REDOX reactions Balancing by the Ion-electron method (acid medium)

REDOX REACTIONS REDOX REACTIONS Balancing by the IonBalancing by the Ion--

electron method electron method

(acid medium)(acid medium)

Chemistry

José Manuel Bélmez MacíasRendered by: David Bélmez Macías

KALIUM academiawww.kaliumacademia.com

(+34) 924 104 283 - 655 840 225

REDOX REACTIONS: Balancing by the ionREDOX REACTIONS: Balancing by the ion--electron method (acid medium)electron method (acid medium)Chemistry

José Manuel Bélmez MacíasRendered by: David Bélmez Macías

KALIUM academiawww.kaliumacademia.com

(+34) 924 104 283 - 655 840 225

WhatWhat are are theythey??REDOXREDOX balancingbalancingTheThe ionion--electronelectronmethodmethodCalculatingCalculating oxidationoxidation no.no.IdentifyingIdentifying ofof semisemi--reactionsreactionsIonicIonic halfhalf--reactionsreactionsBalancingBalancing HalfHalf--reactionsreactionsMultiplyingMultiplying halfhalf--reactionsreactionsGlobal Global ionicionic reactionreactionMolecular Molecular reactionreaction

WhatWhat are are theythey?? A redox process is a reaction in which we find an electron transfer.

José Manuel Bélmez Macías, Rendered by: David Bélmez Macías

In this reactions the element or compound (A) increases its oxidation no. (from n to n+1) so it is said that it has been oxidized.

And then another element or compound (B) diminishes its oxidation no. (from m to m-1) so it is said that it has been reduced.

THIS PROCES THEN COULD BE UNDERSTOOD AS THE ADDITION OF TWO HALF-REACTIONS:

OXIDATION:

REDUCTION:

Where A transfers an electron to B causing its reduction and we would say that A is the REDUCING AGENT

However, if it is B which gets the electron from A, it becomes oxidized, then we would say that B is the OXIDIZING AGENT

INDEX

BalancingBalancing redox redox reactionsreactions The redox reactions adjustment lie in the balance of the chemical equation:

The easiest way is the:

ION–ELECTRON (HALF-REACTION) METHOD

OHKClClMnClHClKMnO 2224 8252162

Before going on werecomend you to download a quick guide of this method in

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BASIC MEDIUMGUIDE ESPAÑOLINDEX

1) Calculate the oxidation numbers of each atom in each chemical.

RULES:

OHKClClMnClHClKMnO 2224

The oxidation state of a simple one-atom ion is the same as its charge

The ionThe ion--electron method electron method (acid medium)(acid medium)


BASIC MEDIUMGUIDE ESPAÑOLINDEX

1) Calculate the oxidation numbers of each atom in each chemical.

RULES:

OHKClClMnClHClKMnO 2224 0

The oxidation state of a simple one-atom ion is the same as its chargeThe oxidation state of a free element is zero



BASIC MEDIUMGUIDEINDEX ESPAÑOL


1) Calculate the oxidation numbers of each atom in each chemical

RULES:

OHKClClMnClHClKMnO 2224 +1 0 +1

The oxidation state of H in its combinations is +1, with the exception of the metal hydrides, where it is -1

The oxidation state of a simple one-atom ion is the same as its chargeThe oxidation state of a free element is zero



RULES:

OHKClClMnClHClKMnO 2224 -2 +1 0 +1 -2

The oxidation state of O in its combinations is -2, with the exception of the peroxides, where it is -1 and in its combination with fluorine, where it will be +2



The oxidation state of a simple one-atom ion is the same as its chargeThe oxidation state of a free element is zeroThe oxidation state of H in its combinations is +1, with the exception of the metal hydrides, where it is -1



RULES:

OHKClClMnClHClKMnO 2224 +1 -2 +1 0 +1 +1 -2

The oxidation state of the alkaline metals in its combinations is +1 (group 1 of the P.S.) and +2 for the alkaline earth metals (group 2 of the P.S.)







RULES:

OHKClClMnClHClKMnO 2224 +1 -2 +1 -1 0 +1 -1 +1 -2

In the case of the binary salts, the halogen elements behaves with -1, the elements in the group of O with -2, the elements in the group of N with -3, C and Si -4 and B -3








RULES:

OHKClClMnClHClKMnO 2224 +1 +7 -2 +1 -1 +2 -1 0 +1 -1 +1 -2

The algebraic sum of all the oxidation numbers of each atom in a neutral compound is zero. If the compound is an ion, this addition will be the same as the charge of the ion


In the case of the binary salts, the halogen elements behaves with -1, the elements in the group of O with -2, the elements in the group of N with -3, C and Si -4 and B -3







2) Identify which element is reduced and which is oxidized.


