Senior High School Final Chemistry Review-Ch16

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1 Oxidation-reduction reaction (Redox reaction) Redox reaction (Ch 16)

1.1 Redox reaction Redox reaction

The physical essence of Oxidation-reduction reaction is electron transfer.

1. Oxidation: The process of losing electrons, oxidation reaction
2. Reduction: The process of obtaining electrons, reduction reaction
3. reducing agent: Reducing agent, oxidize others and are reduced by themselves
4. oxidizing agent: Oxidizing agent, others are reduced by themselves

e.g.
$$\ce{C(s) + O_2(g) ->[ignite] CO_2(g)}$$
During this reaction, $C$ is a reducing agent and $O_2$ is an oxidizing agent.$C$ is oxidized by $O_2$, reducing $O_2$, losing electrons; $O_2$ is reduced by $C$, oxidized $C$, and obtaining electrons.

1.2 oxidation number: electron transfer from the elemental state to the current state.

This part needs to be given to any compound, so that the oxidation number of each element can be judged; and to give an equation, so that the oxidation number can be judged.

e.g.1
$$ \overset{+4}{C}\overset{-2}{O_2}$$
$\overset{+4}{C}$ indicates that the oxidation number of $C$ is +4, and 4 oxidation has been experienced from the elementary state. $\overset{-2}{O_2}$ indicates that the ocxidation number of $O$ is -2, and 2 reduction has been experienced from the elementary state.In terms of numerical value, it is actually “number of electrons that have been lost” (positive is loss, negative is gain).The overall is 0 because both $C$ and $O$ have become like this from a single substance (electrical neutral).

Note that it should be distinguished from the charge of the ions. The charge of the ions refers to the positive charge of several units of the chestnut’s current state, and the oxidation number refers to the electron transfer from the elemental state to the current state.
e.g.2
$$\overset{+2}{Mg^{2+}} \quad \overset{-1}{Cl^-}$$
$\overset{+2}{Mg}$$ is the oxidation number of $Mg$. So far, this state has lost two units of electrons and has experienced two oxidation. The $2+$ of ${Mg}^{2+}$ is the charge of the ion, which means that the current ${Mg}^{2+}$ has 2 units of positive charge.Similarly, $-1$ of $\overset{-1}{Cl}$ is the oxidation number of $Cl$, while $-$ of ${Cl}^{-}$ is the charge of ions, which means that the current state of ${Cl}^{-}$ has a negative charge of 1 unit.

e.g.3
$$\overset{0}{Zn} \quad \overset{+2}{Zn^{2+}} \quad \overset{+2}{Zn}\overset{+6}{S}\overset{-2}{O_4} \quad \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}(aq) \quad \overset{+1}{K_2}\overset{+6}{Cr_2}\overset{-2}{O_7}(aq)$$

1.3 Electronic transfer analysis and balance (4’*2)

e.g.

1. $\ce{\overset{0}{Zn}(s) + \overset{+1}{H}_2SO4(aq) -> \overset{-2}{Zn}SO4(aq) + \overset{0}{H}_2(g)}$
   
   In this process, $Zn$ loses two units of electrons, $H$ obtains one unit of electrons, but a total of two $H$ obtains electrons, so the total reaction electrons are conserved.Among them, $Zn$ is oxidized by $H_2SO_4$, $H_2SO_4$ is reduced by $Zn$; $H_2SO_4$ is an oxidant, and $Zn$ is a reducing agent.
2. $3\ce{\overset{0}{Cu}(s) + 8H\overset{+5}{N}O3 -> 3\overset{+2}{Cu}(NO3)2(aq) + 2\overset{+2}{N}O(g) +4H2O(l)}$
   
   In this process, 3 $Cu$ loses two units of electrons, a total of 6 units of electrons are lost, 2 $N$ obtains three units of electrons, a total of 6 units of electrons are obtained, and the total reaction electrons are conserved.Among them, $Cu$ is oxidized by $HNO_3$, $HNO_3$ is reduced by $Cu$; $HNO_3$ is an oxidant and $Cu$ is a reducing agent.
   
   It should be noted that during this process, some N did not undergo electron transfer.In this process, only two $N$ undergo electron transfer, while the other six $N$ do not undergo electron transfer, their oxidation number has not changed, and is still +5.This is because the oxidation number of $N$ in $HNO_3$ is +5, and the oxidation number of $Cu(NO_3)_2$ is also +5, so these six $N$ have not undergone electron transfer.These six $N$ are only in different matters in this process, without undergoing electron transfer, and their oxidation number is not changed.
3. $\ce{\overset{0}{Cl}_2(g) + 2NaOH(aq) -> Na\overset{-1}{Cl}(aq) + Na\overset{+1}(Cl)O(aq) + H2O(l)}$
   
   In this process, one $Cl$ gets one unit of electrons, and the other $Cl$ loses one unit of electrons, and the total reaction electrons are conserved.
   It should be noted that in this process, ***Cl undergoes electron transfer and self.In special circumstances, it is possible that the same substance transfers electrons by itself.In this case, $Cl$ itself is both an oxidizing agent and a reducing agent.

1.4 Application of redox reactions (Chemical equations need to be paid attention to)

Blast furnace iron smelting

$$\ce{\overset{+3}{Fe}_2O3(s) + 3\overset{+2}{C}O(g) ->[high temp] 2\overset{0}{Fe}(s) + 3\overset{+4}{CO}_2}(g)$$
$$\ce{C(s) + O2(g)->[ignite] CO2(g)}$$
$$\ce{C(s) + CO2(g) ->[high temp] 2CO(g)}$$
$$\ce{CaCO3 ->[high temp] CaO + CO2}$$
$$\ce{SiO2(s) + CaO(s) ->[high temp] CaSiO3(l)}$$

Note: Using CO can increase the reaction area

$\ce{C(s) + O2(g)->[ignite] CO2(g)}$, $\ce{C(s) + CO2(g) ->[high temp] 2CO(g)}$ These two steps can generate $CO$ to restore $Fe$.
$\ce{CaCO3 ->[high temp] CaO + CO2}$, this step has two functions.On the one hand, $CO_2$ can be generated, and then coke can be added to generate $CO$ to reduce $Fe$; on the other hand, the generated $CaO$ can react with $SiO_2$ to generate $CaSiO_3$. The function of this step is to turn the stones in iron ore into liquid to facilitate flow out, and improve the purity of iron (if it is solid, it is difficult to get the stones away from the iron ore).

Note that it is necessary to distinguish. The first three chemical equations above are Redox reactions, but the last one is not:
$$\ce{SiO2(s) + CaO(s) ->[high temp] CaSiO3(l)}$$
During this process, the oxidation number of $Si$ and $O$ has not changed, so this process is not a Redox reaction.This reaction is the reaction between acidic oxides and alkaline oxides to form salts (ionic compounds, but liquids at high temperatures).

Detergent

$$\ce{Cl2(g) + 2NaOH(aq) -> NaCl(aq) + Na\overset{+1}{Cl}O(aq) + H2O(l)}$$
In this reaction, $\overset{+1}{Cl}$ lacks electrons, has strong ability to grab others’ electrons (strong oxidizer), has good sterilization ability, and has good decontamination effect (directly destroys the stain structure).

Metal corrosion and protection

$$\ce{3Fe(s) + 2O2(g) ->[high temp] Fe3O4(s)}$$
Through this reaction, a dense oxide layer is formed on the surface of the iron, called baked blue.At the same time, applying a layer of paint is also a good way to prevent metal corrosion. See Ch 1 for details.

Chemistry and Bioluminescence (Luminol)

When the light is shining, electron transfer (electron transition) will definitely occur.