Why does Al make the negative electrode of galvanic batteries composed of Al, Mg and sulfuric acid s

Nonsense, how can you make a negative electrode, an acidic solution.

Concentrated sulfuric acid is not to passivate MG, but it is good to passivate AL.

When sodium hydroxide is used as an electrolyte, aluminum is used as a negative electrode because aluminum reacts with NaOH and mg does not react.

Does the landlord want to ask Al--Cu--the galvanic battery of concentrated nitric acid, here because Al is passivated, the Cu that is weaker than him is the negative electrode.

PS: What's the problem, Al is passivated in the presence of concentrated sulfuric acid. You didn't say concentrated sulfuric acid, I just said it to 1L.

There is no reason for this, how can there be a negative electrode in an acid solution.

Hehe, aside from what we discussed earlier, both can be used in the general exam, but if you want to write the positive and negative electrodes, it is best to write it as 6h+ +6e = 3h2, that is, the electrons of the positive and negative electrodes, you have to unify the number, probably because the answer is given so when judging, so try to ask for it to be written, not to write it like this, but it depends on luck if you don't give it.

Let's not talk about it, let's put it aside. I followed what you said, aluminum is the negative electrode. 3e-3 and 2 Least common multiple (It's called least common multiple.) It's 6.

You are mistaken, when the sulfuric acid solution is the electrolyte, magnesium is the negative electrode, and when sodium hydroxide is the electrolyte, aluminum is the negative electrode.

This seems to be aluminum as a positive electrode.

When using an alkaline solution as an electrolyte, aluminum is the negative electrode.

That's concentrated sulfuric acid, because concentrated sulfuric acid will passivate the mg, so the mg is inactive. So al is the negative.

Passivation? If it is passivated, it will not conduct electricity, how can it still constitute a battery?

(+)2H2O + 2E = H2 +2OH - (-AL - 3E + 4OH- = [AL(OH)4]- (can also be written as ALO2-).

I don't know if the electron gain and loss are required to be equal, if it is required, the positive electrode reaction coefficient is multiplied by 3, and the negative electrode is multiplied by 2.

Or: Total reaction: 2al+2naoh+2h2o=2naalo2+3h2 anode:

2Al-6E-+4OH-=2ALO2-cathode: 6H2O+6E-=6OH-+3H2 In NaOH solution, Al can react as a negative electrode, but Mg cannot react as a positive electrode.

(+)2h2o + 2e = h2↑ +2oh-

al-3e + 4oh- = [al(oh)4]- (can also be written as alo2-).

I don't know if you require the electron gain and loss to be equal, if you do, you can multiply the positive electrode reaction coefficient by 3 and the negative electrode by 2.

Total reaction: 2al+2naoh+2h2o=2naalo2+3h2 negative: 2al-6e-+4oh-=2alo2-positive:

6H2O+6E-=6OH-+3H2 In NaOH solution, Al can react as a negative electrode, but Mg cannot react as a positive electrode.

Negative electrode: 2al+8oh- -6e-=2alo2- 4h2o

Positive electrode: 6h2o+6e-=3h2+6oh-

Hope it helps you o(o

In galvanic cell reactions, generally speaking, the active electrode loses electrons at the negative electrode and becomes the negative electrode, but there are exceptions, such as the example you proposed, although Mg is more active than Al, however, Al can react with NaOH and lose electrons, while Mg cannot, so Al is the negative electrode.

It can be seen from this that to determine whether one pole is positive or negative, it is not necessary to look at the active type of the electrode, but to see which pole is more likely to react under certain circumstances.

Of these materials, only aluminum can undergo redox reactions with sodium hydroxide. Although magnesium is more metallic than aluminum, it cannot react with sodium hydroxide.

Working principle: cathode (mg): 2h20+2e-=h2 +2oh-, magnesium is only used as a conductive material, which can be replaced with graphite, silver and other materials that do not react with sodium hydroxide and can conduct electricity.

Negative electrode (Al): Al-3E- +4OH-=(AlO2)- 2H20.

Because galvanic cells require a spontaneous redox reaction, whoever loses electrons first is the negative electrode.

Because the magnesium and sodium hydroxide solution cannot spontaneously carry out redox reactions, aluminum is the negative electrode.

