High School Chemistry Galvanic Cell Electrode Reactive Formula

Take your time and don't rush.

1. Any aqueous solution will have H+, Oh-, H2O present.

2. The chemical reaction formula of the battery reaction you mentioned is like this: PBO2 + PB + 2H2SO4 = 2PBSO4 + 2H2O (that is, the sum of the two formulas you wrote), the one on the left is the reactant, and the one on the right is the product.

Someone asked, only PBO2+PB changes in price, why write 4H,2SO42-,PBSO4?

The answer is this: PBSO4 is solid, so to speak, because it is solid, this reaction can take place. After he became solid, he left the liquid phase, and the reaction was not reversed, so the reaction continued.

From the perspective of the ionic reaction formula (the battery reaction formula is also an ionic reaction formula), to form a solid, it must be left and cannot be deleted.

3. Learn to analyze. Any battery reaction, you have to figure out what its original reaction is. This is very important, it is the foundation of the battery.

For example, the battery reaction formula you mentioned, the essential reaction is PBO2 + PB+2H2SO4 = 2PBSO4 + 2H2O. After getting this reaction, the first step is to turn it into an ionic formula: PBO2 + PB + 4H + 2SO4 = 2PBSO4 + 2H2O, at this time, if there are ions on both sides, they will be eliminated (for example, HCl solution + NaOH solution reaction, H+Cl+Na+OH=Na+Cl+H2O, after the two sides are eliminated, it becomes H+OH=H2O).

Analyze the battery reaction in three moves:

Step 1: Write out the reactive ionic formula:

pbo2 + pb + 4h + 2so4 = 2pbso4 + 2h2o

Step 2: After writing the ion reaction formula, the second step is to increase the price and reduce the price of each separation.

Price increase: PB-2E = PBSO4 (solid).

Price reduction: PBO2+2E = PBSO4 (solid) + 2H2O (that oxygen atom turned into water).

Step 3: Complete the ions (look at the right side of the formula, it is very based on the ionic formula, and the left side is missing what is missing).

Price increase: PB-2E+SO42-=PBSO4

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

The most important thing is the ionic formula of the total reaction. If you don't understand this, you'll be confused. If you figure this out, you can sing: Mom doesn't have to worry about my studies anymore...

1 galvanic cell reaction is an ordinary redox reaction that releases energy, and what reaction occurs depends on the substance given (electrode material, electrolyte solution, etc.).

Example 1 galvanic cells composed of copper, zinc and dilute sulfuric acid, only zinc and dilute sulfuric acid can react, the reaction formula zn + 2h+ = zn2+ +h2

Example 2 Galvanic cells composed of iron, copper, and ferric chloride solutions, Fe is more active than Cu, and Fe ferric chloride reacts with it.

Reactive formula Fe + 2Fe3+ = 3Fe2+

Example 3 Another example is lead-acid battery, PBO2, PB, H2SO4 three substances, according to PB+2 valence stability, so +4 valence has strong oxidation, 0 valence has reduction, is a centering reaction, sulfuric acid provides acidic conditions, so that the generated substances can exist stably in sulfuric acid solution. The reaction formula is.

pbo2 + pb + 4h+ +2(so4)2- = 2pbso4 + 2 h2o

2 Take a lead-acid battery as an example, let's talk about the reaction conditions.

PBO2, PB, H2SO4 are stable according to PB+2 valence, so +4 valence has strong oxidation, 0 valence has reduction, it is a centering reaction, sulfuric acid provides acidic conditions, if you write the product as PB2+, it is not possible to coexist with (SO4)2-).

If it is written as PBO2 + PB + 2H2O + 2(SO4)2- = 2PBSO4 + 4OH- no, OH- does not coexist with H+].

It can be seen that the substances that provide the reaction conditions (also called the reaction environment) also participate in the reaction, which often limits the type of product. That is, you write out that the two organisms must coexist stably with the substances that provide the conditions for the reaction.

3 Electrode reaction equation.

Since it is called an equation, it must be balanced; Since there are reaction conditions, the product must coexist stably with the substance that provides the reaction condition.

