請幫手..有關氧化還原問題

小莓。改

KMnO4 :
in acidic condition
MnO4(-) + 8H(+) + 5e(-) => Mn(2+) + 4H2O
Since its O.N. changes form 7 to 2, in which oxidizes 5 modes of electrons by 1 mode KMnO4, so it is strong.
in neutral and alkaline : OC but still it is a relatively strong O.A.
MnO4(-) + 4H(+) +  3e(-)  => MnO2 + 2H2O
As the O.N. changes from 7 to 4, therefore it is weaker and slower than KMnO4/H+
=======================
For NaNO3,
it is a oxidizing agent IF ACIDS ARE ADDED AND CONCENTRATED.
Due to its great solubility, if you can have sufficient H+ to activate, it just like HNO3(conc.)
=======================
For SO3(2-)
It is a "weak" reducing agent:
SO3(2-) + H2O => SO4(2-) + 2H+ + 2e-
=======================
Reminder:
Electron itself is a strong reducing agent!
and O.R is just like Acid and Base,
[High tendency O.A] + [High tendency R.A] => [Low tendency O.A] + [Low tendency R.A]
which is a interesting chemical nature
so everying around can be R.A or O.A, if you have thought deeply.

Zend

some conceptual errors had made.
but anyway, your stuff can solve header's problem

小莓。改

知錯能改, 焉不能為大賢乎?
請指教, 因有時想得太急。

Zend

1. 英文問題 mode- mole
2.多e- transfer/accept =/= strong/weak oxidizing/reducing agent
3. in very basic condition, MnO4- is reduced to MnO4 2- , and it has a higher reduction potential than MnO4- at acidic condition under standard conditions.
4. Your reminder, if it is used for mnemonic.. that's fine
by in reality, e- is not considered as an agent.
think it yourself.
how to reduce e- ?
how to oxidize e-?
if it is a strong reducing agent, by redox properties , when some species is reduced, then the reducing agent must be oxidized in order to maintain charge balance and ON balance.
OR is like acid and base. although there are some definitions of acids and bases depending on e- pair transfer, we should notice that they are different in nature.
5. everything can be OA or RA.
what about noble gases? or can S in H2SO4 be oxidized to give an ON of +7 or even higher?
or can hydrogen ion be oxidized to form a special stuff?

小莓。改

For 1,
Sorry for my bad english, I am only a F.5 ...
For 2,
This is a ratio-tendency problem, although this should be explained by elergy level, still generally it is true for most F.5 to F.7 agent
For 3.
You had said "In very basic condition"
Thanks for your reminder that "BASIC" is important
MnO4(-) => MnO4(2-) => MnO2 + .O:O. =>MnO2 + O. O. [lone pairs are not drawn]
But simpily reduce into MnO2 is the expected result in the test tube.
For 4,
Ya, this 'e- is a reducing agent' is supposed to be a nature of it.
It is a O.R. basics and also a agent.
e-[1/2] + e- [-1/2] => e:e [1/2,-1/2]
And the AB relationship and the OR relationship problem
ABis a specific reaction in Or only, AB involves several redox that we simpily ignored it.
But this is true for H.T. O.A/R.A to L.T. O.A/R.A,
else using activation energy thoughts can still solve your problem
For 5,
I am sorry that you are only focusing on 'seen' situation.
You should know that in high-energy chemistry or physics, everything can be happened IFF you apply enough energy into it.
Although there are several rules that we can't break or we don't know now.
For example, have you seen or heard H2CO4?
O4?
HClO5?
H2XeO7?
Are they REALLY IMPOSSIBLE?
They may exist for a few nanoseconds or fewer still they exists. Like what A = 100+
For me, without any supportive evidences, don't say something impossible even teachers told you that.

Zend

a F.5 cannot spell mole correctly?

for 2, the reaction between H2O2 and KMnO4  is one of the examples.
and more e- doesn't mean it is strong
let say , Cl2 and Cr2O72-/H+
let us compare their reduction potential under standard conditions.
you will find that although each Cl2 molecule only accept two e- from the reducing agent, but it has a higher E value

Cl2(g) + 2 e− ⇄ 2 Cl−          +1.36        
Cr2O7− − + 14 H+ + 6 e− ⇄ 2 Cr3+ + 7 H2O          +1.33
Although you may argue that it is still possible that Cl2 cannot be reduced easily because of the activation barrier, the reason for showing these data is to indicate that there are no relationship between e- transfer and the strength of stuff.

3. I am not sure. In fact it does exist.  although it may undergo further decomposition.
at least I had an experiment showing that the colour of MnO4 2- is green.
and there are E emf data for MnO4- +e- -->MnO4 2-. The value is quite high, as I remember

For 4, do you mean spin quantum no?  ok can u give me some [e-]^2 (s) in packet form?
the agent here means chemical. I think very few people would say that electron is an agent in a chemical process.

Let say, Cr2O72- +14H+ +6e- -->2Cr3+ + 7H2O
if you treated electron as an "oxidizing/reducing agent" , then this itself is a redox reaction.

AB involved redox, can you use the neutralization between NaOH and HCl as an illiustration?

At least in the level of CE, we don't regard Acid-Base reaction as an redox reaction, because there are no ON change in each atom of the species before and after the reaction

for q5 , Molecular orbital theory can distinguish whether a combination is allowed or not by noticing its bond order. but this is clearly out of scope .
but in general, for H2SO4 , you can't make a speices with S ON= +7 under normal situation
and generally ON is defined in such a way that the number given is the tendency of e- obtained/released.
for something "chemically stable" species, you can't do S with ON=+7

For me, without any supportive evidences, don't say something impossible even teachers told you that.
<=那書是如何教的，我倒想知

and stop posting out-c stuff here.

乂雲影乂

亞硫酸鈉(SO32-)是一個還原劑 ，因為oxidation number of sulphur 由 +4(in SO32-) 變 +6(in SO42-)，
佢自己氧化，然後將高猛酸甲(MnO4-)還原，由+7(in MnO4-) 變 +2(in Mn2+)。