Electrode reaction formula of high school chemistry compulsory 2

I don't think it's necessary to memorize the electrode reactive formula.

You have to remember the chemical reaction.

Then it is clear that the negative electrode loses electrons and acts as a reducing agent, and the valency increases.

The positive electrode gets electrons to act as an oxidant and the valency decreases.

There is no need to remember everything, the electrode reaction is oxidized while reducing at the same time, just remember the redox relationship between the redox and its strength, and the general chemical reaction can cope with it.

It's useless for you to memorize, because the questions are ever-changing. The main thing is to understand the principle.

First of all, write out the equation of the reaction, and then judge the oxidant reducing agent and which is the positive electrode reaction and which is the negative electrode reaction (as long as you remember that the negative electrode is an electron-losing reaction), just disassemble the reaction, if it is acidic, there will be no hydroxide ions, and there will be no hydrogen ions in the alkaline environment, and neutrality is possible, and it is impossible to say, or for example, take the hydrogen and oxygen fuel cell as an example, if it is potassium hydroxide as the electrolyte solution, then.

Cathode: O2+4E

2h2o=4oh-

Negative electrode: 2h2-4e

4oh-=4h2o

Because it is an alkaline electrolyte, what is produced or participated in the reaction is OH- if sulfuric acid is used as the electrolyte, then.

Cathode: O2+4E

4h+=2h2o

Negative electrode: 2h2-4e

4H+ is an acidic electrolyte, so the generation or participation in the reaction is H+, and sometimes it will encounter potassium carbonate solution as the electrolyte.

Positive electrode: O2+2CO2+4E==2(CO3)2-Negative electrode: 2H2+2(CO3)2-

4e==2h2o+2co2

Hydrogen—oxygen. Fuel cell.

There are two types: acid and basic.

In the acid solution, the negative electrode reaction formula is: 2H2-4E-==4H + positive electrode reaction formula: O2+4H++4E-==2H2O electrolyte solution.

It is the negative electrode reaction formula of alkali solution: 2H2 + (4OH)-4E = 4H20 Cathode: O2 + 2H2O + 4E-==4OH - methane fuel cell.

Alkaline) negative electrode reaction formula: CH4

oh-8e-

The positive reaction formula of CO37H2O is: 2O2

4h2o8e-

OH-methanol fuel and wild batteries (alkaline).

Negative pole noisy ridge: 2ch

3oh4h2o+

16oh-=12h2o+

12e-2co3

Positive electrode: 3O2 in the shed

12h2o12e-

12oh-

Because there are only two forms of aluminum in molten salt, there is no presence of Al3+ or Cl- in the solution.

The Al element of the aluminum electrode loses electrons and binds to AlCl4- to become Al2Cl7-

The equation is: al+7alcl4--3e-=4al2cl7-

The negative electrode occurs as an oxidation reaction, and the electrode reaction can be written according to the change in valence. In addition, most of the cathode electrode reactions of gas fuel cells are the reactions of oxygen absorption corrosion, and after the positive electrode reaction is written, the negative electrode reaction can be obtained by subtracting the positive electrode reaction from the total reaction.

In this reaction, you can know the oxidation reaction of methanol according to the change of valency, and the generated carbon dioxide immediately reacts with the base, and the 2 moles of methanol molecules react and lose 12 molar electrons, so that the electrode has 12 units of positive charge before the reaction, and after the reaction, 2 moles of carbonate ions with 4 molar units of negative charge are generated, in order to make the reaction before the reaction also have 4 moles of negative charge, so 16 moles of hydroxide ions should be added to the electrode reaction before the reaction, so the negative electrode reaction: 2ch3oh-12e -+16oh-two 2co3 2-+12h2o

Cathode reaction: 3O2 + 12E -+6H2O = 12OH -

You look for substances that have oxidation reactions and reduction reactions.

The oxidation reaction is CH3OH, which becomes CO32-CH3OH+8OH--6E-=CO32-+6H2O, and the reduction reaction is O2, and the O2 becomes OH-O2+2H2O+4E-=4OH- under alkaline conditions

Step 1: If the negative electrode is written: first find the reducing agent and the corresponding oxidation products, and the second step: mark the valency of the valence element, and determine the number of electrons gained and lost according to the change value of the valence and the number of valence.

Step 3: Distribute the conservation of charge and atom according to the surrounding medium (hydrogen ions, hydroxide ions and water are commonly used in middle schools).

There are three common types of questions:

1。Two electrodes, one electrolyte, are given, like the copper-zinc galvanic cells in the textbook. This electrode reaction is the easiest to write, and the general rule is that the negative electrode itself loses electrons to dissolve, and the positive electrode is a cation in the liquid to obtain electrons to form an element.

We also usually use this phenomenon to judge the positive and negative electrodes of galvanic batteries: the dissolved pole is negative, and the positive electrode is the gas produced or the mass increases).

2。Fuel cell: The law is that flammable gases react at the negative electrode, and oxygen is passed through at the positive electrode. The common test is hydrogen and oxygen fuel cells, and when writing electrode reactions, pay attention to the electrolyte given by the topic:

1) If the electrolyte electrode reaction is made with sulfuric acid solution, the electrode reaction is: negative electrode: 2h2-4e-=4h+

Cathode: O2+4E

4H+ = 2H2O (originally OH- was generated, but this is an acidic solution, and the OH- generated should react with H+ in the solution).

2) If KOH solution is used as the electrolyte: negative electrode: 2H2-4E-

4OH-=2H2O (as above, the generated H+ cannot exist in an alkaline solution, it must react with OH-).

