Redox reactions in high school chemistry, redox reactions in high school chemistry

In this reaction, there is only the electron gain and loss of nitrogen element, the valency has risen and decreased, so C is wrong, the increase of nitrogen in metadimethylhydrazine is oxidized, it is a reducing agent, so a is wrong, the reaction is the recombination of atoms, so there is an endothermic and exothermic process, C is wrong, D is left, of course, there are eight electrons transferred from two N2O4 to N2 D is correct.

Redox reaction: There is a reaction in which the valency of the elements increases and decreases.

High school chemistry knowledge point 1, whether a substance is used as an oxidant or a reducing agent in the reaction mainly depends on the valency of the element.

Active non-metallic elements, such as F2, Cl2, Br2, O2, etc. Oxides, oxygen-containing acids and oxidates when the elements are in **, such as mNO2, NO2; concentrated H2SO4, HNO3; KMNO4, KCLO3, FeCl3, etc.

Application: Mark the valency of the constituent elements of each substance in the chemical equation, as long as the valency of one element has changed, it can be said that the reaction is a redox reaction.

Formula: Raise the oxygen reductant, reduce the oxidant (valence rises, electrons are lost, oxidation reaction occurs, oxidation products are obtained by oxidation, and the reducing agent is used as a reducing agent in the reaction; The valency decreases, electrons are obtained, a reduction reaction occurs, and the reduction product is obtained by reduction, which is used as an oxidant in the reaction).

Answer: First of all, explain the correct system.

The answer should be d, not c.

A false baife - du + 3 valence Fe, by nano2 n ——- 3 valence N, electrons have been transferred, is oxygenation reduction DAO reaction.

b False, is the generation of NH3

c False, Fe3O4 can be rewritten as Fe2O3*FeO (i.e., a composition of ferrous oxide and ferric oxide) Fe 2Fe———3valence Fe loses a total of 6 electrons, Fe———2valence Fe loses 2 electrons. 3Fe – >Fe3O4 iron loses a total of 8 electrons. So the gram loses electrons in total.

d For nano2, +3N———3 valent N(NH3), nano2 is the oxidizing agent.

a.React.

The iron of Bai1 is obviously valened, so it is a redox DU reaction BThis process uses zhi

Or DAO produces back pollutants such as sodium nitrite, ammonia, and so on. Answer: Fe from 0 to +2 to +6 respectively (sodium ferrote and sodium ferrate) so yes.

The nitrogen atom of d sodium nitrite is reduced from +3 to -3 and is reduced to act as an oxidant.

Is it wrong .........?

a False, it is a redox reaction, because there is a valence change, and the iron rises from 0 to +2 valence.

This third equation is not right, there are two fe products in reaction physics, and there are three fe in the product, and the number of atoms is unequal.

A increase in the valency of iron and a decrease in the valency of nitrogen are redox reactions.

b produces NH3 and nano2 is used to produce pollution.

d. The valency of nitrogen element is reduced to oxidant.

(Cl2) is the oxidizing agent, (SO2) is the reducing agent, (S) element is oxidized, (Cl) element is reduced, (H2SO4) is the oxidation product, (HCl) is the reduction product.

Secret: Commonly known as the three-character classic.

Oxidizing reducing agent (elevated valency, losing electrons, being oxidized, is a reducing agent, oxidation product).

It is also reduced to oxidant (if the valency is reduced, electrons are obtained, and it is reduced, which is an oxidant, a reduction product).

After understanding this three-character scripture, no matter how difficult the redox reaction is, it can be done.

It is 3BRF3 + 5H20 = 9HF+HBRO3 + O2 + BR2, where looking at the O element, there are only 2 O atoms whose valency changes from -2 to 0 (that is, the number of O atoms that are oxidized is 2), these 2 are from H2O, that is, the water of 2 is oxidized, and at the same time, 3 BRF3 are all involved in the gain and loss of electrons, but 1 becomes HBRO3, the valency is increased and oxidized, and 2 become BR2 and are reduced. That is, 2 H2O and 1 BRF3 reduce 2 BRF3, according to the electron gain and loss (O gets 4 electrons, BR gets 2 electrons), it can be seen that 2 H2O reduces 4 3 BRF3 so the water is oxidized, there is reduction, there is.

First of all, the chemical equation is wrong should be 3BRF3 + 5H20 = 9HF+HBro3+O2 +BR2——— the reaction equation can be regarded as two parts, one is the oxidation of H2O by BRF3, and the other part is the self-redox of BRF3 into BR2 and BRF5, (BRF5 has no electron transfer when hydrolyzed into HBro3 and HF), and these two reactions react in a certain proportion.

So there is a redox reaction of BRF3 with water 4BRF3 + 6H2O==2BR2 + 3O2 +12HF

So it's easy to see clearly.

According to the above ratio H2O:BRF3=3:2, so BRF3 reduced by water has M = amount of substance * molar mass =

The reactive formulas of the questions are not balanced.

(1) Determine what kind of gas is generated

The formation of two gases, a total of 3 elements, Al, O, N, may be NO, NO2, or O2 of the two, because the reaction conditions are high temperature, 2NO+O2=2NO2, the reversible reaction is carried out in the direction of the generation of NO2, so one of the gases is determined to be NO2

The other gas may be O2.

There is also an Al element left, which generates a stable Al2O3

So the equation (unbalanced): Al(no3)3 = Al2O3 no2 O2

2) Trim: n of no3)- is +5 valence, no2 is +4 valence, loss of 1 e-, o of no3)- is -2**, o2 is 0**, and 4 e- are obtained.

Electron e-, the relationship between gain and loss is 1:4, so al(no3)3 is preceded by 4, and the rest is the problem of conservation of elements, so al2o3 is preceded by 2, no2 is preceded by 12, so 4 al(no3)3=2 al2o3 12 no2 o2

The front 4al(no3) is 36 o, = the back 2 al2o3 +12 no2 is a total of 30 o, and there is still 6 o, so o2 is preceded by 3, and 4 al(no3)3=2 al2o3 12 no2 3 o2

It has been verified that the relationship between no2 and o2 corresponds to the question stem is 4:1, so the answer is: 4 al(no3)3=2 al2o3 12 no2 3 o2