High School Electrochemistry, High School Electrochemistry Questions

Electrochemistry is the science that studies the interrelationship between electricity and chemical reactions. The interaction of electrical and chemical reactions can be done by batteries or by high-voltage electrostatic discharge, which is collectively referred to as electrochemistry, which is a branch of electrochemistry called discharge chemistry. Therefore, electrochemistry often refers specifically to the "science of batteries".

The battery is composed of two electrodes and the electrolyte between the electrodes, so the research content of electrochemistry should include two aspects: one is the study of electrolyte, that is, electrolyte, including the conductive properties of electrolyte, the transport properties of ions, the equilibrium properties of the ions involved in the reaction, etc., among which the physical and chemical research of electrolyte solution is often called electrolyte solution theory; On the other hand, there is the study of electrodes, i.e., electrode, which includes the equilibrium properties of electrodes and the polarization properties after energizing, that is, electrodes and electrolyte boundaries....

The negative electrode loses electrons, and the positive electrode gains electrons.

As long as you look at the correspondence, the substance with the increased valency of the element is the negative electrode, write the chemical formula, subtract the electrons, write an equal sign, and then write the corresponding product. Then mix water, h+, and oh- on both sides to achieve atomic conservation. Note that oh- is not present in acidic solutions and H+ is not present in alkaline solutions.

Take potassium iodate as an example:

Negative electrode: potassium iodide - electron + hydroxide = water + potassium iodate.

Cathode: Water - electrons = hydrogen + hydroxide.

Just trim it. It's best to check after writing that the positive and negative reactions should add up to just eliminate the electrons and hydroxides, exactly the same as the total reaction formula given.

In the galvanic cell reaction, the negative electrode undergoes an oxidation reaction (electron loss) and the positive electrode undergoes a reduction reaction (electron gain), and it is necessary to pay attention to whether the electrolyte solution will react with ions. For example, zinc-manganese dry battery, the negative electrode zn-2e-=zn2+. But since the electrolyte solution is alkaline, the correct negative equation is Zn+2OH-2E-=ZN(OH)2.

For the positive electrode, just subtract the negative equation from the total equation.

There are a few rules to keep in mind when writing an electrolytic equation. "Oxidation of the anode, reduction of the cathode". anodic ion discharge sequence; Metals (except Au, Pt) >S2->I->Br->Cl->OH->Oxygenates》F-.

Cathode ion discharge sequence: AG+>HG+>Fe3+>Cu2+>H+(water separation)>Pb2+>SN2+>Fe2+>Zn2+>H+>Al3+>Mg2+>Na+>Ca2+>K+

(1) As (arsenic) is located in the fourth period va group of the periodic table, associated with the similarity of elements in the same group, n p as

Taking P as an example, the solution is alkaline when Po4 ions are hydrolyzed, so the solution is alkaline due to the hydrolysis of Aso4 ions, so the pH is >7

2) Add an appropriate amount of concentrated sulfuric acid to cup B, and find that the pointer of g is shifted to the right. "g is a sensitive galvanometer whose pointer is always biased towards the positive side of the power supply"Therefore, it is said that the electrons run from C to D, that is, the I ion undergoes an oxidation reaction, loses electrons, forms I2, encounters starch, and turns blue.

Note that the concentrated sulfuric acid oxidizes the low-valent ASO3 ion, and the ion undergoes a reduction reaction, so ASO3 + H2SO4 (concentrated) = ASO4 (self-trimmed) (3).

Because the purpose of this experiment is to plating a layer of zinc on the surface of the iron sheet, it is necessary to retain the iron sheet, so the iron sheet is at the cathode, otherwise it will be oxidized.

In this cell, the anode material is oxidized and an oxidation reaction occurs. A reduction reaction occurs at the cathode. Therefore, the anode should use zinc sheets, which lose electrons during the reaction, are oxidized into zinc ions, and flow to the cathode through the electrolyte solution to obtain electrons and undergo a reduction reaction.

It is gradually reduced to zinc, so a layer of zinc appears on the surface of the iron sheet.

Anode: Zn-2E===Zn2+

Cathode: 2E+Zn2+===Zn

Because the anode metal zinc in the electrolytic cell is oxidized and dissolved into the solution, and the reduction reaction that occurs at the cathode should be the reduction of metal ions and zinc ions to metal zinc-metal precipitation. The cathode metal does not react, it is just a conductor.

Remember, the cathode is the protected pole, and because the cathode is connected to the negative pole of the power supply, it is the electrons that flow in. Therefore, it is extremely difficult for the iron rod to lose electrons after it becomes a cathode. The anode is the electron outflow electrode, so the zinc lost electrons are oxidized to zinc ions.