REDUCTION OXIDATION




3) Write down the half-reactions in the ionic mode


REDUCTION OXIDATION

*¡¡CAREFUL!!: Only the acids, hydroxides and salts are written the ionic mode

The electric charge (q) comes from the addition of oxidation numbers, then as an example in the premanganate ion:

MnO4+7 -2

q=7+4·(-2)=-1 MnO 4-

Reduction:

Oxidation:

MnO 42+-

MnCl - Cl2




4) Balance each half-reaction


REDUCTION

Reduction:

Oxidation:

MnO 42+-

MnCl - Cl2

A) First, elements which are not hydrogen or oxygen are balanced

2

OXIDATION






REDUCTION

MnO 42+-

MnCl - Cl2

B) Oxygen is balanced by adding H2O where necessary

2 OH24

OXIDATION



Reduction:

Oxidation:





REDUCTION

MnO 42+-

MnCl - Cl2

C) Hydrogen is then balanced by adding H+ where necessary

2OH2 4H + 8

OXIDATION




Reduction:

Oxidation:





REDUCTION

MnO 42+-

MnCl - Cl2

D) Then balance the charges by adding electrons (e-) so that at the end the truth charge is the same in both parts of the reaction

2OH2 4H + 8 5e-

2e-

OXIDATION

THIS STEP IS NORMALLY CONFLICTIVE BUT IT IS NOT DIFFICULT AT ALL, WE JUST NEED TO BARE IN MIND THAT THE ELECTRONS ARE PLACED ON THE LEFT IN THE REDUCTION AND ON THE RIGHT IN THE OXIDATION AND THAT EACH ONE APORTS A NEGATIVE CHARGE. SO THAT A SIMPOLE EQUATION CAN BE PLANTED:

MnO 42+-

Mn OH 2 4H + 8 xe-M nO 42+-

Mn OH 2 4H + 8 xe-

x

x=5


C) Hydrogen is then balanced by adding H+ where necessary



Reduction:

Oxidation:

BASIC MEDIUMGUIDEINDEX


ESPAÑOL

5) Multiply each half-reaction with a coefficient for it to have the same number of electrons


MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5


REDUCTION OXIDATION

Reduction:

Oxidation:



6) Add up boths semi reactions


MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl222 5


REDUCTION OXIDATION

Reduction:

Oxidation:



7) Turn the ionic reaction into molecular


THIS STEP IS NORMALLY CONFLICTIVE, HOWEVER IT IS QUITE EASY IF WE BARE IN MIND THAT THE IONIC

REACTION IS ALREADY ADJUSTED, SO THEN WE ADD THE IONS WE NEED IN THE SAME CUANTITY AS BOTH PARTS

OF THE REACTION

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl222 5


REDUCTION OXIDATION

Reduction:

Oxidation:





EACH PERMANGANATE ION (MnO4-) CARRIES ONE POTASSIUM ION (K+) IN THE ORIGINAL REACTION

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl222 5


REDUCTION OXIDATION

Reduction:

Oxidation:





EACH PERMANGANATE ION (MnO4-) CARRIES ONE POTASSIUM ION (K+) IN THE ORIGINAL REACTIONAs in each ionic reaction there are two permanganate ions, we add two ions potassium (K+) to each part of the reaction

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl222 5

K +2 K +2


REDUCTION OXIDATION

Reduction:

Oxidation:





THE PROTONS (H+) AND THE CHLORIDE IONS (Cl-) MUST MAKE UP THE HYDROCHLORIC ACID, AND IT IS USEFUL TO LEAVE THE ACIDS TO THE END

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +

22 52 2


REDUCTION OXIDATION

Reduction:

Oxidation:





EACH Mn2+ ION CARRIES TWO CHLORIDE IONS (Cl-) IN THE ORIGINAL REACTION

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +

22 52 2


REDUCTION OXIDATION

Reduction:

Oxidation:





EACH Mn2+ ION CARRIES TWO CHLORIDE IONS (Cl-) IN THE ORIGINAL REACTION

As in the ionic reaction there are two Mn2+ ions, we add four chloride ions (Cl-) to each part of the reaction

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +Cl -4 Cl -422 5

2 2


REDUCTION OXIDATION

Reduction:

Oxidation:





EACH POTASSIUM ION (K+) CARRIES ONE CHLORIDE ION (Cl-) IN THE ORIGINAL REACTION

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +Cl -4 Cl -422 5

2 2


REDUCTION OXIDATION

Reduction:

Oxidation:





EACH POTASSIUM ION (K+) CARRIES ONE CHLORIDE ION (Cl-) IN THE ORIGINAL REACTION

As in the ionic reaction there are two potassium ions (K+), we add two chloride ions (Cl-) to each part of the reaction

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +Cl -4 Cl -422 5

2 Cl -22Cl -2


REDUCTION OXIDATION

Reduction:

Oxidation:





THE REST OF SPECIES: WATER (H2O) AND CHLORINE (Cl2) BOTH IN THE ORIGINAL AND IN THE IONIC REACTION ARE THE SAME SO THAT THERE IS NO NEED TO ADD ANYTHING

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +Cl -4 Cl -4 Cl -Cl -22 5

2 222


REDUCTION OXIDATION

Reduction:

Oxidation:






MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +Cl -4 Cl -4 Cl -Cl -22 5

2 222

OHKClClMnClHClKMnO 2224 2 16 2 2

NOW WE GROUP UP THE IONS IN ORDER TO FORM THE SPECIES OF THE ORIGINAL REACTIONS

OHKClClMnClHClKMnO 2224 OHKClClMnClHClKMnO 2224 OHKClClMnClHClKMnO 2224

REDUCTION OXIDATION

Reduction:

Oxidation:





REDUCTION

Reduction:

Oxidation:

MnO 42+-

MnCl - Cl22

OH2 4H + 8 5e-2e-

OXIDATION

( )·2( )·5

MnO 42+- Mn OH2 8H +16 Cl -10 Cl2

K + K +Cl -4 Cl -4 Cl -Cl -22 5

2 222

OHKClClMnClHClKMnO 2224 2 16 2 25 8

IN THE CASE OF THE MOLECULAR SUBSTANCES WE PRESERVE THE COEFFICIENTS

OHKClClMnClHClKMnO 2224 OHKClClMnClHClKMnO 2224 OHKClClMnClHClKMnO 2224




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