What are the classic questions about redox reactions? Please attach the answer, thank you!!

Example 1] Analyze the transfer of electrons and the rise and fall of valence in the following redox reactions, which substance is an oxidant and which substance is a reducing agent? (1) Zn+H2SO4===ZNSO4+H2 (2)Fe+2HCl===FeCl2+H2 Analysis: According to the relationship between electron transfer and valence rise and fall in redox reactions, and the definition of oxidants and reducing agentsAnswer:

Example 2] Analyze the valence relationship in the following redox reaction, indicate the direction and total number of electrons, and indicate the oxidant and reducing agent, oxidation product and reduction product. (1)2kmNO4K2MNO4+MNO2+O2 (2)2Fe2O3+3C4FE+3CO2 (3)3NO2+H2O====2HNO3+NO(4)HGS+O2===HG+SO2 (5)2H2S+SO2===3S+2H2O Analysis: In this question, several representative redox reactions are selected, and the valency changes are analyzed, the direction and total number of electrons are expressed, and the redox relationship is analyzed.

Regardless of the redox reaction, the idea and process of answering the above questions are: marking the valency of each element that changes in valency before and after the reaction; Determine the electron transfer situation, use a single line bridge or a double line bridge to indicate the direction of electron transfer or the gain and loss of electrons, and indicate the total number of electrons on **; Oxidants, reducing agents, oxidation products and reduction products are analyzed according to the direction of electron transfer or the gain and loss of electrons. Answer:

KMno4 is both an oxidizing agent and a reducing agent; K2, MnO4 and MnO2 are reduction products, and O2 is an oxidation product. Fe2O3 is an oxidizing agent and C is a reducing agent; Fe is a reduction product and CO2 is an oxidation product. Part of NO2 is an oxidizing agent, and the other part NO2 is a reducing agent; No is a reducing product, Hno3 is an oxidation product, O2 is an oxidizing agent, and HGS is both an oxidant and a reducing agent. SO2 and HG are reduction products, and SO2 is an oxidation product.

SO2 is an oxidizing agent and H2S is a reducing agent; Part of the S is a reduction product and the other part of the S is an oxidation product. (From "New School Cases of Quality Education").

1) R2O8N-: Mn2+ = 5:2 (5 R2O8N- and 2 Mn2+ involved in the reaction).

2) Mn2+ is oxidized to MnO4-, which changes from 2 valence to 7 valence, that is, 7-2, which rises by 5 valence, and 2 Mn2+ participate in the reaction.

Therefore, the valency increases to 2*(7-2)=10

3) Then let the chemical valence of x as r in r2o8 n-, and the valency of r in ro4 2- after the reaction is +6, THE X-6 valence is reduced, and there are 5 R2O8 N- ions involved in the reaction, and 2 R in each ion. Therefore, the valency is reduced by 5*2*(x-6).

4) Increased valency at the time of reaction = decreased valency.

That's it.

A: The valence of Cu has not changed, and the valence of S has decreased, so it is not a redox product.

b: The number of electrons transferred is 20

The valency of D:S decreased, and the valency of Fe increased.

After eliminating the terms that did not change in valence, the reaction can be approximated as 5fes2===7cu2s+3feso4Here only the Fes2 valency actually changes.

It not only rises but also decreases, so it is both an oxidizing agent and a reducing agent. D false. The oxidation product is FeSo4 and the reduction product is right. A false. There should be 21 electron transfers, b wrong.

1. The valency of iron and sulfur has increased and is oxidized; The valency of copper and oxygen is reduced and reduced. So the elements that are reduced are copper and oxygen. Analyzing the electron transfer situation, assuming that 1mol of this reaction is carried out, that is, 8mol of copper is generated.

Iron loses 4mol electrons, sulfur loses 16*6=96mol electrons, oxygen gets 21*2*2=84mol electrons, copper gets 8*2=16mol electrons, and the number of transferred electrons is: 100mol, then for every 1mol of copper generated, the number of transferred electrons is: 100 8=.

2. In Cu2O, copper is +1 valence, oxygen is -2 valence, Cu2S is also +1 valence, and sulfur is -2 valence; The valency of Cu is 0 valence, in SO2, sulfur is +4 valence, and oxygen is -2 valence. Then the valency of copper decreases and the valency of sulfur increases, then the substance containing +1 valence copper is the oxidant, and the oxidant is cuprous oxide and cuprous sulfide.

This kind of topic mainly focuses on the change of valence on OKL. However, if the valency increases, it is oxidized, and when it decreases, it is reduced.

1.。It can be seen in the reactants of the product.

cu :+1 0

fe:+2 +2,+3s：-2 +4

o：0 -2

Then cu o is restored. In the whole 1mol reaction, 8molCu (x1) and 42molo (x2, valence state change is 2) to be reduced, and 8 + 42 x 2 = 92 mol electrons are transferred to form 8 mol of copper.

Answer: The element being restored cu o is transferred.

2.It is still a change in the valence state of all cu from +1 to 0, all of which are reduced and reduced, and both reactants are oxidants.

, galvanic cell 2Fe2++Cl2=2Fe3++2Cl- electromotive force is, known: Cl2 Cl-)=, Fe3+ Fe2+)=, where Cl2 and Cl- are in the standard state. In this galvanic cell, what is C(Fe3+) C(Fe3+).