Example 4 A galvanic cell composed of aluminum, carbon rods, and potassium hydroxide solution.

The chemical reaction can only be reacted with a solution of aluminum and potassium hydroxide.

The ionic equation for the reaction is

Negative electrode: Al loses electrons, but under alkaline conditions, Al3+ cannot exist, to react with OH-, then let him generate the reaction of the substance (ALO2)-, the negative electrode reaction AL-3E + 4OH- = (ALO2)-2H2O [from this we can see how OH- provides basic conditions].

For the positive electrode, H2O is the oxidizing agent (where H is +1 valence, and its oxidation is effected), and the equation is.

2h2o + 2e = h2 + 2oh-

When H+ or OH- in H2O participates in the electrode reaction, it can be directly written as H+ or OH- in the electrode equation, and it can not be written as H2O.

See whether the cation generated by the negative electrode reaction can coexist with the anion in the electrolyte solution, if not, the anion in the electrolyte solution should also be written into the electrode reaction formula of the negative electrode. For example, in the galvanic cell composed of Al Cu NaHCO3 solution, the Al3+ generated by the loss of electrons due to the loss of Al can react with HCO3: Al3++3HCO3 = Al(OH)3 +3CO2, so the reaction on the aluminum part (negative electrode) is:

Al3e +3HCO3 = Al(OH)3 +3CO2 instead of just writing Al3E =Al3+.

When the reactive substance on the positive electrode is O2 (oxygen corrosion), pay attention to the properties of the electrolyte solution. When the solution is alkaline, H+ cannot appear in the electrode reaction formula; When the solution is acidic, OH cannot appear in the electrode reaction.

For the reversible battery reaction, it is necessary to see the direction of "charging and discharging", the discharge process is to reflect the principle of the galvanic battery, and the charging process is to reflect the principle of the electrolytic cell.

Electrolytes are often taken into account in electrode reactions, such as in aqueous solutions, acidity should be matched with hydrogen ions to balance the charge, and alkaline is hydroxide.

Dear students, how to judge the valence change of lithium and cobalt if you know the electrode reaction formula of lithium-ion batteries? How many electrons are gained or lost? Positive and negative poles?

How do you trim charge conservation? Are there any follow-up reactions? What steps and principles are followed?

(1) The positive, negative electrodes are written together, and the gain and loss of electrons must be consistent--- which is a customary way to write!

2) The electrode reaction formula is combined to obtain the chemical equation, and the electrons gained and lost by the positive and negative electrodes must be consistent!

3) It is not a mistake of principle to formulate the electrode reaction to the simplest ---; But it's not necessary!

There are three common types of questions:

1。Two electrodes, one electrolyte, are given, like the copper-zinc galvanic cells in the textbook. This electrode reaction is the easiest to write, and the general rule is that the negative electrode itself loses electrons to dissolve, and the positive electrode is a cation in the liquid to obtain electrons to form an element.

We also usually use this phenomenon to judge the positive and negative electrodes of galvanic batteries: the dissolved pole is negative, and the positive electrode is the gas produced or the mass increases).

2。Fuel cell: The law is that flammable gases react at the negative electrode, and oxygen is passed through at the positive electrode. The common test is hydrogen and oxygen fuel cells, and when writing electrode reactions, pay attention to the electrolyte given by the topic:

1) If the electrolyte electrode reaction is made with sulfuric acid solution, the electrode reaction is: negative electrode: 2h2-4e-=4h+

Positive electrode: O2+4E-

4H+ = 2H2O (originally OH- was generated, but this is an acidic solution, and the OH- generated should react with H+ in the solution).

2) If KOH solution is used as the electrolyte: negative electrode: 2H2-4E-

4OH-=2H2O (as above, the generated H+ cannot exist in an alkaline solution, it must react with OH-).

