Solve the problem of chemical redox reaction in high school

The S in Na2SO3 is +4 valence, which is oxidized, and the valency increases, which can only be increased to +6 valence. There is a total of Na2SO3, then the electrons are transferred.

Total number of electrons gained = total number of electrons lost.

That is, per molxO4-, the price is reduced by 3, which was originally +7, so it becomes +4, choose A.

Mo+4 s becomes +6 s lose Momo +7 x gets Momo electrons become +4 valence x

The question is a little unclear: Is the substance with x a -1 valent acid root?

The amount of the substance of Na2SO3 is.

The S valency of Na2SO3 is +4, which acts as a reducing agent and is oxidized to +6-valent SO42-.

So, 1mol of Na2SO3 is oxidized, and the amount of the substance that loses electrons is 2mol, so the amount of the substance that loses electrons is.

The amount of the substance of XO4 is:

obtained, the valency of x decreases by 3

If it is XO4-, the X valency is +7, and if the drop 3 is +4XO4, the X valency is +8, and the drop 3 is +5

Push in turn:

In the first equation, Fe2+ is oxidized by Cl2 to Fe3+, according to the oxidation of the oxidant is greater than the oxidation product, so the oxidation of Cl2 is greater than Fe3+, and so on, the second equation can be launched: Fe3+>i2, and the third equation: ClO->Cl2, therefore, it can be concluded that the most oxidizing is:

clo-

The principle of left strength and right weakness. According to the first reaction, the oxidation of Cl2 is stronger than that of Fe3+, and the second reaction shows that the oxidation of Fe3+ is stronger than that of I2, and the oxidation of Cl- is stronger than that of Cl2

As for the principle that the left is strong and the right is weak, it can be roughly considered this way. The reducing property of the oxidant is stronger than that of any kind of substance obtained, and the reducing property of the reducing agent is stronger than the reducing property of any kind of substance obtained. So choose B

bBecause the oxidation of the oxidant is greater than the oxidation of the oxidation product.

Choose B, the oxidation on the left side of the equation is stronger than on the right.

b, the oxidation of the oxidant is greater than the oxidation product.

1.Oxidation removes iodine ions without affecting ferrous and chloride ions, so the oxidant used can oxidize I-, but not Fe2+.

Cl2 and KMno4 can oxidize Fe2+ and I-, which is not desirable.

HCl cannot oxidize either I- and Fe2+ and is not desirable.

FeCl3 can oxidize I- to generate Fe2+ by itself, which is desirable.

Therefore, if the mass ratio of C and Na2SO3 is 71:42, the ratio of the quantity of matter is 71:42 126=2:1

Na2SO3 is oxidized to form Na2SO4, and the sulfur element is bivalent. Because the ratio of the amount of chlorine to sulfur is 2:1, the price of chlorine is reduced.

So the reaction equation is: 2Naclo3+Na2So3+H2SO4=2Na2So4+H2O+2ClO2

Therefore, C is chosen

2.Conservation of electrons gained and lost: n(Naclo3):n(Na2SO3)=2:11 The total number of electrons lost in Na2SO3 is 2 electrons, and the total number of electrons obtained in 2 Naclo3 is 2 electrons.

The valency of Naclo3 after the change is +4 valence, that is, C1Using the comparison law of strength and weakness in the redox reaction: it is necessary to add an oxidant to remove only I-, that is, its oxidation should be weaker than Fe3+, that is, C

1 C can not oxidize ferrous ions, exclude a and b, to find oxidizing, exclude d, so choose c

The quantity ratio of 2 c substances is 2 to 1, and the sulfite is oxidized and rises by 2, so the chlorine element needs to be reduced by 1, so C is chosen

1.Both potassium permanganate and chlorine have strong oxidizing properties, which will oxidize ferrous ions to iron ions ab. Hydrochloric acid does not react when added to the solution. Don't choose, choose C for the first question

2.The ratio of the amount of matter obtained from n = m m m is only 2:1....The question type of push is the problem of redox price increase and price reduction. Sodium sulfite should be bivalent to sodium sulfate. Then from the proportion, it can be known that the chlorine price is reduced by 1, and c is selected

O3 2ki H2O===2KOH I2 O2 O3 is the oxidizing agent and Ki is the reducing agent, so it is.

1 2 SiO2 2C Si 2Co SiO2 is the oxidizing agent and C is the reducing agent, so it is.

1 2 SiO2 3C SiC 2Co in one C is the oxidizing agent, and the other two C are the reducing agent, so it is.