Solution: e= Cl2 Cl-)-Fe3+ Fe2+)

fe3+/fe2+)=фθ cl2/cl-)-e =

fe3+/fe2+)=фθ fe3+/fe2+)+lg c(fe3+)/c(fe2+)

lg c(fe3+)/c(fe2+)=[ фfe3+/fe2+)-fe3+/fe2+)]/

c(fe3+)/c(fe2+)=

Add enough C2O42- and Ca2+ to the silver electrode Ag+ Ag, and after the reaction reaches equilibrium, C(Ca2+) = calculate the electrode potential of the electrode at this time Ag+ Ag). key：

KSP( AG2C2O4) = KSP CAC2O4 = Data from "Inorganic Chemistry" of the Normal College

1) c(ca2+)=

c(c2o42-)=

2) ksp( ag2c2o4)= = c(ag+)^2 *c(c2o42-)

c(ag+)^2 *

c(ag+)=

ag+/ag)=фθ ag+/ag) +lg [ c(ag+)/c0 ]

--The standard electrode potential is found in an electrode system composed of metal (Ag) and two refractory salts containing the same anion (Ag2C2O4 and Cac2O4) and water and cation (Ca2+) contained in the second refractory salt.

i.e. known: c(ca2+)=1mol l

c(c2o42-)= 1 =

2) ksp( ag2c2o4)= = c(ag+)^2 *c(c2o42-)

c(ag+)^2 *

c(ag+)=

ag+/ag)=фθ ag+/ag) +lg [ c(ag+)/c0 ]

(ag+/ag)=фθ ag+/ag) +lg [ c(ag+)/c0 ]

n(e)=

n(nh2oh)=25*

So the number of electrons lost per n is n(e) (nh2oh) = according to the calculations, we know that the product is a mixture of 1 and that there is definitely n2 (only one electron is lost, if there is more than one electron, n(e) will exceed 2).

So the oxide is N2, N2O

For , in IBR, i is +1 valence, BR is -1 valence (I is more reducible than BR), product HIO, I is still +1 valence, BR in HBR is still -1 valence, and for the whole reaction, the valency of H and O does not change. Therefore, it is not redox release.

For , in thiosulfate ions, the valence state of S is +3 valence, and the valency of elemental sulfide is decreased, and the valency of sulfur dioxide is increased. Therefore, it is a redox reaction.

For , the valence of CR in the dichromate ion is +6 valence, and the valency of CR in chromium oxide is also +6 valence as known from the structure of chromium oxide given in the title. So this reaction is not a redox reaction. (Explanation:.)

O in hydrogen peroxide is -1 valence, when the reaction is carried out, 5 of the 7 O atoms in the dichromate ion enter the generated 5 molecules of water, the remaining two enter two molecules of chromium oxide, and the 8 oxygen atoms in hydrogen peroxide are divided into two equal parts into two chromium oxide molecules. ）

For , according to the hypofluoric acid structural formula, the valence state of O is 0, the family state of F is -1, and the same is true in the product, so the reaction is not a redox reaction. (Analysis: O's electron-stealing ability is stronger than H, so that H and O electron pairs are biased towards O, and O atoms are negative-charged, but F's electron-stealing ability is stronger than O, so that the electron pairs between F and O are biased towards F, which is equivalent to transferring the electronegativity of O to F, so O shows 0 valence, F shows -1 valence, and H shows +1 valence.) ）

The answer to this question is d. But strictly speaking, 2 and 4 are redox reactions because the direction of the electron cloud shifts is changed.

1.In iodine bromide, the electron cloud moves away from iodine and approaches bromine; The electron cloud in the water is far away from hydrogen and close to oxygen; After the reaction, it is still like this, so 1 is not a redox reaction.

2.In thiosulfate, the electrons between the two sulfur atoms are originally biased towards one side, and after the reaction, some of the sulfur atoms' electrons are not offset. So 2 is a redox reaction.

In the same way, 3 is not a redox reaction. And in 4, the electrons of oxygen in hypofluoric acid are biased towards fluorine, while the electron cloud in oxygen is not offset, so it is actually a redox reaction. It's just that the redox is judged by the valence of this question, so the answer is d.

The easiest way to tell is to look at 4

In hypofluoric acid, because f is the most electronegative, f is -1 valence (which is why other hypoic acids are non-oxygen central atoms are positive and f is -1), which can also explain the arrangement of this molecule, the general hypoic acid is h-x-o(n)) and hydrogen is naturally +1 valence, so o can only be 0 valence.

That is to say, there is no valency change in reaction 4, and all options with 4 are wrong. then you can get d correct.

D, 3Cu + 8Hno3 (dilute) = 3Cu (NO3) 2 + 2 No + 4H2O

If there is a reduction in the reaction, (1) the mass of the oxidized Cu = 3*64*

2) Mass of co-consumed nitric acid =

Option D (1) Mass of oxidized Cu: Mass of reduced HNo3 = 3 64:2 63

Mass of oxidized cu =

2) Mass of co-consumed nitric acid: mass of reduced HNO3 = 8:2 mass of co-consumed nitric acid =

According to the title: 1) a, b, and c are all compounds;2) A and B both contain X elements;3) The reaction generates elemental x, which can be inferred to be a "medium-sized" redox reaction. That is, the valency of element X in compounds A and B is "one high and one low" (one is higher than 0 valence and one is lower than 0 valence), and the two work together to form X elemental (valency is 0).

Since a metallic element only presents a positive valence state in a compound and there can be no negative valence state, it is certain that x is not a metallic element;Only non-metallic elements can be both positive and negative valence in compounds, and the centering reaction can occur under certain conditions. For example: 2H2S+SO2=3S+2H2O, 4NH3+6NO==5N2+6H2O, so X should be non-metallic, that is, the answer to this question is B.

Select B, according to the reaction, it is found that A+B-X+C This reaction is the centering reaction in the redox reaction.

X is a kind of elementality, which means that the X element has a change in valency, X has different valency at least in A, B, and because the valency of X is 0 valence, there must be a negative valence of the X element in A, B, and there is no metal element that shows negative valence, so it is a non-metallic element B.