Bonus points Chemistry Master Advance! Determine the strength of the oxidation of the following subs

Oxidizing: acidic potassium permanganate, sodium peroxide, nitric acid, n-ferric ions.

When its argon gas is used as protective gas, when sodium reacts with sulfur, it can only be burned quietly, and when the system is open, sodium reacts with sulfur and burns violently or even burns**, in addition to sodium sulfide, sodium thiosulfate, sodium sulfate and other oxygenates are generated.

Conclusion: 1. When sodium reacts with sulfur, oxygen in the air can have complex chemical reactions with sodium and sulfur at the same time.

2. The discussion of "sodium and sulfide are combined or even occur to form sodium sulfide" in the textbook of the People's Education Edition does not grasp the essence and main contradiction of the problem, because the author of the textbook does not know that when sodium and sulfur react, oxygen in the air can have complex chemical reactions with sodium and sulfur at the same time.

Therefore, in a closed environment, the reaction between sodium and sulfur is not more violent than the reaction with oxygen, and sodium ignites in oxygen to try, which is definitely more intense than sulfur.

Acidic potassium permanganate, the oxidation of nitric acid is about the same, depending on the concentration.

Acidic potassium permanganate, nitrate, sodium peroxide, and positive ferric ions.

Acidic potassium permanganate, nitric acid, sodium peroxide, trivalent iron.

The reaction of sodium with sulfur I remember no**hey (+

The easier it is for an element to capture electrons, the stronger its oxidation, for example, f is the most oxidizing element, the fewer the number of electron layers of the atom, the tighter the attraction force of the outer electrons in the center of the nucleus, so that the atom is difficult to lose electrons, and it is easy to get electrons, that is, it is highly oxidized.

The strength of oxidation and reduction is essentially the strength of the particle's tendency to gain or lose electrons.

There are two directions to determine this trend, one is to compare according to experimental facts, such as weak force, and the other is to analyze according to theory, which is divided into two ways, one is qualitative analysis, using atomic radius, valency, electronic structure, and so on for quantitative comparison. Or solve the Schrödinger equation for the system, but this equation is too complex to find an exact solution, only an approximate solution.

The strength of chemical oxidation is mainly determined by the nature of the substance, which is manifested in the ability of the electrons of the substance, and the easier it is to obtain electrons, the stronger the oxidation.

We can use the electrode potential of the substance to compare the magnitude of oxidation, and the Nernst equation to calculate the electrode potential of the substance.

Summary. The strength and weakness of the oxidation and reduction of different valence substances formed by the same element by using valency comparison are: the most important state of the element only has oxidation, the lowest valence state of the element only has reduction, and the intermediate valence state of the element has both oxidation and reduction.

We'll be happy to answer your questions.

The strength of the oxidation and reduction of different valence substances formed by the same element is as follows: the most important state of the element only has oxidation, the lowest valence state of the element only has the perturbation of the loss of the group, and the intermediate valence state of the element has both oxidation or reductivity.

We hope you find it helpful.

The oxidizing property of O2 is stronger than that of S

O2 is more oxidizing than I2

The oxidation of I2 is stronger than that of S

The overall answer is A

Look at the valency of the element, for example, potassium permanganate manganese element valency +7 sulfuric acid sulfur sulfur has a valency of +6 potassium permanganate is more oxidizing than concentrated sulfuric acid.

The strength of the reduction depends on which valency is lower.

The reason is that the stronger the oxidation, the easier it is to get electrons, and the stronger the reduction, the easier it is to lose electrons.

Oxidation, a method for judging the strength of reducibility.

1) Oxidant (oxidizing) + reducing agent (reducing) === reducing product + oxidizing product --- reducing product.

Electrons are obtained, the valency decreases, they are reduced, and a reduction reaction occurs.

Reducing agents --- oxidation products.

Electrons are lost, valency increases, is oxidized, and oxidation reactions occur.

Oxidation: Oxidizing agent" oxidation product.

Reducibility: Reducing agent" reducing product.

2) The oxidation can be judged according to the oxidant and reducing agent in the same reaction: oxidant and reducing agent.

Reducibility: Reducing agent》 Oxidant.

Hello! From the chemical equation given in the question, it can be seen that (Attached: the oxidation of reactants as an oxidant is always greater than that of the product) in the first equation FeCl3 is used as the oxidant, so the oxidizing FeCl3 > CuCl2 in the first equation, the second equation Kmno4 is used as the oxidant, so the oxidizing Kmno4 > Cl2 is the oxidant, and the third equation Cl2 is the oxidant of the reaction, so the oxidizing Cl2 >FeCl3.

So the answer is C. (Attached: The oxidant has oxidizing properties, and the reducing agent has reducing properties.)

The oxidant is reduced in price to gain electrons, and the reducing agent is increased in price and loses electrons, so don't be confused. ）

The oxidation of the oxides in the reactants is greater than that of the oxidation products, FeCl3> CuCl2 in the first equation, KMNo4>Cl2 in the second equation, Cl2 >FeCl3 in the third equation.

It follows the principle of forced weakness, with strong oxidants preparing weak oxidants and strong reducing agents preparing weak reducing agents.

In A, it is known from reaction (4) that i2 can prepare S, so it should be i2>S, and the error is made at the beginning.

Item B is correct.

In term c, it can be seen from the reaction (2) that Cl2 can prepare iron ( ), so it should be Cl> iron ( ) is also wrong at the beginning.

In item d, it can be seen from reaction (3) that iron ( ) can prepare elemental iodine, so it is iron ( ) i2, and the middle part is wrong.