How to utilize charge transfer to equation trimming

For example, mno2+hcl---mncl2+cl2+h2o

The charge should be conserved when Mn decreases by 2 and Cl rises by 1, so the ratio of the amount of Mn to Cl is 1:2, but there are some chlorine elements that have not been oxidized, which should be considered by the ion equation.

mNO2 + 4H+ +2Cl-===Mn2+ +Cl2 + 2H2O, at this time, the ratio of the amount of Mn to Cl is 1:2.

Trimming method: The balancing method of redox reaction generally uses the method of equal total number of positive and negative valence changes, that is, the valency rise and fall method. The key to the valency lifting method to trim the redox reaction:

1. Correctly mark the valency of the variable elements;

2. The price change should be "counterpart"; That is, when the redox reaction occurs between substances of the same element with different valence states, it is necessary to pay attention to the absence of "valence crossing".

3. Balancing from substances containing "full valence elements";

4. The total number of electrons gained and lost should be equal.

Steps of valence lifting:

1. Indicate the valency of the variable valence element. 2. List the number of changes in valence.

3. Find the least common multiple. According to the principle of equality of the total number of electrons gained and lost, find the coefficient applied that increases the total number of valencies equal to the decrease of the total number of valences.

4. Balance the substance of the full valence element. The coefficients from step 3 are used to balance the species of the full valence element.

5. Balance the rest of the matter according to the principle of conservation of atomic type and number and conservation of ionic charge.

Example analysis: Example 1, Fe2O3+ CO - Fe+ CO2

The valency of the element that indicates the valence.

Lists the number of changes in valency.

fe2o3+ co—— fe+ co2

Find the least common multiple.

fe2o3+ co—— fe+ co2

is the coefficient of the corresponding substance).

Trim the substances of the full valence element (1) Fe2O3 + 3) Co – Fe + Co2

Trim the rest of the substance: (1) Fe2O3 + 3) Co (2) Fe + 3) CO2

Another example is KBR + H2SO4 (concentrated)— K2SO4 + SO2 + BR2 + H2O

br: -1 0 up 1

s: +6 +4 drop 2

The least common multiple is 2, but the s of H2SO4 in the reactant is not full valence (there is also K2SO4 +6 valence S in the product).

So the lifting method is trimmed:

br: -1 0 up 1 2

S: +6 +4 Drop 2 1 1 should be matched on the SO2 of the product, 2) KBR + H2SO4 (concentrated) — K2SO4 + 1) SO2 + BR2 + H2O

The rest of the trim is observed:

2) KBR+2)H2SO4 (concentrated) 1)K2SO4+1)SO2+1)BR2+1)H2O

The valency increases and loses the redox agent". The corresponding is "reducing the valency to obtain a reducing oxidant". The specific usage is:

1) Mark the valency of the element in the upper part of the element where the valency of the element changes, and distinguish whose is higher and whose is lower.

2) Connect the same elements with lines to find and mark the number of higher charges or the number of lower charges.

3) Find the least common multiple and multiply it by the number of charges that increase or decrease, respectively.

4) Trim: Write the least common multiple of each before their respective chemical formulas (i.e., coefficients). And pay attention to whether these elements with valency changes are equal before and after the chemical change, and in general, if they are not equal, they are integer multiples.

5) Cooperate with the observation method to balance others, such as water and insoluble matter generated.

The essence of the redox reaction is that electrons are transferred during the reaction, and the total number of electrons obtained by the oxidant (or the total number of reduced valency of the elements) must be equal to the total number of electrons lost by the reducing agent (or the total number of increased valency of the elements), according to this principle, the chemical equation of the redox reaction can be balanced.

Trim steps:

1) Marked price: Correctly mark the valency of elements whose valence has changed before and after the round out reaction.

2) Column Changes: List the values of the increase and decrease of the valency of the elements.

3) Find the total number: find the total number of increases and decreases in the valency of the element, and determine the stoichiometric number of oxidant, reducing agent, oxidation product and reduction product.

4) Matching coefficient: the stoichiometric number of other substances is leveled by observation method.

5) Careful inspection: Using the three principles of "conservation" (i.e., conservation of mass, conservation of electrons of gain and loss, and conservation of charge), check whether the equation of the balancing hand core is correct one by one.

Cuprous sulfide reacts with sufficient concentrated nitric acid to form sulfuric acid, nitrogen dioxide, copper nitrate and water as an example.

1. First write out the reflexes and products Cu2S + Hno3 - H2SO4 + NO2 + Cu(NO3)2 + H2O

2. Mark the valence.

3. List the valency changes.

Elevated s -2 to +6 change 8

cu +1 to +2 change 1 2 copper 2 = 2 total 10 decrease n +5 to +1 change 1 10

4. Write the coefficient and balance other substances.

5、cu2s+14hno3 =h2so4+10no2+2cu（no3)2+6h2o

You have 7 apples, I have 2 apples, he has 3 apples, a total of 12 apples.

I'll give you an apple, give him an apple, and now you have 8 apples, I don't have an apple, he has 4 apples, and the total is still 12 apples.

This balancing method is based on the most basic theorem, which is that "matter does not disappear out of thin air, nor does it create out of thin air".This is called conservation of valence.

When an element loses electrons, it behaves as normal, and as many electrons as it loses, it carries as much valence as it loses. In the same way, when an element gets electrons, it behaves as negative valence, and as many electrons as it gets, it carries as many negative valences.

And the total number of electrons lost by the first element must be equal to the total number of electrons gained by the second element.

The gain and loss of electrons is manifested in the valency of the elements.

Therefore, if the valency of one element changes during the reaction, it means that the valency of another element must have changed accordingly. And these two elements can be the same.

