Senior 3 Chemistry, Electrolytic Cell, High School Chemistry Primary Cell Electrolytic Cell

Electrolytic sulfuric acid is the electrolysis of water in the solution, so the amount of H+ (hydrogen ions) in sulfuric acid is unchanged.

In the solution of electrolyzed NaCl, the position of Cl- (chloride ion) is replaced by OH- (hydroxide), so the amount of OH- = the amount of Cl in chlorine =

The amount of H+ and OH- in the two solutions is exactly equal, neutralizing the !!

So, the pH of the solution = 7

The total electrolytic reaction of the sodium chloride solution is.

2nacl＋2h2o＝2naoh＋cl2↑＋h2↑←

That is, NaOH is generated

Electrolytic sulfuric acid can be regarded as electrolyzed water, so the amount of sulfuric acid remains unchanged as NaOH and H2SO4 are completely neutralized after mixing the two solutions, and the mixed solution pH=7

Electrolytic sulfuric acid solution belongs to the type of electrolytic strong acid, which is essentially electrolyzed water.

Cathode: Hydrogen is released.

Anode: Releases oxygen from water.

Electrolyzed NaCl solution.

Cathode: Hydrogen is released.

Anode: Chlorine gas is released.

Finally, NAOH is generated

Sulfuric acid is neutralized with NAOH just right.

Finally ph 7

After the electrolysis of the original ions in the solution, this part of the hydrogen is equivalent to the hydrogen in the electrolysis of water, and the residual is hydroxide, that is, so the pH is 12

In fact, the electrolysis is HCI gas, and H2 is equivalent to pH=13

You go to the chemistry bar and ask, and someone will give you the answer. Thank you Happy New Year.

1.First of all, judge the positive and negative electrodes, and then look at the products after judging them, and write the gain and loss of electrons first.

For example: alkaline zinc-manganese battery Total reaction formula: Zn 2mNO2 2H2O=2mNOh+2OH- The electrolyte is KOH

Negative electrode: You can write zn-2e= first, and because the product zn(oh)2 in the total reaction formula is insoluble in water, so there is zn-2e zn(oh)2 At this time, it is found that two oh- - need to be added to the reaction formula to balance, so write the negative reaction formula: zn-2e+2oh-=zn(oh).

In the same way, the positive electrode reaction: first write 2mno2 +2e- 2mnooh because the electrolyte is koh solution alkaline, so there are only two possibilities: h20 and oh- because oh- is not conserved when added, so choose h2o to get 2mno2 +2e- +2h2o =2mnooh+2oh-

I'm also studying and hope to help you-. -

1.For such a dilute HCl solution, the H+ ionized by the water itself is not negligible. In fact, at this time, the H+ in the solution is basically ionized by water.

The specific calculation method is: total C(H+) in solution = C(H+) ionized by HCL + C(H+) ionized by water

The former = 10 -13 and the latter = c(oh-) in solution, because the amount of H+ and Oh- ionized by water is always the same.

Therefore, the equation c(h+)=10 -13+c(oh-) is obtained, and combined with the ionic product formula of water, the equation can be solved to obtain c(h+), which is very close to 10 -7 numerically. That is, at this point the solution pH = 7

2.The characteristic of galvanic cells is that the oxidation reaction and the reduction reaction occur in different places. In copper-zinc galvanic cells, Zn is oxidized into Zn2+ into solution. Cu2+ is reduced to Cu and precipitated from solution.

Essentially, the galvanic cell reaction can occur when Zn is inserted in any solution, and it can also be inserted in CuSO4. However, direct contact between Zn and CuSO4 solution will cause a direct displacement reaction, which interferes with the galvanic cell reaction and consumes energy, so it is generally not done.

3.Theoretically, the concentration of the electrolyte is constant, and the cathode is precipitated as much as the anode is dissolved. However, in practice, there are always some impurities in the Cu of the anode, such as Fe, etc., which are more active than Cu, and they will not precipitate at the anode after they are dissolved into the solution, but will be replaced by the precipitation of Cu2+ in the solution.

The unit mass of these impurities is smaller than that of Cu, so the amount of anode dissolution (including Cu and these impurities) will always be less than the amount of cathodic precipitation during actual electrolysis, and the concentration of CuSO4 in the electrolyte will also decrease slightly.

The two poles of the galvanic battery are called the positive and negative electrodes, and the negative electrodes lose electrons; The electrolytic cell is called the cathode, with an external power supply, and the power supply is facing the anode (loss of electrons).

