Chemical reaction in the battery, why is there a chemical battery

The first one is that it is all oxidation.

Zinc and carbon are oxidized and lose electrons The lost electrons move in a directional way to form an electric current.

Actually: CH4 + 2O2 = ignition = CO2 + 2H2O It is very difficult to design as a battery electrons move in air or vacuum.

It's hard. So the second one do not write ethanol.

Burn. The second writes li+mno2=limno2 Lithium is an alkali metal, which is very reactive and easily loses electrons and is oxidized by **manganese.

It depends on what kind of battery it is.

The main component of the positive electrode of the lithium-ion battery is licoo2, and the negative electrode is mainly when charging.

Positive reaction: licoo2 li1-xcoo2 + xli+ +xe-

Negative reaction: C + xli+ + xe- clix

Total cell response: LiCoO2 + C Li1-XcoO2 + Clix

The nickel-cadmium battery uses Ni(OH)2 as the positive electrode, CDO as the negative electrode, and lye (mainly KOH) as the electrolyte.

ni(oh)2 –e + oh- →niooh + h2o

The reaction that occurs at the negative electrode:

cd(oh)2 + 2e →cd + 2oh-

The total reaction was: 2Ni(OH)2 + CD(OH)2 2NiOH + CD+ 2H2O

When discharged, the reaction is reversed NiooH + H2O + E Ni(OH)2 + OH-

cd + 2oh- +2e→cd(oh)2

When charging, with the increase of NiOOH concentration, the concentration of Ni(OH)2 decreases, the potential of the positive electrode gradually increases, and with the increase of Cd, the decrease of CD(OH)2, the potential of the negative electrode gradually decreases, when the battery is fully charged, the potential of the positive electrode and the negative electrode reach an equilibrium value, and the difference between the potential of the two is the charging voltage of the battery.

NiMH batteries use the same Ni oxide as the positive electrode, hydrogen storage metal as the negative electrode, and lye (mainly KOH) as the electrolyte, and when the NiMH battery is charged, the positive electrode reacts as follows: Ni(OH)2E + OH-NIOOH + H2O

Negative reaction: MHN + NE M + N 2H2

When discharged, the positive electrode: NiooH + H2O + E Ni(OH)2 + OH-

Negative electrode: m + n 2h2 mhn + ne

The reverse reaction of the above reaction occurs when discharging.

In a chemical battery, the direct conversion of chemical energy into electrical energy is the result of spontaneous chemical reactions such as oxidation and reduction within the battery, which are carried out on two electrodes. The anode active material consists of a reducing agent with a negative potential and stable in the electrolyte, such as reactive metals such as zinc, cadmium, lead, and hydrogen or hydrocarbons.

The cathode active material is composed of oxidants with positive potential and stable in the electrolyte, such as metal oxides such as manganese dioxide, lead dioxide, nickel oxide, oxygen or air, halogens and their salts, oxygenated acids and their salts, etc. Electrolytes are materials with good ionic conductivity, such as aqueous solutions of acids, alkalis and salts, organic panicles or inorganic non-aqueous solutions, molten salts or solid electrolytes.

When the external circuit is disconnected, although there is a potential difference (open-circuit voltage) between the two poles, there is no current, and the chemical energy stored in the battery is not converted into electrical energy.

When the outer circuit is closed, a current flows through the outer circuit due to the potential difference between the two electrodes. At the same time, inside the battery, due to the absence of free electrons in the electrolyte, the transfer of charge must be accompanied by the oxidation or reduction reaction at the interface between the active substances and the electrolyte at the two poles, as well as the material migration of reactants and reaction products.

The galvanic cell reaction is a redox reaction.

A device that converts chemical energy into electrical energy is called a galvanic battery. In galvanic cells, the chemical reaction that takes place is called redox reaction. Electrons flow out from one pole where the oxidation reaction occurs, and the substance that acts as a negative electrode and loses electrons is a reducing agent.

Electrons flow to the positive electrode through the negative electrode of the galvanic cell through the wire, and the electrons are obtained by oxidizing the agent on the positive electrode, and a reduction reaction occurs. Galvanic cells convert chemical energy into electrical energy through chemical reactions.

