Chapter 18.2 - Balancing redox equations Redox reactions: reduction – oxidaton reactions

Chapter 18.2 - Balancing redox equations

Redox reactions: reduction – oxidaton reactions.

Reduction: lowering of positive charge (increase of negative charge)

Oxidation: increase of positive charge (lowering of negative charge)

If an electron is transferred, thenthe negative charge must decrease somewhere (where the electron comes from), and it must increase somewhere else (where the electron goes),

hence reduction and oxidation must occur together.

There is a balance in such changes, and reduction – oxidaton reactions provide good examples for balancing chemical equations.

Redox Reactions, Acid, Base, Paul G. Mezey

Chapter 18.2 Balancing Oxidation-Reduction Equations [19.1]

Chapter 18: Electrochemistry

• Chapter 18: Electrochemistry

• 18.2 Balancing Oxidation-Reduction Equations [19.1]

• (Note: This section is usually covered with a very brief review of oxidation numbers and oxidizing/reducing agents, as covered in Chemistry 1010, which is found in section 4.9 of the Tro text).


Redox reactions are easier to balance if we understand where the electrons are coming from and where they are ending up.

Oxidation numbers help us figure this out.

Redox Reactions, Acid, Base , Paul G. Mezey


Oxidation numbers

In a very simplistic way, the basis of oxidation numbers is the following idea:

Each (single) chemical bond between any two atoms X and Y is formed by two electrons.There are three possibilities: 1. X attracts electrons more than Y2. Y attracts electrons more than X3. X and Y attract electrons exactly the same way (important: this happens only if X = Y)

Case 1. Even if the attraction is only slightly stronger by X than by Y, we pretend that the entire electron pair belongs to X. Case 2. Similarly to case 1, we pretend that the entire electron pair belongs to Y. Case 3. We assign one electron to X, and the other electron to Y.

If we carry out this formal assignment of electrons for each bond of themolecule or ion, then the “charge” obtained on each atom is the oxidation number of the given atom.



A numerical measure of how strongly atoms attract electrons is electronegativity

We may phrase the concept of oxidation number in terms of electronegativity:

In calculating oxidation numbers the two electrons in a bond are completely assigned to the more electronegative element,

UNLESS

the bond is between two atoms of the same element, in this case the electrons are shared equally so that one electron is assigned to each of the atoms.



Based on this simple idea, we can follow a set of rules to assign oxidation numbers .

We go through the set of rules until we find the FIRST rule that applies to our specific atom in the compound or ion of interest.

Oxidation number rules (Rule 1)

Atoms of pure elemental compounds (e.g. metals, solid carbon, O2 gas, Br2 liquid, I2 solid, etc.) have an oxidation number of ZERO

Oxidation number rules (Rule 2)

Monatomic ions (like Mg2+, Li+, F-, S2-, etc.) have an oxidation number equal to the charge

Oxidation number rules (Rule 3)Fluorine, as the most electronegative element, will ALWAYS have an oxidation number of -1 EXCEPT in F2 where it has an oxidation number of ZERO (Rule 1)



Oxidation number rules (Rule 4)Oxygen, as the second most electronegative element, will usually have an oxidation number of -2 UNLESS it is bonded to another oxygen or

fluorine

Oxidation number rules (Rule 5) Hydrogen, will have an oxidation number of +1 unless it is bonded to a metal atom, where it will have anoxidation number of -1, or if it is bonded to another H atom, where it will

have an oxidation number of 0 (Rule 1).

Oxidation number rules (Rule 6)Halogens (Cl, Br, I, and At), generally have an oxidation number of -1EXCEPT when bonded to F, O, or halogens of the same type or above it

on the periodic table.

Oxidation number rules (Rule 7)The sum of the oxidation numbers for ALL the atoms in a compound or ion MUST ADD UP to match the total charge on the compound (zero) or ion (ion charge).



Apply the rules in the order given.

Any atoms not specifically covered in the rules can usually be assigned

oxidation numbers by applying Rule 7 and some logic.

If in any doubt, please remember that the main idea is to artificially

assign the entire bonding electron pair to the more electronegative

atom in each bond. If all bonding electron pairs are distributed this way,

then the charge obtained on each atom becomes the oxidation number.

(Of course, the charge on each atom includes the protons in the

nucleus as well as all the electrons of the atom).



Problems:

Assign oxidation numbers to every atom in the following compounds and ions:

S8 LiH

TiO2 H2O

H2O2 HSO4-

Cr2O72- CaCO3



Using oxidation numbers, the balancing of redox chemical equations becomes simpler.

