How to write the more complex electrode reaction formula in high school chemistry

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 loss 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 it is possible to neutral, and it is impossible to say, or for example, take the hydrogen and oxygen fuel cell as an example, if potassium hydroxide is used as the electrolyte solution, then the positive electrode: O2 + 4E + 2H2O ==4OH - Negative electrode: 2H2 - 4E + 4oh- ==4H2O Because it is a basic electrolyte, it is all oh- if sulfuric acid is used as the electrolyte, then the positive electrode:

O2 + 4E + 4H+ ==2H2O Anode: 2H2 - 4E ==4H+ Because it is an acidic electrolyte, it is H+ that is generated or participated in the reaction

2h2+2(co3)2- -4e==2h2o+2co2

It's good to memorize a few that are often written, and if you read a few more questions, you will naturally understand.

Electrode-reactive writing.

Principle: Electrode reactions are basically redox reactions, which should follow the conservation of mass, electron and charge. In addition to this, follow the following:

1.Additive principle: the two electrode reaction formulas are added, and the total reaction formula of the battery is obtained after the electrons are removed. Using this principle, the total reaction formula of the battery is subtracted from the known reaction formula of one electrode to obtain the reaction equation of the other electrode.

2.Principle of coexistence: CO2 cannot exist in alkaline solution, and there will be no H+ to participate in the reaction or generation; Similarly, in acidic solutions, there will be no OH to participate in the reaction or be generated, and there will be no presence of carbonate ions.

According to this principle, the form in which a substance gains and loses electrons in different media environments is different. We can write according to the acidity and alkalinity of the electrolyte solution to determine whether H2O, OH, H is on the left or right side of the equation.

At the same time, there is 1Write the electrode reaction according to the battery device diagram 2Write the electrode reaction according to the total cell response3Writing of the electrode reaction of the secondary battery 4Writing of fuel cell electrode reactions, etc.

Friends, you can watch it directly.

The most classic method: the writing of the electrode reaction equation of electrochemistry: (1) write the total reaction equation (2) see who does the positive electrode and who does the negative electrode (that is, the electron loss is the negative electrode, and the electron gain is the positive electrode) (3) indicate the ion destination (that is, the valency of an element in the reactant, to the valency of the product of an element, whether it rises or decreases is indicated by arrows in the total reaction equation) (3) Write the reaction equation of the positive and negative electrodes, rule (1) (to write the negative electrode standard, The same is true for the positive electrode) write the reactant in the total reaction equation in which the reactant loses electrons, after writing how many electrons are lost (the minus sign is used for the loss of electrons, and the plus sign is used to get the electrons, that is, the reactant in the total reaction equation that the reactant loses electrons minus the electrons he loses), and then write the electrons of the reactants that lost electrons in the total reaction equation to the "who" of the product of the total reaction ("who" It refers to the item that loses electrons in the reactant to the product formed by the change in the valency of the reactant) and then writes it out.

It's written on the reactant side, and this one is written on the product side. (2) Since the left and right charges in the written electrode equation are not conserved, it is necessary to find out the products in the total reaction equation (premise: find out the products in the total reaction equation to be the closest to the elements in the electrode equation just written, and disassemble it into ions, of course, the disassembled ions should also be closest to the elements in the electrode reaction equation), and then carry out charge conservation.

There are special cases where the "secondary step" is used (the second step refers to, for example, an electrode reaction containing a "salt bridge", which is the use of the OH in the salt bridge for charge conservation. (3) Since the left and right elements are not necessarily conserved after the conservation of charge, the elemental balancing is carried out by the product of the total reaction. (Select the product of the total reaction to select the element closest to the electrode reaction for trimming, note:.)

This time, the product of the total reaction is selected without dismantling. ）

It's over, I'm exhausted. This is the only law that writes the electrode reaction equation. Exclusive invention. It is suitable for writing general electrode reaction equations. It can be simplified but not invented yet, I hope it helps you. Hope spreads widely. Hehe.

First write the general equation, then actively lose electrons to make the negative electrode, which is oxidized (referred to as negative oxygen activity), and then write the one pole, which can be subtracted.

You can send the topic ** up, generally on the basis of the original chemical equation divided into two, the negative electrode plus the electron, the positive stage plus the loss of electrons.

Positive electrode: 2O2+ 4H2O+8E- == 8OH-, the total reaction minus the positive electrode reaction yields the negative reaction: CH4+10 OH- —8E-==(CO3)2-+7H2O

The positive electrons of the galvanic cell lose electrons at the negative electrode. The anode of the electrolytic cell loses electrons, and the cathode gains electrons.

Galvanic type, the electrolyte is dilute sulfuric acid or copper sulfate.

The electrolyte is NAOH

3.For lead-acid batteries, PB is the negative electrode, PBO2 is the positive electrode, and the electrolyte is dilute sulfuric acid.

4.Hydrogen-oxygen fuel cells, hydrogen is the negative electrode, and oxygen is the positive electrode.

5.In methanol fuel cells, KOH is the electrolyte, which does not produce carbon dioxide, but carbonate.

6.In ordinary alkaline batteries, Zn is the negative electrode, Mno2 is the positive electrode, and the lye is the electrolyte, which generates Zn(OH)2 and MNO(OH).

Basically, there are only these types, and the others are derived, focusing on the first three, but the latter ones are also tested.

Determine the two poles of the following 4 groups of galvanic cells, the electrode reaction equation, will you do it?

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

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.

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