Because Mg does not react with NaOH solution, only Al can react with NaOH to produce hydrogen.

So al is the negative pole.

Electroslider reaction: Al- 3E- +4OH-==Al[(OH)4]- or Al- 3E- +4OH-==AlO2- +2H2O

Working principle: positive electrode (mg): 2h20 + 2e-=h2 +2oh- negative electrode (al): al—3e- +4oh-=(alo2)- 2h20

Among Al, Mg, and NaOH, only Al reacts with NaOH solution, so Al is the negative electrode and Mg is the positive electrode.

It depends on your electrolyte. The acidic electrolyte mg is used as the negative electrode. Alkaline on the contrary.

Hope it helps.

Can you add a question, huhu

In Mg-Al-HCL, both Mg and Al can react with HCl, but Mg is more active than Al activity, so in fact, only Mg is oxidized by HCl, so the negative electrode is Mg at this time

Whereas, in Mg-Al-Naoh, Mg does not react with NaOH and only Al is oxidized by H2O, so Al is the negative electrode.

Al will turn into metaaluminic acid.

It depends on the composition of the electrolyte solution.

Generally speaking, because mg is more active than al, magnesium is the negative electrode;

But if the electrolyte solution is a sodium hydroxide solution, magnesium does not react, while aluminum reacts with it, so aluminum does the negative electrode.

One of the important conditions of the galvanic cell is that the total reaction can be carried out spontaneously redox reaction, when magnesium aluminum dilute sulfuric acid constitutes the galvanic battery, magnesium and aluminum can react with dilute sulfuric acid, magnesium is more likely to lose electrons, so magnesium is used as the negative electrode; When the magnesium-aluminum sodium hydroxide solution constitutes the galvanic battery, only aluminum can react with the sodium hydroxide solution, so aluminum is used as the negative electrode.

I remember seeing an article in a magazine that used experiments to refute this view, but most of the questions I do at the moment are based on this answer. If you are interested, you can do some experiments.

Negative electrode: Electrons are lost, and oxidation reactions occur.

Generally, the negative electrode itself loses electrons).

Positive electrode: electrons are obtained, and a reduction reaction occurs.

It is generally positive in solution.

Ions gain electrons at the positive electrode, but it may also be o electrons at the positive electrode (oxygen corrosion), or the positive electrode itself has electrons).

Total Reaction (Battery Reaction) = Positive Reaction + Negative Reaction.

Because of aluminum with sodium hydroxide.

The reaction produces sodium metaaluminate.

Lose electrons, pretend to be a negative electrode.

Electrode antilimb command: al+3e

4oh⁻alo₂⁻+2h₂o。

Magnesium does not react with sodium hydroxide, so it is the positive electrode, and the electrode reaction: 2h o + 2e = h + 2oh.

Justification according to the total reaction: 2

al+2naoh+2h₂o=2naalo₂+3h₂

Extended Materials. Only one electrode material is involved in the reaction:

The metal electrode itself participating in the reaction is the negative electrode, and the other electrode is often the positive electrode, the negative electrode is the metal that participates in the reaction and the positive electrode is the electron obtained by the particles in the medium solution (the reaction is generally hydrogen evolution, oxygen absorption, Cu analysis, AG, etc.). The positive electrode reaction has the following rules:

1. When a strong acidic solution is used as the medium solution, the hydrogen evolution reaction generally occurs at the positive electrode.

2. When a near-neutral solution is used as the medium solution, the oxygen absorption reaction generally occurs at the positive electrode.

3. When the alkaline solution is used as the medium solution, the positive electrode generally undergoes oxygen absorption reaction.

4. When the salt solution containing inactive metals is used as the medium solution, the inactive simple metals (Cu, Ag, etc.) are precipitated from the cathode.

Aluminum reacts with sodium hydroxide to form sodium metaaluminate, which is the negative electrode of the hall, and the electrode reaction: ai-+3e-

4OH-ALO2-+2H2O, magnesium does not react with sodium hydroxide, so it is the positive electrode, and the electrode reacts: 2H2O+2E=H2+2OH-.