The main steps in balancing redox equations:

1.Using oxidation numbers, identify what is oxidized (what loses electrons) and what is reduced (what gains electrons).2.What are the products after the oxidation and reduction take place?3.Is the redox reaction done under acidic or basic conditions?

The information obtained from these 3 steps results only in an unbalanced skeleton equation where we know generally what reactants and products are specifically involved in the electron transfer (redox) process.




Examples

Examples of skeleton equations with oxidation numbers shown:

2- 4 3 2- 3 2- 6

232

42272 gCOaq Craq OCaq OCr

1 3 1 2- 3 2- 5 0

3-4

-3 aqNHs OHAlg NOs Al


How to proceed further?

Brake up the problem to two formal half-reactions :

We break the skeleton reaction into two unbalanced half-reactions where the

oxidation half-reaction has an atom where the oxidation number becomes more positive

and the

reduction half reaction has an atom where the oxidation number becomes more negative.



Example: Half-reactions from skeleton reaction


Oxidation half-reaction:

Reduction half-reaction:

2- 4 3 2- 3 2- 6

232

42272 gCOaq Craq OCaq OCr

2- 4 2- 3

-2

242 1e lose should atom C each gCOaq OC

3 2- 6

3272 e 3 gain should atomCr each aq Craq OCr


Steps for balancing half-reactions in ACIDIC solution:


1. Balance all atoms EXCEPT H and O in each half reaction:

2. Balance O atoms by adding water to the side missing O atoms:

oxidation gCO 2aq OC 2242

reduction aq Cr 2aq OCr 3272

O!for balancedalready gCO 2aq OC 2242

(l) OH 7aq Cr 2aq OCr 232

72


Steps for ACIDIC solution


3. Balance H atoms by adding H+ to the side missing H atoms:

Oxidation half-reaction

Reduction half-reaction (l) OH 7aq Cr 2aq H 14aq OCr 2

3272

H!for balancedalready gCO 2aq OC 2242




4. Balance charge by adding electrons to the side with more total positive charge:

6 is charge total

23-

12 is charge total

272 (l) OH 7aq Cr 2e 6aq H 14aq OCr

-

zero is charge total

2

2- is charge total

242 e 2gCO 2aq OC


Reduction half-reaction




5. Make the number of electrons the same in both half-reactions by multiplication, while avoiding a fractional number of electrons:

-2

242 e 6gCO 6aq OC 3

(l) OH 7aq Cr 2e 6aq H 14aq OCr 23-2

72


Reduction half-reaction




6. Add the half reactions together and then simplify by cancelling out species that show up on both sides:

-

23

2

-272

242

e 6(l) OH 7aq Cr 2gCO 6

e 6aq H 14aq OCraq OC 3

(l) OH 7aq Cr 2gCO 6

aq H 14aq OCraq OC 3

23

2

272

242

Added together

Simplified (should have NO extra electrons!)




7. Confirm that the reaction is balanced in number of atoms and total charge on both sides of the arrow.

If the reaction stoichiometry can be simplified by division without giving fractional coefficients, you can simplify further:

(l) OH 7aq Cr 2gCO 6

aq H 14aq OCraq OC 3

23

2

272

242


Steps for balancing half-reactions in BASIC solution:


First follow steps one to seven as seen in the case of acidic solution.

8. Add the same number of OH- groups as there are H+ present to BOTH sides of the equation:

added

23

2

added

272

242

aq OH 14(l) OH 7aq Cr 2gCO 6

aq OH 14aq H 14aq OCraq OC 3


Steps for BASIC solution


9. One side of the reaction has BOTH OH- and H+ present in equal amounts.

Combine these together to make an equal amount of water:

aq OH 14(l) OH 7aq Cr 2gCO 6

aq OH 14aq H 14aq OCraq OC 3

23

2

(l) OH 14 becomes

272

242

2


Steps for BASIC solution


10. Simplify by cancelling out an equal number of water from each side until one side has no water and confirm that the reaction is balanced in number of atoms and total charge on both sides of the arrow:

aq OH 14aq Cr 2gCO 6

(l) OH 7aq OCraq OC 33

2

2272

242


Problem


Balance the following unbalanced redox skeleton equation in BASIC solution

1 3 1 2- 3 2- 5 0

3-4

-3 aqNHs OHAlg NOs Al


End of section



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Chapter 18.2 - Balancing redox equations Redox reactions: reduction – oxidaton reactions