The principle of galvanic cells is redox reaction, where the anode undergoes an oxidation reaction and the cathode undergoes a reduction reaction. Therefore, you should first know what the reaction principle is.

The main points are as follows: 1) The oxidation reaction occurs at the anode, and the reduction reaction occurs at the cathode, and the positive electrode connected to the anode is the negative electrode.

2) When writing the electrode reaction formula, first write the pole that you think is correct and simple, and then subtract it from the total reaction formula.

3) Pay attention to the acidity and alkalinity of the reaction environment. For example, in an acidic environment, no hydroxide can appear, and no hydrogen ions can appear in an alkaline environment.

Typical examples are: 2H2O+4E-+O2=4OH- (alkaline environment) 4H++4E-+O2=2H2O (acidic environment).

In the electrolytic cell, the anode loses electrons and undergoes an oxidation reaction, if it is an active electrode, the electrode participates in the reaction, you give an example, which metal is the anode? You didn't make it clear, if iron is used as an anode, it means that iron loses 2 electrons to form divalent iron. If copper is used as an anode, it is copper that loses electrons into divalent copper.

Cathodic reaction, electrons, reduction reaction, cations in the solution have copper ions and H ions, comparatively, the oxidation ability of copper ions is stronger than hydrogen ions, therefore, the cathode is copper ions to obtain electrons to generate copper.

The total reaction is the addition of the cathode and cathode reactions.

option b, which is an electrolytic cell, chloride ions are discharged at the anode, losing electrons and generating chlorine gas; Hydrogen ions are discharged at the cathode to get electrons and produce hydrogen.

So A is the anode and B is the cathode. The statement of "positive and negative electrodes" in the question is also wrong, the electrolytic cell says "cathode, cathode", and the galvanic battery says "positive and negative electrodes".

option d, since the two electrodes are made of the same material, does not constitute a galvanic battery.

b d is not wrong.

Because according to the diagram, K2 is turned off and K1 is turned on, and the electrolytic cell is formed.

Electrolysis of the aqueous solution of NaCl yields H2 and H2 at the cathode and anode, respectively.

The solubility of Cl2Cl2 is greater than H2, so the left gas is.

h2, then power supply A is the negative pole.

After electrolysis for a period of time, there is H2 on the left side, Cl2 on the right side, and an electrolyte solution, turning on K2 and turning off K1 will form a galvanic cell, similar to a fuel cell, the total reaction of the battery is.

H2 + Cl2 + 2NaOH = 2NaCl + H2O because. H2 has naoh on this side).

1。（1）2i-

Cu2+=Electrolysis==CuI2

2) The anode starts to put H2

Indicates that Cu2+ has fully responded.

When the i-reaction is complete.

cu2+ only reacted.

Moreover. cu2+

2cl-==electrolysis===

cucl2↑

At this point, cl- only reacted.

So. The anode electrode reaction equation is.

2cl-2e-

Cl2 3) The anode starts to put Cl2

Description I-Reaction completely.

At this time, Cu2+ has not finished responding.

The cathode electrode reaction equation is.

cu2+2e-

cu2.Tong equal amount of electricity.

It shows that the number of electrons gained and lost in the two reactions is equal.

So. The amount of the substance obtained is equal to that of ag and hg.

The anode is all about getting oxygen.

The amount of matter is definitely equal.

The ratio of the amount of reduced silver nitrate to mercurous nitrate (silver nitrate): n (mercurous nitrate) = 2:1

Illustrate the chemical formula of mercury nitrate as:

hg2(no3)2

Option d if you still have doubts.

Hi look for me. Details.

(1) The law of electron loss on the anode.

If it is an inert electrode, the anion in the solution loses electrons, and the stronger the reduction of anions, the more easily lost electrons, and the discharge sequence of anions is: S2->i >br >Cl >OH >oxylate ions; In the case of active electrodes, these metals first lose electrons into the solution, where the other ions in the solution no longer lose electrons.

2) The law of electrons obtained on the cathode.

The cathode can only obtain electrons from the cations in the solution, and the stronger the cation oxidation, the easier it is to obtain electrons, and the cation discharge sequence is generally: AG+>HG2+>Fe3+>Cu2+>H+(acid)>Fe2+>Zn2+>H+(salt solution).

3) Variable value of metals.

