Find a small redox problem in high school chemistry

The valency of I-ion I has been reduced to its lowest valence -1 and cannot be reduced, so there is no oxidation.

The lowest price of the element is only reducing, the most ** is only oxidizing, and the intermediate valence is both oxidizing and reducing.

Simple non-metallic anions have no oxidation due to their lowest valence.

The valency of some metal cationic elements is not the most **, so it is both oxidizing and reducing. Such as Fe2+.

zn + fe2+ = zn2+ +fe，2fe2+ +cl2 = 2fe3+ +2cl-。

1.The valence state of oxidizing metal cations will decrease, such as ferric ions are reduced to ferrous ions, and ferrous ions can also be oxidized to ferric ions, at which time ferrous ions are reducible, so metal cations are not necessarily only oxidizing.

2.Simple non-metallic anions such as Cl- can only be oxidized to Cl2 and are therefore only reducible.

How are i-ions oxidizing?

If so, what is the product being reduced? i-ion i is already the lowest price of -1, how to reduce the price if it is reduced? ‘

Metal cations are both oxidizing and reducing because of the presence of ions such as Fe2+, which do not reach the highest positive valence.

Iodine ions are only reducible, and the outer shell is already full.

Solution: choose a by the title, oxidation: co

oclfei so reducibility: co

CL reducing CLCO

Therefore, item c is correct.

D item oxidizing Fe

iReduction ife

Therefore, item D is correct.

mol NO3- is all converted to nitrogen and ammonia as products.

Because the volume ratio of nitrogen to ammonia is 4:1

can find the build.

The amount of n2 substance is =

The amount of NH3 substance is =

Charge balance according to redox.

Each NO3- gives 5 electrons to generate 1 2 N2 molecules.

So 1 N2 molecule needs to get 10 electrons.

then the molecule needs to get an electron.

Each NO3- gives 8 electrons to generate 1 NH3 molecule.

Then the NH3 molecule, an electron needs to be obtained.

So no3- needs to get an electron.

Maintain the balance according to the charge.

And only Al loses electrons:

Each Al loses 3 electrons, and to lose an electron, the amount of matter that requires Al is = do not understand hi

The chemical equation is not one less.

10al+6no3-+4oh-=10alo2-+3n2+2h2o

I calculated that it was equal to grams.

In redox reactions, oxidant + reducing agent = reducing product + oxidation product oxidation: oxidant oxidation product.

Reducibility: Reducing agent" reducing product.

A small amount of KI solution was taken in a test tube, bromine water was added first, shaken, then CCL4 was added, and then it was allowed to stand after shaking, and the lower liquid was observed to be purple-red. Reaction: br2 +2ki = 2kbr + i2, so oxidation: a br2>i2

2Fe2 br2===2Fe3 2br So oxidation : C br2>Fe3

Starch turns blue when exposed to elemental iodine.

If the solution turns blue, it indicates that elemental iodine is formed, which can prove that Fe3 is more oxidizing than I22FeCl3 + 2Ki = 2FeCl2 + I2 + 2kCl

Let's look at experiment 1 first: Br2 in bromine water reacts with I-, oxidizes it, and then produces I2, at this time, adding CCL4 is a non-polar solvent, which can dissolve I2, so the lower liquid shows the purple color of iodine molecules.

Experiment 2: Br2 in bromine water can oxidize Fe2+ to become Fe3+, and after adding SCN-, it can form a complex with Fe3+ to replace Br-, red is the color of Fe(scn)3.

Therefore, the first question is the oxidation of 2-valent iron ions by bromine molecules.

In the second question, the Br2 oxidation of I- from Experiment 1, the Br2>I2 was obtained; Oxidation of Fe2+ by Experiment 2 BR2 yields BR2>Fe3

The third question is, if it is known that Fe3 is more oxidizing than I2, then there must be Fe3 that can react with I- and oxidize it to I2, then it can be reacted with reagents 1 and 3, and starch needs to be used in order to indicate whether I2 exists. I think there is something wrong with your answer, you should put the ki solution and starch solution in the test tube now, find that it is colorless, prove that there is no i2, and then add fecl3 solution** and turn blue, which proves that i2 is indeed produced by the reaction.

In chemistry, it is generally forced to be weak. For example, strong acid to weak acid, strong alkali to weak alkali, strong oxidant to oxidant, and so on.

In addition, there is the problem of chemical energy, when there are no external heating, pressurization, electrification and other conditions, after the chemical reaction, the total chemical energy is reduced, and the more active electrons in the outer shell of the atom always tend to the lower energy state. The atoms of matter in the lower energy state are more tightly bound to each other, more stable, and not easy to decompose, and if they need to decompose, the outside world needs to provide higher energy to break the chemical bonds.

1) 14HNO3 3Cu2O 6Cu(NO3)2 2NO 7H2O2) acidic oxidation.

3) and 4) Concentrated nitric acid is used, and the product is formed by nitrogen oxide in the dinuclear part of the modified land.

Because the number of iron in a chromium ferrite molecule is greater than that of chromium, the number of iron ions in the liquid must be greater than the concentration of chromium ions, and the reflected solution is a salt solution, without considering hydrolysis, the anion is only equal to the cation, so the nitrate concentration in the solution after reflection is the largest, then it is ranked as C(NO3-)>C(Fe3+)>C(Cr3+), which is the first question.

