How do you balance the reaction equation of redox in high school chemistry?

If you use algebra, sometimes an equation can have as many as 5 unknowns, and the equation has only 4, which is the worst method;

The general balancing method is based on the conservation of electrons gained and lost, that is, the rising valency of the oxidized element multiplied by a coefficient is equal to the decreasing valency of the reduced element multiplied by a coefficient.

That's right, but it's still very troublesome if you're not proficient in it, so after you understand the method, you have to do more questions so that you can fully master it!!

Trimming of the redox reaction equation.

1. Trim principle: the first is the conservation of electrons, followed by the conservation of charge, and finally the conservation of atoms.

2. Trim steps: standard price change - column change - find the total number - match coefficient - fine inspection. According to the total number of the increase of the valence of the reducing agent and the decrease of the valence of the oxidant, the coefficient in front of the reducing agent and the oxidant is determined, and the coefficient of the oxidation product and the reduction product is determined correspondingly, and the coefficient of other ions (generally the coefficient of hydrogen ion and hydroxide ion) is determined by the conservation of charge on the left and right sides of the equation, and finally the coefficient in front of other substances is determined according to the conservation of atoms.

3. Trim method:

Forward balancing method: if the oxidized and reduced elements are two different reactants, the balancing is usually carried out from the left side of the equation, especially the centering reaction;

Reverse balancing method: For oxidants and reducing agents are the same substance, that is, their own redox reactions or there is only one reactant, especially for disproportionation reactions, it is generally easier to balance the products from the right side of the equation.

Upstairs has already said trimming for this equation.

There are many ways to balance chemical equations, but the main ones are preliminary balancing according to the rise and fall of valence, and then final balancing according to the conservation of elements....If there are three valence elements that can use the zero-price balancing method, the teacher should have talked about them all...

Whistleblowing 1The simplest: It can be conserved according to atoms, that is, the total number of atoms before and after the reaction is equal, and the corresponding number of atoms is equal.

2.The most effective: according to the conservation of electrons (which can be understood as the rise and fall of valence), because the essence of redox reaction is the gain and loss of electrons, and the gain of electrons must lead to the loss of electrons, and the conservation of electrons is used to make sure that it is right.

It's useless to say too much, let me give you an example.

cu+ hno3= cu(no3)2+ no2↑+ h2o

When encountering such a problem, we first observe who lowers the price of reactants and who increases the price.

cu(0)--cu2+,,hno3(+5)--no2(+4)

Therefore, Cu is the loss of electrons, Hno3 is the gain of electrons, according to the conservation of electrons, the coefficient of No2 should be 2 times that of Cu, write 2 first. Cu + Hno3 = Cu(No3)2 + 2NO2 + H2O (the coefficients of Cu and No2 are determined).

However, it should be noted that Hno3 not only provides N of No2, but also No3-, so according to the conservation of atoms, the coefficient of Hno3 is 2+2==4, and the coefficient of H2O is 2

The final trim is.

cu+ 4 hno3= cu(no3)2+ 2 no2↑+ 2 h2o

fes2+( hno3=( fe(no3)3+( so2+( no+( h2o

About this equation.

Let's start by marking their valence changes:

Fe from +2 to +3 (loss of 1e-) and S from -1 to +4 (loss of 5e-) both of these are reducing agents.

N from +5 to +2 (to get 3E-), it is also worth noting that Hno3 provides the N atom of No3- in addition to the N atom of No, this part of Hno3 is not used as an oxidizing agent, but only as an acid, and in the case of nitric acid with double action, the coefficient of the reactant is generally inverted by the coefficient of the product.

Now we are starting to make use of the conservation of electrons, trimming.

Fe from +2 to +3 (loss of 1e-) and S from -1 to +4 (loss of 5e-), so 1mol of reducing agent loses a total of 1+2 5=11mole-

n from +5 to +2 (to get 3e-), 1mol oxidant to get 3mole-, take the least common multiple, is 33

Therefore, first FES2 is matched with 3, NO is matched with 11, and then according to the conservation of atoms, FE(NO3)3 is matched with 3, and SO2 is matched with 6, so the coefficient of HNO3 is 3 3+11=20, and HO2 is 10

So the final answer is.

