simultaneous
Post by f***@hotmail.comequations.
aMnO4(-) + bSCN(-) + cH(+) ---------> dMn(2+) + eHCN + fSO4(2-) + gH2O
and relating the coefficients together
say a = d ---(1)
4a = 4f + g ---(2)
b = f ---(3)
and many more.
Please check that whether the system of equations is correct or not.
There is no harm in asking again, as there are variety of answers in
the post you referred to, many suggest zero and some other -2.
Several years ago I wrote a computer program for this kind of problems.
Just entering the reactants at one side and entering the reaction
products at the other side. Then equations are determined for each
element, and for the total charge. This gives a homogeneous equation of
the form Ax=0, where A is a matrix, depending on the numbers in the
formula's (e.g MnO4(-) gives a 1 in A for Mn, a 4 for O and a -1 for
the charge). The vector x is the vector of unknown coefficients a, b,
c, etc. By determining the null-space of A, the system can be solved.
If the null space of A is one-dimensional, we get solutions of the form
k*v, where k can be chosen freely (this is reasonable, because chemical
equations can be scaled by some factor without changing their meaning),
and v is any element from the null-space of A.
The problem with this is that the null-space need not be
one-dimensional. For the following reaction, it is two-dimensional:
a Cu + b HNO3 ---> c NO + d NO2 + e Cu(2+) + f H2O
When the dimension of the null-space is higher than one, then this
indicates that reaction products can be exchanged for each other. My
experience is that most reactions, beyond the simple schoolboys
chemistry have a null-space of higher dimension and no determinate
answer can be derived. For the example above, NO and NO2 can appear in
any ratio, depending on reaction conditions, like temperature,
concentration, etc.
The reaction you describe above has a 1-dimensional null-space. I'm
quite sure that determining the null space of A yields exactly the same
equations as I posted earlier. When other compounds are allowed in the
equation (e.g. S2O6(2-), (CN)2 or HOCN) then the null-space becomes
higher dimensional.
The sad thing is that the solution from higher dimensional null spaces
does not tell anything about how much of the reaction products is
formed. For higher dimensional null spaces of dimension N, any solution
k1*v1 + k2*v2 + ... kN*vN satisfies the equation. E.g. for the equation
above with copper and HNO3, one can simply choose (k1, k2) such that no
NO appears, but one can as easily choose another (k1, k2) such that no
NO2 appears.