The electron gain and loss of oxygen atom in potassium superoxide is why the molecular formula is K2

Here oxygen is not -2 valence but -1 valence, and potassium is +1 valence so it is K2O2

This should be an ionic compound, oxygen gets electrons, and potassium loses electrons.

The reduced element is (N, S), the oxidized element is (C), the oxidizing agent is (Kno3, S), the reducing agent is (C), the oxidation product is (CO2), and the reduction product is (K2S, NO2).

Oxidants are substances with reduced valency in chemical reactions, which are reduced to produce reducing products. In this problem, the valency of n changes from +5 to +4, and the valency of s changes from 0 to -2.

A reducing agent is a substance with an increased valency in a chemical reaction, which is oxidized to produce oxidation products. In this problem, the valency of c changes from 0 to +4.

I suggest you take a look.

The oxidant gains electrons, the valency decreases, and it is reduced to produce a reduction product.

The reducing agent loses electrons, the valency increases, and it is oxidized to form oxidation products.

The oxidant is Kno3 and S, and the reducing agent is C. The elements to be reduced are n and s, and the elements to be oxidized are c. The oxidation product is CO2, and the reduction product is K2S and NO2.

+5 0 0 -2 +4 +4

2k n o3+ c ＋ s === k2 s + 2 n o2↑ +c o2 ↑

n is reduced by 1 in kno3 n is reduced··· KNO3 is an oxidant S that reduces 2-valent S and is reduced··· S is the oxidant C increases 4 valent C is oxidized··· c is a reducing agent.

2Na2O2+2H2O=4NaOH+O2 This is before the reaction, O is -1 valence, during the reaction, part of it increases, O2,0 valence is generated; Partially decreased, becoming -2 valence.

Up 1, down 1

That is, 2 O's in Na2O2, 1 is raised and 1 is lower, i.e. 1 O transfers 1 electron to another O

That is, in 1 Na2O2, 1 electron is transferred.

So, the coefficient of equation 2, i.e., transfers electrons 2, from the o in Na2O2 to O, i.e., so transfers electrons.

Both oxygen atoms in the Na2O2 atom are one valent. Only one of the oxygen atoms in the H2O atom is bivalent.

N, S is reduced, C is oxidized, oxidizing agents: Kno3 and S, reducing agent C

1) In 2FeCl3

2ki═2fecl2

2kcl+i2

In the reaction, the valency of iodine element changes from -1 valence to 0 valence, when the valency increases, it is used as a reducing agent, Ki is oxidized and undergoes an oxidation reaction, the valency of iron is reduced from +3 valence to +2 valence, and the valency is reduced to be used as an oxidant, and the reducing product FeCl2 is obtained by reduction

So the answer is: i; fecl2

2) The ionic equation for this reaction is: 2Fe3+

2i-2fe2+

i2 so the answer is: 2fe3+

2i-2fe2+

i23) oxidant FeCl3 in redox reactions

The oxidation is greater than that of the oxidation product i2

So the answer is: ;

4) The valency of iron element changes from +3 valence to +2 valence to obtain electrons; The valency of iodine element changed from -1 valence to 0 valence, and electron transfer was marked with a two-line bridge: <>

So the answer is: ;

In the reaction of 2FeCl3+2Ki 2FeCl2+2KCl+I2, the valency of element I increases, the lost electrons are oxidized, Ki is the reducing agent, the valency of the Fe element decreases, the electrons are reduced, and the FeCl3 oxidant, so the answer is: I; Lose; Oxidation; fecl3；ki．

To put it simply, the stronger the oxidation of the oxidant, the weaker the reduction of its corresponding reduction products.

The valency of Fe element in the reaction decreases, then FeCl3 is the oxidant, and the oxidation of the oxidant is greater than the oxidation of the oxidation product, and the oxidizing FeCl3 I2 is reduced, then Cl2 is the oxidant, and the oxidation of the oxidant is greater than the oxidation of the oxidation product, and the oxidation of the oxidizing Cl2 FeCl3 is reduced, and the valency of the Mn element in the reaction is reduced, then KMNO4 is the oxidant, and the oxidation of the oxidant is greater than the oxidation of the oxidation product. Oxidizing Kmno4 Cl2, that is, the oxidation strength is Kmno4 Cl2 FeCl3, then there is Fe2+ and I- coexistence in a solution, and I - should be oxidized to remove I- without affecting Fe2+ and Cl-, and Fe2+ and Cl- can be oxidized when selecting substances with strong oxidation to remove impurities, so the weak oxidant FeCl3 is selected to remove impurities without affecting Fe2+ and Cl-, and no new impurities are introduced, so C is selected

1) In 2FeCl3

2ki═2fecl2

2kcl+i2

In the reaction, the valency of iodine element changes from -1 valence to 0 valence, when the valency increases, it is used as a reducing agent, Ki is oxidized and undergoes an oxidation reaction, the valency of iron is reduced from +3 valence to +2 valence, and the valency is reduced to be used as an oxidant, and the reducing product FeCl2 is obtained by reduction

So the answer is: i; fecl2

2) The ionic equation for this reaction is: 2Fe3+

2i-2fe2+

i2 so the answer is: 2fe3+

2i-2fe2+

i23) oxidant FeCl3 in redox reactions

The oxidation is greater than that of the oxidation product i2

So the answer is: ;

4) The valency of iron element changes from +3 valence to +2 valence to obtain electrons; The valency of iodine element changed from -1 valence to 0 valence, and electron transfer was marked with a two-line bridge: <>

So the answer is: ;