For example, the simplest, O2+2H2==2H2O

Oxygen has a valency of 0 before the reaction and -2 after the reaction, then oxygen gains two electrons for each oxygen atom in this reaction.

Hydrogen has a valency of 0 before the reaction and +1 after the reaction, then hydrogen loses one electron per hydrogen atom in this reaction.

Then, when hydrogen and oxygen react, the electrons obtained by oxygen are provided by hydrogen. So oxygen needs two electrons, so it needs two hydrogens, and each hydrogen provides one electron for oxygen. So the amount of oxygen:

The amount of hydrogen = the number of electrons lost by hydrogen: the number of electrons gained by oxygen = 1:2

Therefore, in the trim of hydrogen combustion in oxygen, because one unit of oxygen contains two units of oxygen atoms, it needs 4 units of hydrogen atoms, which means two units of hydrogen to generate two units of water.

That's how it's trimmed.

Conservation of transfer electron number trim chemical equation.

It is important to emphasize that this can only be used for redox reactions.

By the nature of the redox reaction we can know that the oxidant gets the electrons and the reducing agent.

The number of lost electrons is the same, so we can balance based on that.

For example: kio3+ki+h2so4 k2so4+i2+h2o (no trim).

This is a typical conversion reaction.

From the two-line bridge method, we know that the oxidant is kio3, the reducing agent is ki, and both the reduction product and the oxidation product are i2, and we also know that ki loses 1e and kio3 gains 5e, then their least common multiple.

is 5, so the coefficient of ki is 5, the coefficient before kio3 is 1, and the coefficient before i2 is equal to (5+1) 2=3, according to the conservation of potassium ions, the coefficient of K2SO4 is 3, the coefficient of H2SO4 conserved by sulfate ions is 3, and the coefficient of water obtained by the conservation of hydrogen ions is 3

You can get the equation after trimming:

1kio3+5ki+3h2so4==3k2so4+3i2+3h2o

This method is a very useful method, and there are many tricks in it, which I can't say at once, so I will talk about the actual topic later.

Hope you are satisfied.

This method is suitable for more complex ionic equations (redox reactions), which are more complex with general methods, but simpler from the point of view of ion transfer (valence rise and fall). This method is to observe the number of electrons gained and lost by ions before and after the reaction of the compound, and the stoichiometric ratio of the two substances is obtained by balancing the electrons gained and lost, and then the balance of the equation is completed by setting the unknowns.

Example: Reaction of potassium permanganate and concentrated hydrochloric acid.

mno4- +h+ +cl- =mn2+ +cl2 + h2o

1.Let's look at the permanganic acid heel first, where MN is +7 valence, but in manganese dichloride it is +2 valence.

So mn:+7 +2--- get 5 electrons (drop back).

2.Looking at the chlorine atom again, the valence of potassium chloride and manganese dichloride has not changed, only the chlorine gas has changed from -1 valence to 0 valence.

So Cl:-1 0 --- lose 1 electron (liter of oxygen).

3.We have 5 electrons less than x parts of Mn, that is, 5x parts less electrons, and let y parts of chlorine have 1 more electron, because there are 2 chlorine atoms in a chlorine molecule, so a total of 2y parts of electrons are raised.

That is, 5x=2y, so the simplest ratio of x and y is 2:5

4.Then the 2nd generation is added to the original potassium permanganate or the generated manganese dichloride (manganese's ** or generated), and the 5th generation is added to chlorine gas (because only its chlorine valency is changed).

2 n times) mNO4- +H+ +Cl- —Mn2+ +5 N-fold) Cl2 + H2O

Keeping this minimum ratio unchanged, the remainder is balanced by observation, uneven, and then transformed into the size of the coefficient n until it can be balanced.

Finally, we get: 2mNO4- +16H+ +10Cl- = 2mN2+ +5Cl2 + 8H2O

na+h2o==naoh+h2

You are a combination of valence sedan car annihilation high and low to match the leveling, Datuk is not very big....Change the head....Sodium: 0--+1

Hydrogen: +1 - -0

1:1, OK pyrrol closure or, sodium with 2, element conservation, sodium hydroxide with 2, oxygen conservation with water with 2, with hydrogen conservation check whether it is flat can be .........

The so-called electron transfer is the essence of redox reactions.

In practice, the total number of valence rises and falls is equal to the balance and hunger letter.

It's just the ions you give.

Before the reaction. o There is -1 2-2

After the reaction o has 1,0, there should be 2 oh You can mark these valences on the number line, and then according to the principle that the valency changes can be attributed to the center but do not cross each other, judge the combustion wheel to conclude that the change in valency is 1 2 o disproportionated into 1 and 0 valence.

The o of the original 2 is unchanged.

1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 2 0, 1 2, and because O2 is a diatomic molecule, the number of rises is 1 and the total number of rises and falls is equal, so it becomes like this.

o2－ ＋h2o＝h2o2 ＋o2 ＋

There is a negative charge on the left, there is no on the right, definitely add oh, 2, and then use h atom conservation, trim everything.

o2－ ＋2h2o＝h2o2 ＋o2 ＋2oh-

Cumous sulfide reacts with a sufficient amount of concentrated nitric acid to produce sulfuric acid, nitrogen dioxide, copper nitrate and water.

1. First write the reactants and products Cu2S + Hno3 - H2SO4 + NO2 + Cu(NO3)2 + H2O

2. Mark the valence.

3. List the valency changes.

Elevated s -2 to +6 change 8

Cu +1 to +2 Change 1 2 Copper 2=2 Total 10

Decrease n +5 to +1 change 1 10

4. Write the coefficient and balance other substances.

5、cu2s+14hno3 =h2so4+10no2+2cu（no3)2+6h2o