1 2 4HCl (concentrated) Mno2 mnCl2 Cl2 2H2O in Mno2 is an oxidizing agent, and two of the four HCl are reducing agents, so it is 1 2

The answer is D

2, 3, 4 is c. It has to be lowered by oxygen, and the loss of altitude is returned. 2. Among them, the price of silicon dioxide and the price of carbon have increased. Out of 3, one C decreases in price, and two C's increase in price. One MN price reduction in 4 and two HCL price increases.

Kclo3 +6HCl ==KCl +3Cl2 +3 H2O This reaction is a transfer of 5 electrons.

kclo3 +6hcl ==kcl +3cl2 ↑+3 h2o --5e-

x x = electrons.

If generated, the amount of matter to which electrons are transferred is how much mol?Answer.

KClO3 +6HCl ==KCl +3Cl2 +3 H2O In the product 3Cl2, 6 Cl- are oxidation products and 1 is a reduction product. This corresponds to 4cl- more oxidation products than reduction products.

Then when 3cl2 is generated, the oxidation product is 4cl more than the reduction product, and now it is more, the amount of matter. So it can be concluded that the Cl2 produced by the reaction is.

kclo3 +6hcl ==kcl +3cl2 ↑+3 h2o

50%x x=

If the utilization rate of HCl in the reaction is only 50%, when there are more oxidation products than reduction products, what is the concentration of concentrated hydrochloric acid? Answer.

1、kclo3 +6hcl ==kcl +3cl2 ↑+3 h2o；This reaction is a centering reaction (of course, it is also a redox reaction), KCLO3 is an oxidizing agent, HCl is a reducing agent (only 5mol of HCl participates in the redox reaction, and the other 1mol HCl finally generates KCI), the oxidation product and the reduction product are Cl2, in this equation, every 1mol of KCLO3 participates in the reaction, then 5mol of electrons is transferred, that is, 3mol of chlorine gas is generated to transfer 5mol of electrons, so chlorine gas is generated to transfer electrons mol;

2. The amount of chlorine is n=m m=, and the ratio of the number of chlorine atoms (or the amount ratio of the substance) in the oxidation product and the reduction product is 5:1, that is, when there is a 6mol hydrochloric acid reaction, the oxidation product is, and the reduction product is, with a difference of 2mol; From this proportional relationship, it can be seen that when the oxidation product is more than the reduction product, that is, the difference, hydrochloric acid is needed, so the concentration of concentrated hydrochloric acid is c=n v= (where, the reason is that the utilization rate of HCl is only 50%); Hope it helps!

1. Assuming that the reactants KCLO3 and HCl are 1mol and 3mol respectively, 5 chloride ions in 6 HCl parts are oxidized to 0 valence, and the chlorine element in KCLO3 is reduced to 0 valence

2. The oxidation product is from 5 parts of Cl atom in HCL (the other one is not oxidized), and the reduction product is the Cl atom in Kclo3, when the oxidation product is more than the reduction product, it is easy to calculate that the oxidation product is (, the reduction product is, and the amount of the corresponding substance is Hcl participating in the reaction, and the hydrochloric acid concentration is (

1> Answer: Cl2 transfers 5mol of electrons. )

2. Both the reduction product and the oxidation product are Cl2

That is, there is more oxidized Cl2 than reduced Cl2.

From the chemical equation, it can be seen that the reduced Cl2: oxidized Cl2=1:5 yields the Cl2 produced by oxidation and reduction as;

Then according to the chemical formula, calculate that the HCl of the reaction has 3 x=, and then find the concentration=

Pick B. Since io3-oxidizes I- and becomes I2 itself, it can be obtained:

1molH2O2 to obtain 2mol electrons, to obtain 1moli2; 1molio3- to obtain 5mol electrons, to obtain 3moli2; (Oxidation 5mol.)

i-, while becoming i) itself

1mol mno4 - to get 5mol electrons, get.

i2；1molHno3 to obtain 3mol electrons, to obtain i2; To sum up, choose B

Select D for the second question

Analysis: aThe lower the solubility, the easier it is to form a precipitate.

b.The valency of S in CUS is -2, which can be increased to 0 2 4 6, so copper blue is reducible.

c .Zns is intolerant, so the ion equation is Cu(2+) Zns = Cus +Zn(2+) to generate Cus precipitate.

Find the ratio of the amount of oxidation product and reduction product of these two equations, the first: oxidation product N2, reduction product SO2, use the stoichiometric number to compare the second: oxidation product.