There are four indispensable conditions for the composition of galvanic batteries (pole, electricity, circuit, and spontaneity):

1. Two metals with different activity (or the other is a non-metallic conductor) constitute the electrode.

2. Electrolyte solution.

3. Form a closed loop.

4. The redox reaction can be carried out spontaneously.

Judgment of positive and negative electrodes of galvanic battery:

1. According to the electrode material: the more active pole is negative, and the more or less active pole is positive (electrons are lost by reacting with the electrolyte).

2. According to the reaction that occurs at the two poles: the one pole where the oxidation reaction occurs is negative, and the one pole where the reduction reaction is positive.

3. According to the electrode weight gain or weight loss: the pole that dissolves or decreases is negative, and the pole that increases or releases bubbles is positive; (Lead-acid battery (PB-PBO2) can be used as a classic special case: when discharged, because the positive electrode PBO2 changes to PBo4 after the reaction, and the negative electrode PB also changes to PBo4 after the reaction, the weight increases, and the solution solute H2SO4 decreases.)

4. According to the direction of electron or current flow: current direction: positive and negative electron flow direction: negative positive.

5. According to the direction of ion movement in the solution: the pole of anion shift is negative, and the pole of cation shift is positive.

In general, hydrogen ions are involved in acidic conditions, water in neutral conditions, and OH- in alkaline conditions. Here are some of my summaries you can check out.

Electrode-reactive writing.

According to the writing of the galvanic battery device.

Negative electrode: If the negative electrode material itself is oxidized, its electrode reaction formula is in two cases:

One is that the metal cation formed after the negative electrode metal loses electrons and does not react with the components of the electrolyte solution, and the electrode reaction can be expressed as M ne Mn.

The other is the reaction between the metal cation generated after the negative electrode metal loses electrons and the electrolyte solution components, and the electrode reaction at this time should superimpose the reaction of the metal electron loss and the reaction of the metal cation and the electrolyte solution, such as the negative electrode reaction of the lead-acid battery is: PB+2OH 2E 2H2O.

If the anode material itself does not react, such as a fuel cell, when writing the anode reaction formula, the reaction of the fuel cell losing electrons and its products should be superimposed with the electrolyte solution reaction to write, such as the negative electrode reaction of H2 O2 (KOH solution) battery.

h2+2oh－－2e－＝2h2o。

Cathode: The substance that reacts at the cathode is first determined. When the anode material undergoes a spontaneous chemical reaction with the electrolyte solution, the substance that undergoes the electrode reaction on the positive electrode is a certain particle in the electrolyte solution; When the anode material does not undergo a spontaneous chemical reaction with the electrolyte solution, the substance that reacts on the positive electrode is O2 dissolved in the electrolyte solution.

Then write the positive electrode reaction formula according to the specific situation, and consider whether the positive electrode reaction product reacts with the electrolyte solution when writing, and if it reacts, it is also necessary to write the superposition formula.

Write according to the galvanic cell reaction.

Find out the substances that have oxidation and reduction reactions, and determine the positive and negative electrode products.

The electrode reaction formula is written separately using the conservation of charge.

Verification: If the formula obtained by the addition of the two electrode reactions is the same as the original chemical equation, the writing is correct.

The principle is that the redox reaction that takes place spontaneously converts chemical energy into electrical energy.

Cu(OH)2, that is, taken apart, is a CuO and an H2O, respectively, 2CuSO4 + 2H2O = 2Cu+ 2H2SO4 + O2 In the process of electrolyzing copper sulfate, only copper is precipitated and oxygen is released, only copper and oxygen are changed before and after electrolysis, and the solution can be restored to its original state by adding Cuo after electrolysis. However, this question prompts that the solution returns to its original state after adding Cu(OH)2, indicating that not only copper sulfate is electrolyzed but also water is electrolyzed during the electrolysis process.

A Cu2+ 2mol electron is an electron, and an O24mol electron is generated, which is an electron.

Are you sure you pick A?

The electrolytic product is copper and oxygen, with a ratio of 1:1, which can be understood as the remaining unelectrolyzed Cu2+ in the Cuo solution, the H+ generated by the reaction, and the Cu(OH)2 added by SO42- that does not participate in the reaction, which can be understood as CuO+H2O, where the amount of H2O is negligible compared with the amount of water in the solution.

