The chemistry galvanic battery question in high school should have a certain degree of difficulty! T

The questions in high school are generally not difficult, and they generally examine which pole and which substance is precipitated first, ionized first, etc., and can be examined from some aspects. You can also come up with calculation problems, such as how long a certain current will be generated, how much gas will be produced, how much precipitation, or how thick the electroplating film can be produced.

You can judge from the Cu-Zn galvanic cells in the textbook, you have to know the composition of the electrolytic cell, which pole is soluble.

Galvanic battery questions are nothing more than that, and you need to master several typical questions. You can find specific examples in the reference book! I can only make a brief summary of the galvanic battery problem here.

1- If the electrolyte is an acid and the electrode is metal, determine which metallic is stronger, and then determine the positive and negative electrodes and the reactive electronic formula. 2- When the electrolyte is some salt and the electrode is metal, look at the reaction that occurs between the ions and the metal. It is a conductor of the unreacted metal, and the reaction that occurs on it is the cation reduction of the electrolyte.

Re-analysis of the positive and negative electrodes, electronic. 3- There is an example that is very important, that is, when the electrolyte changes when the electrodes are aluminum and magnesium, its positive and negative electrodes also change, and you need to remember that example. I think it's very important.

3-Oxyhydrogen fuel cells are also important and a typical example. You have to be able to write about the reactions that occur at the poles. Oxyhydrogen fuel cells are an application of galvanic batteries.

It is a reaction in which chemical energy is converted into electrical energy. I'm wrong here, and I think you're probably going to get it wrong too.

I think if you have a good grasp of all this, no matter how the question is solved, you should be able to do the basic work of galvanic batteries.

That's all for galvanic batteries! I don't know if I can be been. ）

Because aluminum does not react (passivated) in concentrated nitric acid, while copper can react in concentrated nitric acid, the copper sheet is a negative electrode, and the aluminum sheet is a positive electrode, and the copper sheet will gradually dissolve to form copper ions after connection, and the solution will gradually turn blue, and there will be bubbles on the surface of the aluminum sheet, and the gas will be no

The aluminum sheet is passivated to form a dense alumina protective film, and the copper sheet is dissolved, accompanied by the production of NO2. The copper sheet is the negative electrode, the aluminum sheet is the positive electrode, and the sensitive galvanometer pointer is biased towards the copper sheet.

At the beginning, the aluminum sheet passivates without reaction, the copper sheet dissolves and gas is generated, and as the galvanic cell heats up, the aluminum sheet will react, which is the direction deflection of the sensitive galvanometer.

First of all, Al is passivated in concentrated sulfuric acid concentrated nitric acid, which produces a dense oxide film, which prevents the response, and it is a galvanic cell, so Cu dissolves, and reddish-brown NO2 is produced near Al . After a period of time, nitric acid becomes diluted, and dilute nitric acid can react with Al, and the reducing property of Al is stronger than that of Cu, so Al dissolves, and produces reddish-brown gas NO2 near the Cu pole.

The reaction equation is:

Start: al-pole: 2e-

2no3-4h+=2no2+2h2o

Cu pole: Cu-2e-=Cu2+

After some time:

al-pole: 2al-6e-

2Al3+Cu pole: 6NO3-+6E-+12H+=6H2O+6NO2.

Solve the landlord's questioning.

Because electrons flow through the wires and copper to aluminum, electrons react with nitrate ions on the surface of aluminium and vice versa.

The aluminum sheet is passivated, and the copper sheet reacts.

Reddish-brown NO2 is formed on the surface of the aluminum sheet, the copper sheet is dissolved, and the solution gradually turns blue.

There is an electric current passing through the sensitive galvanometer.

Easy to learn the principle of galvanic batteries High School Chemistry Compulsory 2 It's really easy to learn chemistry.

Because this is the galvanic cell principle, iron is the negative electrode, and carbon is the positive electrode.

1.The iron wire has no rust at the liquid surface of the solution, because the iron loses electrons and becomes ferrous ions at the negative electrode, while the positive electrode produces oh-, both of which react between the two electrodes and do not at the negative electrode.

2.There is an iron hydroxide precipitate at the bottom of the solution (as mentioned earlier, Fe2++2OH-=Fe(OH)2), which reacts between the two poles and is then oxidized by oxygen in the solution to Fe(OH)3, 4Fe(OH)2+O2+2H2O=4Fe(OH)3

3.There is no rust or iron hydroxide attached to the carbon rods, for reasons already described.