Cathode: O2+4E

2h2o=4oh-

3。Dry batteries and lead-acid batteries, etc.: If you want to test a type of battery, the question will generally give a general chemical equation, and first judge the positive and negative electrodes according to the change of the valency of the elements in the formula (the valency of the negative electron loses electrons increases, and on the contrary, the valency decreases to the positive electrode).

pbo2+pb+2h2so4=2pbso4+2h2o

It is easy to see that the negative electrode is pb, and according to the law, the negative electrode itself loses electrons: pb-2e-==pb2+

However, the product in the given total formula does not have Pb2+, but PbSO4, so it is known that the generated Pb2+ is also the same as SO4 in solution

2-binding, it is easy to get the result, pb-2e-+so42-=pbso4. The next positive electrode is easy to do, and the negative electrode can be subtracted from the total reaction. Thanks for adopting! Hehehe!

There is a general principle:

A redox reaction is divided into two halves, one is a half-reaction in which the electrons obtained by the oxidant become a reduction product; The other is a half-reaction in which the reducing agent loses electrons and becomes an oxidation product. The total number of electrons gained and lost in the two halves is equal.

Here are some examples:

CuSO4 +Fe==FeSo4 + Cu ionic reaction equation: Cu2+ +Fe ==Cu + Fe2+

Oxidation half-reacted) electrode reaction formula (negative electrode): Fe - 2E- ==Fe2+

Electrode reaction formula (positive): Cu2+ +2E- ==Cu

This is an example of a galvanic cell, and the situation is similar to that of an electrolytic cell, except that the order of the particles on the electrode where the redox reaction is taking place is clear.

It's not hard.

Whether it is a galvanic cell or an electrolytic cell, an oxidation reaction always occurs on the anode and a reduction reaction always occurs at the cathode. Who should be preferentially involved in the electrode reaction when discharging is the electrode? In fact, this is exactly the same as the general redox reaction, that is, to see who has the strongest oxidation or reduction (considering electrode materials other than inert electrodes, ions in the solution in the electrode area, and even oxides, insoluble salts and other substances) who preferentially participates in the electrode reaction.

So, the cathodic reaction considers who is most likely to gain electrons, and the anodic reaction considers who is most likely to lose electrons.

Example 2Write the electrode equation and total reaction equation for electrolyzing copper sulfate solution with copper as electrode; Then write the electrode equation and total reaction equation for electrolysis of sulfuric acid solution with graphite electrode.

Analysis: When copper is used as the electrode material for electrolysis, it is obvious that copper is most likely to lose electrons in the anode region, and the ability to obtain electrons is the strongest in the cathode region. Therefore, there is an anode and a cathode.

The sum of the two can obtain the total electrolytic equation: Cu (coarse) Cu (electrolytic copper), which is the refining of copper.

Graphite is an inert electrode and does not participate in the electrode reaction, only considering the discharge sequence of the ions in the electrolyte solution, including considering the concentration. Anode (graphite):

Cathode (graphite):

The sum of the two gives the total electrolytic equation:

Three. The relationship between charging and discharging of the battery.

A rechargeable battery (accumulator) is equivalent to a galvanic battery when discharged and an electrolytic cell when charged. The electrode reaction when the battery is discharged and the electrode reaction when it is charged is also reversible (chemical) to each other.

If you understand it, you don't have to memorize it, but if you don't understand it, you have to memorize it.

Just remember the ability of ions to gain and lose electrons.

1 Cathode: 4(OH-)4E- = O2 + 2H2O Anode: 4(AG+)+4E- = 4AG

Total reaction: 4AGnO3 + 2H20 = 4AG (precipitation) + O2 (gas) + 4Hno3 (reaction conditions: electrolysis or energization).

2 Cathode: 4(OH-)-4E- = O2 + 2H2O Anode: 2(H+)+2E-=H2

Total reaction: 2H2O = 2H2 + O2 (Reaction conditions: electrolytic or energized)3 Cathode: 4(oh-)-4E- = O2 + 2H2O Anode: 2(Cu2+)+4E-=2Cu

The following is the reaction of alkaline zinc-manganese dry cell (followed by acid zinc-manganese dry cell reaction).

The cathode reacts with cathode:

Mno2 H2O e Mno(OH) OH Mno(OH) has a certain solubility in alkaline solution.

mno(oh)＋h2o＋oh－→mn(oh)4－mn(oh)4－＋e→mn(oh)42－

The negative electrode is the anodic reaction:

zn＋2oh－→zn(oh)2＋2e

Zn(OH)2 2OH ZN(OH)42 The total cell response is:

zn＋mno2＋2h2o＋4oh－→mn(oh)42－＋zn(oh)42－

Acid zinc-manganese dry cell reaction:

The cathode is cathode, and manganese is reduced from tetravalent to trivalent.

2mNO2 2H2O 2E 2mNO(OH) 2OH The negative electrode is the anode, and the zinc is oxidized to divalent zinc ions:

The total cell reaction of Zn 2NH4Cl Zn(NH3)2Cl2 2H2E is:

2mno2＋zn＋2nh4cl→2mno(oh)＋zn(nh3)2cl2

This is only one and a half reactions, which is half of the entire redox reaction. In the whole reaction, Zn loses 2mole- and converts to Zn2+, and the other side of Mno2 gets electrons and converts to Mno(OH). In an alkaline environment, Zn2+ will be converted into Zn(OH)2, so part of the reaction formula of Zn is Zn 2Oh- Zn(OH)2 2E- [it seems that the right side should be written as zinc hydroxide, and there is no need to split it into zinc oxide and water].

Total Reaction Formula - Positive Reaction Formula = Negative Reaction Formula.