The huge question about redox reactions is hasty and urgent

It's super important to figure out the redox products of common substances!!

For example, the reduced product of Fe3+ is Fe2+

The oxidation product of SO2 is SO42-

The reduction product of KMNO4(H+) is Mn2+ or something like that.

This makes it easy to write the equations.

The reactants and products are basically written, and finally look at the environment of the solution, just add H2O or H+ or OH-, which is super easy

And your teacher must have something to help you summarize these products, you go to find the papers.

If you don't understand, you can ask

Redox reaction, if you want to learn it well, you have to do the questions. Moreover, you have to learn and skillfully apply methods to balance chemical equations (you must have been taught by the teacher), and then find problems to practice.

(1) Potassium chlorate is catalyzed by manganese dioxide to generate potassium chloride and oxygen, and the chemical equation of the reaction is: 2kclo3mno

KCL+3O2, the valency of Cl element in the reaction decreases from +5 valence to -1 valence, and the valency of O element increases from -2 valence to 0valence, and it can be seen from the change of valency that the direction and number of electron transfer can be expressed as <>, so the answer is: <>

2) Reaction 2kmNO4

16HCl (concentrated) 2KCl + 8H2

o+5cl2

Mn gets electrons, Cl loses electrons, the reaction is transferred 10E-, the "two-line bridge method" means that the direction and number of electron transfer of the reaction is <>, when there is 5mol chlorine gas generated under the standard condition, the electron transfer number of the reaction is 10Na, when there is chlorine gas generation under the standard condition, the electron transfer number of the reaction is Na

So the answer is: <>na

3) Reaction Na2

In S+I2=2Nai+S, the oxidant is iodine element, and the oxidation product is S, so oxidation I2

S, reaction O2

4hi═2i2

In 2H2O, the oxidant is oxygen, and the oxidation product is I2, so oxidizing O2

i2, so the oxidation order is: O2

i2 s so the answer is: o2

i2 s or s i2 o2

4）nahso4

It is an acid salt of a dibasic strong acid, which can be understood as fully ionized When the solution is neutral after the reaction, its reaction formula is: 2NaHSO4

ba（oh）2

baso4+na2so4

2H2O, then the ionic reaction equation is: 2H++SO4

ba2++2oh-

BaSO4+2H2O, at this time the solute in the solution is only Na2

SO4, add BA(OH)2

The ionic reaction equation is: BA2++SO4

baso4, so the answer is: 2h++so4

ba2++2oh-

baso4+2h2o；so4

ba2+=baso4

5) Sodium bicarbonate is weakly alkaline, and reacts with hydrochloric acid to form sodium chloride, carbon dioxide and water, and the reaction ion equation is HCO3

H+=H2O+CO2, so it can be ** hyperacidity, so the answer is: HCO3

h+=h2o+co2

The redox reaction is that the stronger oxidant + stronger reducing agent produces the weaker oxidant and the weaker reducing agent, so the oxidation magnitude is obtained from the equation.

Acid potassium permanganate》Cl2>Fe3+>I2

Therefore, if Fe2+, I- and Cl- coexist in a solution, I- should be oxidized to remove I- without affecting Fe2+ and Cl-

The oxidation must be greater than I2, but so it is FeCl3 and C

Cl2 and KMno4 have strong oxidizing properties and will oxidize Fe2+, so HCl cannot go out of I-

So choose C (which can be derived from the equation above).

Thank you for adopting.

When 10 mol of electrons are transferred, the amount of I2 produced is 6 mol.

Analysis: 5ki + kiO3 + 3H2SO4 = 3I2 + 3K2SO4 + 3H2O

Among them, the oxidant is kio3 and the reducing agent is ki. In this reaction, the ratio of the amount of oxidant and reducing agent is 1 5, iodine element is both an oxidation product and a reduction product, six moles of iodine atom in three moles of iodine element, one mole ** in potassium iodate, five moles ** in potassium iodide.

5mol potassium iodide was used as a reducing agent, and the iodide ions increased from negative monovalent to zero valence, and 5mol electrons were lost; 1mol potassium iodate is used as an oxidant, and the iodine in the iodate group is reduced from positive pentavalent to zero valence, and 5mol electrons are obtained. When 5mol of electrons is transferred in the reaction, 3mol of iodine is generated. So when 10 mol of electrons are transferred, the amount of I2 produced is 6 mol.

You can write both reactions as ionic reactions, and the equations are both.

2 (Fe3+) + 2 (I-) = I2 + 2Fe2 + is the machine rotten and the cavity is the ferric ion of the iodine ion, the iodine ion oxidation of the hunger shirt into the iodine element, and the jujube is reduced to the divalent iron ion.

According to the reductive I-Fe2+

So the reaction can be carried out.

o changed from zero valence to -2 valence, and a total of 44 electrons were transferred.

Fe changes from +2 to +3, transferring a total of 4 electrons.

s changes from -1 to +4, transferring a total of 40 electrons.

Oxidant O2 reducing agent Fes2

Fe2O3 and SO2 are both oxidation and reduction products.

The direction of electron transfer is S (-2 valence) 40mol to 0 (0 valence) and Fe(+2) to O (0 valence) 4mol

Single-line bridge writing.

The total number is 44mol (write 44e).

The oxidizing agent is O2

The reducing agent is FeS2

The oxidation products are Fe2O3 and SO2

The reduced products are SO2 and Fe2O3

Before the reaction, S in FeS2 is -1 valence, Fe is +2 valence, O in O2 is 0 valence, after the reaction, Fe in Fe2O3 is +3 valence, and S in SO2 is +4 valence, then the electron transfer direction should be from Fe and S to O

There are 4 +2 valence Fe and 8 -1 valence S on the left side of the reaction formula, 4 +3 valence Fe and 8 +4 valence S on the right, Fe and S transfer a total of 44 electrons to O, the oxidant is O2, the reducing agent is FeS2, and the Fe2O3 and SO2 on the left are both oxidation products and reduction products.

44mol

The oxidizing agent is O2

The reducing agent is FeS2

The oxidation product is Fe2O3

The reduction product is SO2

c nai .A sufficient amount of Cl2 was introduced into the solution of 1mol of Na2SO3 and Febr2, Na2SO3 was oxidized by Cl2 to NaSO4, Febr2 was oxidized to FeCl3 by Cl2, Nai was oxidized to NaCl by Cl2, and Br2 volatilized and I2 precipitated. When the solution is heated and evaporated in air and fully burned, the I2 will be volatilized, and FeCl3 will be completely hydrolyzed into iron hydroxide at high temperature, and it will become Fe2O3 after full burning.

This question examines a comparison of oxidation strengths. Cl2 can oxidize I-, Br-, So3 2-, Fe2+ ions. It is then oxidized to iodine, bromine, sulfate, and ferric ions, respectively.

And after heating and evaporating and burning thoroughly. Iodine and bromine will be volatilized. Ferrric ions are hydrolyzed into iron hydroxide.

After burning, it becomes ferric oxide. So. I choose C!

Select fe2o3 na2so4

A sufficient amount of Cl2 is introduced to become NaCl, Na2SO4 and FeCl3, but the iodine element and bromine element will be volatilized when heated, and FeCl3 will be hydrolyzed, and the following reaction occurs: FeCl3+3H2O = Heating = Fe(OH)3+3HCl2Fe(OH)3=Heating = Fe2O3+3H2O