Because the title says that it is an acidic environment, the positive reaction should be: O2 + 4H+ +4E- = 2H2O, the negative reaction is: 2C2H5OH - 4E- = 2CH3CHO + 4H+, and the total reaction is:

2c2h5oh + o2 = 2ch3cho + 2h2o。

As for why the positive reaction cannot be written as O2 + 4E- +4H+ = 2H2O under acidic conditions, it is because there can be no large amount of OH- under acidic conditions, O2 electrons and water generated by OH- immediately neutralized OH- +H+ = H2O, to merge the two reactions of O2 + 4E- +4H+ = 2H2O and OH- +H+ = H2O into O2 + 4H+ +4E- = 2H2O.

Because under acidic conditions:

The positive reaction formula should be: O2+4E-+4H+=2H2O cannot generate OH-.

Happy New Year.

Select option D. The device constitutes a galvanic cell. Because zinc is more active than copper, zinc loses electrons as a negative electrode, copper is a positive electrode, and Cu does not have h activity "Liveliness can be consulted in the order table of metal activity".

So the positive electrode is H+ and the electrons become H2 (hydrogen)...In the whole galvanic cell reaction, the amount of SO4 does not change, nor does the amount of solution. So the concentration hasn't changed.

b Zinc is used as the negative electrode in the zinc-copper dilute sulfate primary battery, which loses electrons and shows positive valence, so the sulfate goes to the negative electrode.

In the stem information: ch3ch2oh-2e-=2x+2h+Wrong, it should be changed to ch3ch2oh-2e-=x+2h+

Based on this atomic conservation, it can be inferred that x is CH3CHo, and it can be inferred that the galvanic battery works in an acidic environment, and if it is alkaline, there will be no H+ generation. Therefore, the positive reaction should be O2 + 4H+ +4E- = 2H2O

The total reaction was: 2ch3ch2oh+o2=2ch3cho+2h2o.

The electrolyte is acidic, so B cannot be present even for OH-.

Referring to acidic hydrogen-oxygen fuel cells, so the negative electrode reaction should be O2 + 4H+ +4E- = 2 H2O

If you choose d, this kind of problem basically verifies that the charge is conserved, the valency is correct, and the product is reasonable, that is correct.

Great, you're sorry for your chemistry teacher! In galvanic cells, the cations in the electrolyte move to the positive electrode, and the anions in the electrolyte move to the negative electrode. (The situation of fuel cells is different, and it is impossible to tell for a while).

Reason: The essence of galvanic cells is that redox reactions occur. The active metal of the negative electrode undergoes an oxidation reaction, loses electrons and the electrons flow to the positive electrode through the wire, so there are a large number of electrons on the surface of the positive electrode, which attracts the cations in the solution to move to the positive electrode, and at the same time, the active metal of the negative electrode loses electrons and forms a large number of metal cations near the negative electrode, and the positively charged metal cations attract the negatively charged anions in the electrolyte to move to the negative electrode.

You may also ask why the cations produced by the negative electrode do not move towards the positive electrode, because different ions move at different rates).

It's the opposite. In galvanic cells, the cations move to the positive electrode and the anions move to the negative electrode. The battery is inherently a closed loop.

The internal circuit of the battery is the internal circuit, and the current direction of the internal circuit is from the negative electrode to the positive electrode, the cation is consistent with the current direction, and the anion is opposite to the current direction. It can also be explained by the principle of attraction of opposite-sex charges.

The cations in the galvanic cell should move to the positive electrode, and the anions should move to the negative electrode, because the positive electrode material gains electrons, the ions show negative valence, so they attract the cations past.

There is a mistake in your question!!

Your question reverses the knowledge point, the cation in the galvanic cell should move towards the positive electrode, and the anion should move towards the negative electrode.

Your question is wrong, the cations in the galvanic cell should move towards the positive electrode and the anions towards the negative electrode.

When the negative electrode loses electrons, it becomes a cation, attracting the anion to pass.

Negative electrode oxidation produces freely moving cations, and it should be that the anion in the solution moves to the negative electrode following the principle of opposite-phase attraction.

There is an electric current flowing through the closed loop.

Dilute sulfuric acid acts as an electrolyte solution, forming a closed loop, participating in the galvanic cell reaction, the small bulb glows, the sponge turns black, (mNO4)-+e=(mNO4)2- has a closed reflux, and the redox reaction is carried out spontaneously.

The total reaction formula must be superimposed on two and a half reactions, as is the case in any case. So as long as you write about the half reaction, the total reaction will never be wrong.

