Is S an oxidizing agent or a reducing agent? At the time of S O SO

In the case of s+o = so, s is the reducing agent.

Oxidative reducing properties are possessed by oxidant reducing agents.

Oxidant reducing agent is not a property of the substance itself, but is reflected in the redox reaction. The substance corresponding to the element in its most ** state may be oxidizing.

In different reactions, the same substance will have different properties.

For example, if the valence of element A increases and the valence of element B decreases in a certain reaction, then AB is an oxidant and a reducing agent.

Redox is judged directly by the gain and loss of electrons.

The species with elevated valence states in the reaction lose electrons and are oxidized to reducing agents.

The electrons gained by the substances with reduced valence in the reaction are reduced to oxidants.

Rise and lose oxygen, and drop to return oxygen.

Oxidative reducibility is generally determined for the role played by specific substances (atoms, ions, molecules, but generally not elements) in a certain reaction, and is equivalent to the concept of oxidant reducing agent. However, if a substance can exhibit oxidation or reduction in many reactions, then it can also be said to be oxidizing or reducing without pointing out the specific reaction.

Oxidity and reducibility are properties of a substance, which are related to its specific valence state. Take S as an example, the highest is +6 and the lowest is -2, so the elemental S is both oxidizing and reducing, and will be reflected differently in different reactions. Generally speaking, the most ** state of the substance is only oxidizing, provided that you are sure that there is no higher valence state than it.

AB is both an oxidizing agent and a reducing agent.

Oxidation refers to the electron-gaining property of a substance. A substance can be ionic or molecular. For example, sodium sulfide solution, it can be said that sulfur ions are reducing, and sodium sulfide is an oxidizing agent.

Most of the most ** compounds can only exhibit oxidation.

There are many examples of how the same substance can exhibit different properties in different reactions.

It is both an oxidizing agent and a reducing agent.

Reducing agent of CO2+3H2?

Answer: The reducing agent of CO2+3H2 is: C, Co, and H2 as reducing agents. The pure hall formula of the reaction equation is: CO2+3H2 CH3OH+H2O.

Chlorine is oxidized, chromium is reduced, chlorine is an oxidation product, chromium chloride is a reduction product, and 6mol electrons are transferred unidirectionally, which is a total of 12mol

The decrease in valency is the oxidant to obtain the oxidation product, and the increase in valency is the reduction product obtained by the reducing agent.

K2Cr2O7 is an oxidizing agent.

HCl is a reducing agent.

Cl2 reduction product.

CrCl3 oxidation products.

+1+6

k2cr2o7+14h

cl=2kcl+2cr

cl3+3cl2+7h2

o It can be seen that the valency of the three elements of KOH remains unchanged.

So look at cr, cl

k2cr2o7→2cr

Cl3 obtains six E-, electrons are reduced products, oxidant, so K2Cr2O7 oxidant Cr

Cl3 reduction product.

14HCl 3Cl2 loses six E-, electron loss oxidation products, reducing agent, so HCl reducing agent Cl2 oxidation product.

There is a mantra: liter, loss, oxidation, reducing agent. i.e.: valency on (

liter), (lost) electrons, (oxidation) reaction (can also be translated: oxidized), in the reaction as a (reducing agent), of course, the reduced is the reduction product, the oxidized is the oxidation product. Corresponding formula:

Lower, gain, reduce, oxidant. You just need to memorize the previous sentence, which is very useful for identifying redox reactions later.

As questioned: What is an oxidant in K2Cr2O7+14HCl=2KCl+2CrCl3+3Cl2+7H2O, what is a reducing agent, oxidation product, reduction product, and why?

K2Cr2 +6 Valence O7 + 14HCl<-1 Valence》=2KCl+2Cr +3 Valence Cl3+3Cl 0 Valence 2+7H2O

k2cr2o7）

Cr<+6>--CrCl3)Cr<+3> valency decreases, electrons are obtained, they are reduced, oxidants, oxidizing products (CrCl3) (formula: down, get, reduce, oxidant. ）

14HCl<-1 valence".

--cl2 0 valence.

The compound medium is elevated, electrons are lost, oxidized, reducing agent. (Formula: L, L, Oxidize, Reducing Agent.) ）

So this question: oxidant: K2Cr2O7

Reducing agent: HCl

Oxidation product: Cl2

Reduction product: CrCl3

Valency. Elevated is the reducing agent.

It is the oxidant that reduces the absolute valence of the acre, and the double-line bridge is drawn, and the corresponding product of the reducing agent is the reduction product, and the corresponding product of the oxidant is the oxidation product.

The valency of sulfur in H2SO3+I2+H2O2Hi+H2SO4 increases, so H2SO3 is a reducing agent.

The valence of iodine is low and low, so I2 is an oxidizing agent.

So H2SO4 is an oxidation product and HI is a reduction product.

If you don't understand, just ask me).

2co+o2=2co2

C in CO from.

2 valence rises to.

4-valent, is oxidized, so.

CO is a reducing agent.

o2 reduced from 0 valence to.

2 valence, is reduced, so O2 is an oxidizing agent.

CO2 is.

Oxidation products. It is also a reduction product.

The valency of C rises from 0 to +2, so C is oxidized and must be a reducing agent.

The valency of the individual species in SNO2 is not easy to determine, but O is -2 valence, and the Sn atomic cluster is definitely +4 valence.

SN4+ becomes SN after the reaction, the valency decreases, and it is an oxidizing agent.

Conclusion: C is the reducing agent and SNO2 is the oxidizing agent.

C is the reducing agent and SNO2 is the oxidizing agent.

4Fes2+11O2 2Fe2O3+8SO2 can be balanced using the conservation method of electronic gain and loss (iron + 2 valence + 3 valence, sulfur - 1 valence + 4 valence), so the reducing agent is iron and sulfur, and the oxidizing agent is oxygen.

The products contain iron and oxygen, sulfur and oxygen, respectively. So both products are both oxidation and reduction products.

o Valency decreases, electrons are obtained, and they are reduced and are oxidizing agents.

Fe valency increases, electrons are lost, and it is oxidized, and it is a reducing agent.

S valency increases, electrons are lost, and it is oxidized, and it is a reducing agent.

That is, oxygen is the oxidizing agent and ferrous sulfide is the reducing agent.

The oxidizing agent is oxygen because its valency is decreasing.

The ratio of the quantity of matter is 1:4

na2s2o3 + 4 cl2 + 5 h2o == 2 h2so4 + 2 nacl + 6 hcl

The CL valency in chlorine gas decreases, 0 -1, and is an oxidizing agent.

The valency of S in Na2S2O3 is increased, +2 +6, which is a reducing agent, so reducing agent:oxidant = 1:4 (the ratio of the amount of a substance).

The reaction equation is: Na2S2O3+4Cl2+5H2O=Na2So4+8HCl+H2SO4

Or: Na2S2O3+4Cl2+5H2O=2H2SO4+2NaCl+6HCl

In the reaction, the sulfur element in Na2S2O3 rises in price, which is a reducing agent; The price of chlorine in Cl2 is reduced and it is an oxidizing agent. Therefore, the ratio of reducing agent to oxidant is 1:4

The product of the reaction between Na2S2O3 and Cl2 is H2SO4, and what is the ratio of the amount of reducing agent to oxidant substance between NaCl and HCl? Question addendum: The process is clear and urgently needed according to the reaction: Na2S2O3 + Cl2 = H2SO4