Justification according to the total reaction: 2

al+2naoh+2h2o=2naalo2+3h2

Substance: Aluminum reacts with water first, 2Al+6H2O=2Al(OH)3+3H2

Aluminum hydroxide amphoteric and reacts with a strong base: Al(OH)3+NaOH=Naalo2+2H2O

The above two anti-dressing limbs should be superimposed to get: 2

al+2naoh+2h2o=2naalo2+3h2

Extended information: Judgment of positive and negative electrodes of galvanic batteries.

1. According to the electrode material: the more active one is negative, and the less active one 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; (Pb-PBO2 can be used as a classic exception: during discharge, the weight of the positive electrode PBO2 becomes PBo4 after the reaction, and the negative electrode PB2 also changes to PBo4 after the reaction, 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 one pole of anion shift is negative, and the pole of cation shift is positive [3].

Aluminum reacts with sodium hydroxide to form sodium metaaluminate, which is called the negative electrode, and the electrode reacts: a raid bureau-3e-

4OH-ALO2-+2H2O, magnesium does not react with hydrogen and sodium oxide, so it is the positive electrode, and the electrode Zen pole reaction: 2H2O+2E=H2+2OH-

Do you want to ask whether the Mg, Al, and NaOH solutions constitute the original electric side cell of the oak, and Al is the negative electrode; A galvanic cell composed of Fe (or Al), Cu and concentrated nitric acid, with Cu as the negative electrode. Why?

Conditions for judging the negative electrode of galvanic batteries: The electrode reacts spontaneously with the electrolyte solution The more active electrode is the negative electrode. The first.

Mg does not react with NaOH whereas Al reacts with NaOH so Al acts as the negative electrode. The second. Fe or Al is passivated in concentrated nitric acid.

Whereas, Cu can react with concentrated nitric acid. Therefore, cu is a negative void pole.

These two are special galvanic cells.

mg is used as the closed negative electrode of transport, mg-2e- =mg2+

Al is the positive electrode, 2H+ +2E- =H2 (gas).

Total by-cleavage reaction formula: Mg+2H+ = H2(gas)+Mg2+

Negative electrode: electron loss, oxidation reaction occurs (generally the negative electrode itself loses electrons).

Positive electrode: electrons are obtained, and a reduction reaction occurs (generally the cation in the solution gets electrons on the positive electrode, but it may also be o electrons on the positive electrode (oxygen absorption corrosion), or the positive electrode itself gets electrons).

Total Reaction (Battery Reaction) = Positive Reaction + Negative Reaction.

Because aluminum reacts with sodium hydroxide to form sodium metaaluminate, it loses electrons and acts as a negative electrode.

Electrode reaction: AL+3E4OH = ALO+2HO.

Magnesium does not react with sodium hydroxide, so it is the positive electrode, and the electrode reaction: 2h o + 2e = h + 2oh.

The rationale is based on the total reaction: 2 al+2naoh+2h o=2naalo +3h

Aluminum reacts with sodium hydroxide to form sodium metaaluminate, so it is the negative electrode, the electrode reaction: ai-+3e- 4oh- =alo2-+2h2o, magnesium does not react with sodium hydroxide, so it is the positive electrode, and the electrode reaction: 2h2o+2e=h2+2oh-.

The rationale is based on the total reaction: 2 Al + 2NaOH + 2H2O = 2Naalo2 + 3H2

Substance: Aluminum reacts with water first, 2Al+6H2O=2Al(OH)3+3H2

Aluminum hydroxide amphoteric and reacts with a strong base: Al(OH)3+NaOH=Naalo2+2H2O

The above two reactions are superimposed to obtain: 2 Al+2NaOH+2H2O=2Naalo2+3H2

In this galvanic cell, Al is used as the negative electrode. The total reaction is Al and NaOH to produce H2 and sodium metaaluminate.

In principle, the positive electrode is the reaction that gets electrons, and the key question is who gets electrons. 1. The reaction is oxygen to obtain electrons; 2 is the electron obtained by hydrogen.

First of all, it is necessary to consider the ions in the solution to gain electrons at the electrode, and then to consider the first reaction when there are no strong electron ions in the solution.

This is a matter of electronic order, which is not easy to think of.

Therefore, for the general reaction formula of the battery, you can write the total reaction first, and then write the electrode reaction.

May I ask where did o2 come from in your first formula? There is no oxygen in the galvanic cell reactants, did you get it out of **?