3fes2+20hno3=3fe(no3)3+6so2+11no+10h2o

Finally check it with atomic conservation.

Absolute: The trim of the redox reaction should be in accordance with the conservation of atoms and electrons, and the charge should be conserved if the ion equation is preserved.

Its trimming points: 1 valence state 2 column changes 3 matching coefficient 4 watch conservation.

The most important step is step 3, where the gains and losses of electrons must be conserved.

Look at your teachers, they certainly don't use a two-line bridge to trim.

It is just that the change in valency is directly represented below the substance, and then the least common multiple is balanced.

In fact, this can be marked to reactants and products, but it must be marked under the substances with corresponding valence changes, but as in the following reaction, Cu + Hno3 = Cu(No3)2 + No2 + H2O

Cu can be directly labeled, but nitrate Hno3 can not be directly labeled, because the combination of N in the reactant Hno3 is not all changed, N in Cu(NO3)2 is still +5 valence, and only the generation of can be No2 N element is +4 valence, so it is necessary to match the coefficient of No2 first, if the coefficient of nitric acid is first matched, then it contains the N element with no change in valency. Here we apply the equal rise and fall of valency, or the equal gain and loss of electrons to balance the substances with varying valence, and the coefficient of nitric acid should be conserved by atoms, we determine the coefficient of Cu as 1 and the coefficient of No2 2 by using the conservation of electron gain and loss according to the change Cu(0)--Cu2+,,Hno3(+5)--No2(+4) electron gain and loss, and finally is.

cu+ 4 hno3= cu(no3)2+ 2 no2↑+ 2 h2o

These reactions are all of the same class, called partial redox reactions, that is, elements with valence changes, and some products have no valence.

For example, the following reaction:

cu+ hno3= cu(no3)2+ no+ h2o

2kmno4 = k2mno4+ mno2+ o2↑

6hno3+fe===fe(no3)3+3no2↑+3h2o

4hno3+fe===fe(no3)3+no↑+2h2o

30hno3+8fe===8fe(no3)3+3n2o+15h2o

36hno3+10fe===10fe(no3)3+3n2↑+18h2o

30hno3+8fe===8fe(no3)3+3nh4no3+9h2o

And so on, this is a big type, and also the equation focuses on balancing, and you must master the content.

There are so many on it that you can give it a try!

It doesn't matter if it's on top or below the product, it's important to understand why you want to look at the valency of the product.

Redox trimming is not enough to look at the law of conservation of mass, and it is very difficult to balance.

It is also necessary to look at the conservation of charge and the conservation of electron transfer, the latter being what you asked.

It depends on the [change] of the valency, so the difference is known by the valency of element A in the product and the valency of element A in the reactant.

The rest is up to you.

For example, look at the first floor.

The teacher may be handy, or convenient, or there is not enough space, and the formal one is on top, hehe.

High School Chemistry Experiments Redox Reactions.

feo cr2o3.

When positive divalent Fe is oxidized to NaFeO2, Fe loses 1mol of electrons.

The positive trivalent Cr is oxidized to Na2Cro4 and the Cr is positive hexavalent.

A CR loses 3mol of electrons.

So the whole FeO Cr2O3 is seen as a whole losing 7mol of electrons.

1mol of oxygen gives 4mol of electrons.

After that, it is balanced with electron conservation and then mass-conserved trim.

4Feo CR2O3+7O2+20NaOH 8NA2CRO4+4NAFEO2+10H2O Ferrous loses electrons, chromium loses electrons, and oxygen gets electrons that are reduced.

In Feo Cr2O3, positive divalent Fe is oxidized to NaFeO2 and Fe loses 1mol of electrons.

After the positive trivalent Cr is oxidized to Na2Cro4, the Cr becomes positive hexavalent and one Cr loses 3mol electrons.

So the whole Feo Cr2O3 is seen as a whole losing 7mol electrons and 1mol oxygen gaining 4mol electrons.

After that, it is balanced with electron conservation and then mass-conserved trim.