Zinc ions are positively charged, so they move to the cathode, combine with the electrons gathered by the cathode, and reduce them to become zinc elemental matter attached to the surface of the iron rod.

Solution: By the chemical equation: 2NaCl + 2H2O = (electrolysis) = 2NaOH + H2 + Cl2

Let the amount of NaOH species formed be x

You can get the relation:

2naoh---2mole-xx=

then the solution after mixing c(oh-) =

The pH of the solution after mixing is 13

What is the pH of the solution after mixing? The answer is 13How is it calculated?

<> set the resistance of the PB part to X and the current of the trunk circuit to I

In the process of gradually moving up from the lower end B to the A end of the slide P, X continues to increase, and the single-liquid Zen adjustment in the figure increases, so the current of the trunk road continues to increase, so the electrical power of R4 has been increasing, and the B option is correct.

We look at R2, R3, X as a whole, due to the increase of X, this part of the resistance is increasing, and the current through this part of the wax is also increasing, so the voltage at both ends of R2 is increasing, so the electrical power of R2 has been increasing, and option A is making trouble with nuclear dust.

Equation is the expression of the electrical power of r3, it can be seen that as x increases, p3 decreases first and then increases, when x=10 ohms is the minimum value, when x=20 ohms, the electrical power of r3 is equal to the electric power when x=0, and as x continues to increase, the electric power of r3 continues to increase, so when p is at the a end of r1, the maximum electric power of r3 is the d option.

Split R1 into two Qi rolls, a sedan car Xun part for RB and R3 in series and R2 in parallel, and then with the rest of the closed part of the Ra series, R1 = 30 = Ra + RB, the column equation is on the line, very simple.

Hi, isn't it OK to list an equation, what is the difficulty of the college entrance examination questions, that is, the calculation is a little complicated, and the relationship between the variables should be very clear.

1.No, the question tells you that the difficulty has been reduced, and it was impossible to take the college entrance examination before. Because it is mentioned in the textbook that aluminum forms an oxide film on the surface of H2SO4.

2.The question of childishness shows that your electrochemistry really needs to be improved. The reaction process can be analyzed, which can be regarded as the reaction between Fe and H2SO4. In addition, H2O is generally used as a reactant at the negative electrode.

The first question, can not be written like that, Al this substance only has two valencies 0, +3, when electrolysis occurs, Al can only lose three electrons to generate Al3+ ions, this ion will form aluminum hydroxide with OH roots in water, but this substance has a characteristic is instability and precipitation, if this is an acidic solution to continue the reaction, if it is alkaline, then generate precipitation, depending on the specific situation.

The second question, the title said that this is acidic wastewater, that is, it contains a large number of H ions, Fe This substance has two valences, +2, +3, since it is used as a reducing agent, it must be easier to lose two electrons in the outermost shell to generate ferrous ions instead of 3-valent iron ions, why should 3H2O be added, +2 valence is not +3 valence, remember to remember, why do you add 3H2O in the previous question because you want to use the OH root in the water,

The key is that it needs to be trimmed, but the iron one should also be added! Look again!

Oxygen-absorbing corrosion. Metal in.

Acidity. Very weak or.

Neutral solution. air.

. Oxygen.

Soluble in metal.

Surface. For example, the corrosion of steel in near-neutral humid air belongs to oxygen absorption corrosion, and its electrode reaction is as follows:

Negative electrode (Fe): 2Fe

4e2fe2+

Cathode (c): 2H2OO24E

The galvanic corrosion of metals such as 4OH-steel is mainly a necessary condition for oxygen absorption corrosion.

to oxygen. Reduction reaction.

For. Cathodic process.

, known as oxygen reduction corrosion or oxygen absorption corrosion. The necessary condition for oxygen corrosion to occur is that the potential of the metal is lower than that of the oxygen reduction reaction.

One. in more acidic ones.

Solution. It is released when galvanic corrosion occurs.

Hydrogen. This corrosion is called hydrogen evolution corrosion.

Two. in steel.

Goods. It generally contains carbon. In moist air, the steel surface adsorbs.

Water vapour. and forms a thin layer of water.

Membrane. Dissolved in the water film.

Carbon dioxide. After that, it becomes one.

Electrolyte solution.

Increases the number of h in the water. It is composed of countless tiny solutions with iron as the negative electrode, carbon as the positive electrode, and acidic water film as the electrolyte solution.

Primary batteries. Three. Systems in which hydrogen evolution corrosion occurs.

A reactive metal with a negative standard potential.

Most of the metals used in engineering, such as FE

Electropositive metals are generally not attacked, and hydrogen evolution corrosion will occur. But when the solution contains.

Complexing agent. , electropositive metals (e.g., Cu, Ag) may also undergo hydrogen evolution corrosion. This brother.