The second question, according to the molecular formula of the question, then the x can only take 1, other conditions are not true, there can be half a chromium ion in a molecule, and then you can get the molecular formula, which is Fe2+Fe3+[ Cr3+1· O4]. It can be seen that the number of iron elements in the molecular formula is twice the number of chromium elements. Then, according to the law of conservation of charge in solution, there is 3N(Fe3+)+3N(Cr3+)=N(NO3-), according to the conservation of nitrogen, take the amount of nitric acid minus the amount of overflowing gas is the amount of nitrogen in the solution, that is, the concentration of nitric acid in the solution, then according to the charge conservation formula and N(Fe3+)=2(Cr3+), we can get N(Cr3+)=.

There is only one chromium atom in a chromium ferrite molecule, and the two are one-to-one, so the amount of chromium ferrite consumed is.

The third question, this problem can be solved by using redox knowledge, because the reaction of generating no and no2 is the reaction of divalent iron ions and nitric acid, and the ion reflection equations of the two reactions are written respectively, which are all in the book, and then the snake generates no for xmol and no2 for ymol, according to the total amount of gas generated, we can know x+y=, and then according to the correspondence of the reaction formula, according to the content of divalent iron (equal to the amount of chromium ferrite consumed) is 3x+y=, and the simultaneous equation is solved to obtain x = y = , then x: y-1:2

Redox reaction focuses on: 1. Remember the basic concepts: 1. The reaction with electron transfer is a redox reaction, and the electron transfer includes the gain and loss of electrons and the shift of shared electron pairs.

2. Oxidant: a reactant that obtains electrons in the reaction and reduces the valency of the elements contained. 3. Reducing agent:

A reactant that loses electrons in a reaction and increases the valency of the elements it contains. 4. Oxidation reaction: the process of reducing agent losing electrons, resulting in an increase in the valency of the elements contained.

5. Reduction reaction: the process of the oxidant gaining electrons, resulting in the reduction of the valency of the elements contained. 6. Oxidation:

The properties exhibited by the oxidant are actually the ability of the oxidant to obtain electrons. 7. Reducibility: The properties of the reducing agent are actually the ability of the reducing agent to lose electrons.

It should be noted that the strength of the oxidation of a substance refers to the strength of its ability to obtain electrons, not the number of electrons, so it cannot be said that the more electrons it has, the stronger the oxidation. If chlorine gas becomes chloride ion, each chlorine atom can only get one electron, but sulfur becomes sulfur ion, and each sulfur atom gets two electrons, but the oxidation is stronger than that of sulfur. The same goes for reductivity.

8. Oxidation products: the products obtained by the oxidation reaction when the reducing agent loses electrons. 9. Reduction product:

Dude, try to write it yourself.

2. Figure out the representation of redox reactions: use single-line bridge method and two-wire bridge method to analyze the direction and number of electron transfer. The single-line bridge method targets the electron-gaining and electron-losing atoms in the reactant, and the arrow direction is pointing from the electron-losing atom to the electron-gaining atom.

Indicate the number of transferred electrons above the arrow. The two-line bridge method targets atoms of the same element before and after the reaction, and the number of electrons gained and lost is indicated above the arrows. It should be noted that:

The number of electrons gained and lost must be equal. The number of electrons obtained is multiplied by the number of atoms with reduced valency, and the number of electrons lost is multiplied by the number of atoms with increased valency. I should have attached a specific example, but I can't type it with my hands, so please understand.

3. Master the trim of redox equations. Principle: Equal according to the total number of electrons gained and lost.

Steps: Write the chemical formulas of reactants and products - Mark the valency of each element in each chemical formula - Find the value of the increase or decrease in the valency of each element - Find the least common multiple of the elevated value and the decrease value of the valency - determine the measurement number of oxidant, reducing agent, oxidation product and reduction product - Balance the measurement number of other substances by observation method. I can't make an example of it by hand.

4. Master common oxidants and reducing agents and their important reactions (see the third unit of chemistry for details).

The core of learning redox reaction is the concept. Memorize more, practice more, and if you don't understand, ask more teachers for advice, and that's pretty much it. Good luck!

Look at the valency, the valency increases, the electrons are lost (the electrons are negatively charged), and they are oxidized, and the product they get is called the oxidation product, and they are oxidized by themselves and they are the reducing agent The valency of the flowers decreases, and the electrons are obtained, and they are reduced, and the reduction products are obtained, and they are reduced are oxidants.

I feel the same way, we just reviewed that, and it was oxidized, oxidized, and compared with the strength and weakness, ahhh

If you ask me two years ago, I'm still pretty good, but I'm almost a junior now, and I've almost forgotten about it.

Well, you should first memorize the most basic ranking of the oxidation and reduction of the substance, and then judge the oxide and the reduction and so on.

Me too.,I'm going to be a sophomore soon.,I've struggled in games and ** for a year.,Don't be high school.,It's just that this semester has been failed.,I've forgotten all about it like that buddy!

You just need to remember to return oxygen when you drop it, and you can return oxygen when you lose it. As for the discharge order, it is OK to remember the order of metal activity.