That is, the amount of the substance of Cuo that went out is the same as the amount of substance of Cu(OH)2 that is added.

Electrolysis requires electrons.

I think C is the right choice.

There may be differences between different textbooks, and there may be revisions to the same textbooks. For students who use the Renjiao version of the textbook and graduate this year, they must not write, and all galvanic battery reactions cannot be written (except for the more novel galvanic battery in the individual question conditions, which may indicate the medium and its state).

You can think of it this way, the first reaction above is not energized, it is the electricity generated by the galvanic battery, if it is energized, it is the electrolytic cell, and the second reaction is also not ignited, but the origin cell reaction is not unconditional, its power is the potential difference of the electrode material or something more complex, and it does not need to be mastered.

Or you can also think of it this way, the total reaction is the sum of the various sub-reactions together, if the total reaction is written condition, how to write the sub-reaction condition? The sub-reactions can only be added if the conditions are the same!

Absolutely. Because a chemical reaction can only occur under certain conditions, the reaction may not be able to occur if the conditions are not written.

No. This is the battery. Reaction without conditions,.

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

You have to look at this in combination with the total chemical reaction formula of charge, discharge and chemical reaction:

Anode Material: PB Cathode Material: PBSO4 Electrolyte Solution: H2SO4

Discharge process Negative electrode: PB - 2E- +SO42- = PBSO4

Positive electrode: PBO2 + 2E- +SO42- +4H+ = PBSO4 + 2H2O

Total reaction: PB+PBO2 + 2H2SO4 = 2PBSO4 + 2H2O

Charging process Cathode: PBSO4 + 2E- = PB + SO42-

Anode: PBSO4 - 2E- +2H2O = PBO2 + 4H+ +SO42-

Total reaction: 2PBSO4 + 2H2O = PB+PBO2 + 2H2SO4

The part of the lead ion that loses two electrons gets PBO2, and the part that gets the electrons is PB, and the two are on different electrodes, and it is not surprising that there is elemental lead and lead oxide.

The water in the reactant is produced by the discharge process, and the charging and discharging are two opposite processes, and the water will never run out, not to mention that there must be water in the sulfuric acid solution.

PB in PBSO4 is +2 valence.

Losing two electrons becomes +4 valent PB in PBO2

Why did it become 0 price?

The essence of electrolysis is that the anions and cations in the solution move to the anode and cathode respectively under the action of electric current, and accept or lose electrons for redox reaction. When we learn about electrolytic cells, the first thing we learn is the electrolysis of saturated saline. Among them, the reactions that occur at the cathode are:

2h+ +2e=h2 But obviously, there are no hydrogen ions in the solution, the reason for this is that considering the principle of the electrolytic cell reaction (cation moves to the cathode, and the cation is reduced), so when writing the cathodic reaction, the reactant is written as hydrogen ions instead of water (2h20+2e=h2+2oh-), but when the total reaction equation is written, the reactant is written as water, which is exactly this.

Similarly, in the electrode reaction formula of electrolytic copper sulfate solution, -2 valent oxygen is oxidized at the anode, and since according to the principle of the electrolytic cell, it is the anion that is oxidized at the anode, so when writing the electrode reaction formula, hydroxide is selected (not that the solution contains hydroxide). Similarly, when writing the total reaction formula of the electrode, the reactants no longer contain hydroxide (the essence of the reaction is indeed the formation of sulfuric acid as you said) The total equation: 2Cu2+ +2H20==2Cu+4H+ +O2 (because the hydroxyl group in the water is oxidized, there are relatively more natural hydrogen ions, and the solution becomes acidic).

As for the anode mentioned in your question, there are more hydroxides, which can be understood as follows: water contains a very small amount of hydroxide in the solution due to weak ionization, and under the action of electric current, they move to the anode and are oxidized. Then, according to the principle of equilibrium, the water is further ionized to hydroxide, which is then moved to the anode to be oxidized.

Therefore, it is not that there are more hydroxides in the solution, but that the equilibrium is constantly moving (which is the fundamental reason why the total equation does not include hydroxides).

That's basically it, and there's nothing unclear that can be added to it.

What I mean by this is that the amount of hydrogen ions is negligible, so small that it cannot be included in the total equation of the reactiono( o If my answer is helpful to you, remember to take it, thank you.