When Fe is oxidized as the anode, the electrode reaction is: Fe 2e = Fe2+

b Conservation of electrons H+ is 10-3mol L, oh- is 10-3mol L, the volume is 500ml, so 5 10 4mol less, 4oh 4e = 2H2O+O2 So lose 5 10 4mol, and ag++e = ag, so the mass of silver produced is 5 10 4mol 108g mol=

Cathode: AG+ +E-=AG

Anode 2H2O-4E = O2 + 4H+

Because the title says that "there is no hydrogen production at the cathode", the cathode has always been reacting with silver ions, and the anode has not changed.

ph=6 ph=3 The hydrogen ion changes to 10 -3-10 -6 After being brought into the anode, the number of electrons obtained is 10 -3-10 -6 So the mass of silver formed is (10 -3-10 -6) 108=

Glad to answer for you :

The process of causing a redox reaction on the cathode and anode, which causes the current through the electrolyte solution, is called electrolysis 1, and the anode undergoes oxidation reaction, if the electrode is metal, the electrode loses electrons and is reduced, and metal ions are generated into the solution, so the mass reduction is the anode, and the cathode undergoes a reduction reaction, and the metal ions in the solution get electrons to form metal elemental elements attached to the electrode to increase the electrode;

2. Phenolphthalein is used to test whether the solution is alkaline, if it is shown red, it is obvious that the solution is alkaline, that is, there is OH-, this hydroxide is due to the existence of no metal cation in the solution that is easier to obtain electrons than H+, H+ gets electrons to generate hydrogen, and neutral water becomes OH-after losing H+, so there is the above phenomenon;

I hope it can help you, and I wish you good luck in your studies

1. The electrode mass is reduced, indicating that the electrode participates in the reaction, only the metal electrode with high activity will participate in the reaction, if it is the cathode, it will be protected, and the electrode itself will not lose electrons to participate in the reaction, so it must be the anode; As the electrode mass increases, there must be metals precipitated at this electrode, and the electrode for cation precipitation reaction must be the cathode.

2. The phenolphthalein turns red, indicating that there is excess OH- near it, that is, because the reaction at this electrode consumes H+, so it is a cathode.

There is a decrease in the mass of an electrode, which means that it reacts with the electrolyte, the anode loses electrons, the valence of the element increases, and it is oxidized into particles into the electrolyte......The electrons obtained by the transfer of the positive particles to the cathode are reduced to elemental attachment to the cathode, and its mass increases. The phenolphthalein turns red when exposed to alkali, indicating that the electrons obtained by the hydrogen ion at the pole are reduced, and it can only be the cathode.

If both produce gas, the gas at the positive electrode is chlorine, and the cathode is hydrogen. However, the cathode discharge sequence is Cu2+>H+, so the cathode is discharged Cu2+ first, and if all the copper ions in the solution are converted into copper element, it is time to discharge hydrogen ions. In the same way, the anode is also discharged first by chloride ions, and if the chloride ions are discharged, it should be a hydroxide ion discharge.

The relationship between copper ions and chloride ions is not stated, so the resulting volumes may be equal.

1.If there is a power supply, the electrolytic cell is the battery, and if there is no power supply, the galvanic battery. Of course, that can only be judged if the conditions are met. Figure A belongs to fuel cells (galvanic batteries).

2.There is KOH in the total reaction formula, but the actual redox reaction is still methanol and oxygen, so you use 2CH3OH+3O2=2CO2+4H2O minus 3O2+6H2O+12E-=12OH- to get CH3OH+6OH--6E-=CO2+5H2O.The judgment can be thought of like this:

The pH before and after the reaction of the total reaction (methanol and oxygen) is unchanged, that is, it must remain neutral, but the positive electrode water and oxygen get electrons to form OH-, and the pH increases, in order to maintain pH = 7, then OH- must be consumed at the negative electrode. As for KOH, it's just an electrolyte, and if it's an acid, it doesn't react, but it happens to be a base, a base that reacts with the CO2 generated, but it doesn't participate in the actual "redox reaction".

3.Your question is why the cathode doesn't turn H+ into H2 and run away. That's because if H+ all becomes H2 and runs away, OH- will also combine with Mg2+ to form a precipitate, then chloride ions will be left in the solution and become a negatively charged solution, but the solution always has to remain electrically neutral, so only when water gets the OH- combined with Mg2+ generated by electrons to form a precipitate, the electrical property of the solution can remain neutral.