If there is no obvious electron-gaining substance in the solution, it can be considered O2 to gain electrons. Write according to acidity and alkalinity.

If the cathode material is a carbon rod, then C will not get electrons, and if it is a lively element such as Cl2 and O2, it will get electrons.

On the contrary, the positive electrode is the cathode and the negative electrode is the anode. The so-called positive and negative electrodes refer to the movement of electrons, the negative electrode must be the outflow of electrons, that is, the loss of electrons, and the positive electrode is the inflow of electrons, that is, the acquisition of electrons. The cathode and anode refer to the level of electric potential, and the high potential is called the anode, and the low potential is called the cathode.

As the negative electrode loses electrons and is left with a positive charge, the electric potential rises and is therefore the anode. In the same way, the positive electrode is the cathode.

1.It's all about writing total reactions. The first one you said only writes about one pole is actually Zn oxidation at the anode and H+ reduction at the cathode. A reaction equation does not necessarily have to be completely reacted at one pole, but can be reacted by transferring electrons to the other pole through an electrical circuit.

2.In general, the dissolved oxygen reaction is not written, but OH- is oxidized to oxygen.

3.The inert electrode (not necessarily a non-metal, such as Pt) itself is not involved in the reaction, but is reacted on top of ions in solution, such as H+ or OH-.

4.The anode is the pole where the oxidation reaction takes place, and the cathode is the pole where the reduction reaction takes place. So when charging, the anode is the positive electrode, and the cathode is the negative electrode. When discharged, the anode is the negative electrode and the cathode is the positive electrode.

This is actually very simple, first of all, the battery is mainly divided into two types, one is the electrolyte to participate in the reaction, and the other is that the electrolyte only does the conductive medium and does not participate in the reaction. No matter what kind of reaction they are in essence, but the gain and loss of electrons is not a direct transfer, that is, a face-to-face giving, but an indirect transfer of electrons through the circuit connecting the oxidant and the reducing agent. In other words, almost all redox reactions can be used to power batteries.

First of all, the electrolyte participates in the reaction, such as Cu-H2SO4-Zn, this reaction sulfuric acid is used as an electrolyte at the same time, but also as an oxidant, the battery principle is essentially Zn to replace H2, and the chemical reaction formula is the total reaction formula of the battery principle. Therefore, at this time, the reaction formula between the negative electrode and the electrolyte is the total reaction formula; Therefore, it can be said that galvanic cells with only electrolyte participation in the reaction are based on the reaction formula of the negative electrode and the electrolyte as the total reaction formula. The second is that:

The electrolyte only makes a conductive medium and does not participate in the reaction (strictly speaking, it is a redox reaction): this is the case for lead-acid batteries, the same is true for the more popular fuel cells, and basically all batteries that can be charged are. What they all have in common is that the oxidant and the reducing agent are fixed or adsorbed at the positive and negative electrodes, respectively, i.e. the oxidant (not only O2) and the reducing agent are not in direct contact.

It is precisely because the two are separated that they are superimposed with positive and negative formulas. Got it. For example, the negative electrode is dissolved H2 and CH4, so the positive electrode can be used to form a fuel cell with dissolved O2.

The essence of the reaction H2 is the reaction formula + the number of electrons to be burned in O2. Anyway, the negative electrode must lose electrons (the electrons move in the opposite direction of the current), H2 loses electrons to form H+, and C loses electrons is basically CO2. What is produced in the end depends on the electrolyte.

Generally, fuel cells use alkali as the electrolyte. There is no need to ask the first question of the third question, by the way, the carbon rod is only used as an inert material for the electrode. You can use PT too.

Hey. The positive electrode is the anode, and the negative electrode is the cathode, right. It's just an anode, cathode is mainly used in electroplating, electroanodizing and other surface treatment fields, as well as electron beam, etc., cathode ray is actually an electron beam.

Ask for a collection. Hey.

1. In fact, the positive and negative formulas are superimposed, but some can be eliminated left and right. The total reaction formula reflects the substances that are really involved in the reaction.

2. When the solution is acidic, it is hydrogen evolution: when it is neutral or alkaline, it is oxygen absorption (that is, dissolved oxygen). However, there are exceptions.

3. It depends. For example, the Cu in the Cu-H2SO4-ZN galvanic battery can be replaced with a carbon rod, and the reaction formula of the positive electrode is still the same as the original.

4. The positive electrode and negative electrode are the concepts in the galvanic cell, and the anode and cathode are the concepts in the electrolytic cell. In general, it can be memorized in this way.