Positive electrode: O2+4E-

2h2o=4oh-

3。Dry batteries and lead-acid batteries, etc.: If you want to test a type of battery, the question will generally give a general chemical equation, and first judge the positive and negative electrodes according to the change of the valency of the elements in the formula (the valency of the negative electron loses electrons increases, and on the contrary, the valency decreases to the positive electrode).

pbo2+pb+2h2so4=2pbso4+2h2o

It is easy to see that the negative electrode is pb, and according to the law, the negative electrode itself loses electrons: pb-2e-==pb2+

However, the product in the given total formula does not have Pb2+, but PbSO4, so it is known that the generated Pb2+ is also the same as SO4 in solution

2-binding, it is easy to get the result, pb-2e-+so42-=pbso4. The next positive electrode is easy to do, and the negative electrode can be subtracted from the total reaction. Thanks for adopting! Hehehe!

Because there are only two forms of aluminum in molten salt, there is no presence of Al3+ or Cl- in the solution.

The Al element of the aluminum electrode loses electrons and binds to AlCl4- to become Al2Cl7-

The equation is: al+7alcl4--3e-=4al2cl7-

1 Cathode: 4(OH-)4E- = O2 + 2H2O Anode: 4(AG+)+4E- = 4AG

Total reaction: 4AGnO3 + 2H20 = 4AG (precipitation) + O2 (gas) + 4Hno3 (reaction conditions: electrolysis or energization).

2 Cathode: 4(OH-)-4E- = O2 + 2H2O Anode: 2(H+)+2E-=H2

Total reaction: 2H2O = 2H2 + O2 (Reaction conditions: electrolytic or energized)3 Cathode: 4(oh-)-4E- = O2 + 2H2O Anode: 2(Cu2+)+4E-=2Cu

Galvanic Battery: Negative Electrode: Usually the negative electrode metal loses electrons.

If the solution is alkaline. It depends on whether the metal will react with hydroxide after becoming an ion.

Cathode: The electron is obtained by the metal slag ions in the electrolyte.

Electrolytic cell: anode.

If it is an inert electrode.

Then it depends on the order in which the ions in the solution are discharged.

S ions, SO3 ions, I ions, BR ions, CL ions, oh roots, oxyacids (such as nitrate, sulfate), > F ions.

These ions lose electrons.

Galvanic battery electrode reactive writing skills:

According to the principle of galvanic cells, it can be obtained:

Negative electrode: electron loss occurs oxidation reaction positive electrode: electron gain reduces reaction.

To write the electrode reaction formula accurately, the most important thing is to grasp the quasi-total reaction, we can further write the electrode reaction formula through the total reaction, that is, through the total reaction to determine the oxidation and reduction of the substance (one of the conditions of the galvanic battery is the spontaneous redox reaction), find the substance before and after the reaction, take it out separately to make reactants and products, and then see if there are any missing elements, and make up the possible reactants (such as H is not enough, H+ is added under acidity, Alkaline undercompensated H2O or OH-, etc.) to separate the oxidation and reduction reactions, combine the reaction environment, and finally balance according to the conservation of charge, and the polar reaction can be obtained.

The topic is: a new type of battery, with NaBH4 (B's valency is +3 valence) and H2O2 as raw materials, the battery can be used as a power source in an airless environment such as deep-water exploration, and its working principle is shown in the figure below. The following statement is true.

The electrode reaction formula on the A electrode is: BH4- +8OH- -8E- = BO2- +6H2O

The electrode reaction formula on the b pole is: H2O2 + 2E- +2H+ = 2H2O

c For every 3mol H2O2 consumed, transfer 3mol E-

d When the battery is working, Na+ moves from the B pole region to the A pole region.

Option A From this question, we can know that the principle of galvanic battery is the redox reaction of NaBH4 (the valency of B is +3 valence) and H2O2, with NaBH4 as the reducing agent and H2O2 as the oxidant. Therefore A is the negative electrode and B is the positive electrode. (As can be seen from the figure, H+ cannot occur in alkaline environments).

A is true and B is false, H+ cannot occur in alkaline environment, it should be H2O2 + 2E-= 2OH-

C error, for every 3molH2O2 consumed, transfer 6 mol E-

D is false, when the battery is working, the A pole loses electrons, and the B pole gets electrons, which will produce an increase in OH -,, negative charge, and the Na+ moves from the A pole region to the B pole region when the battery is working.

Hope it helps.

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