Oxidation, Reduction and Redox

Question 1

Define oxidation number/oxidation state

Answer 1

Oxidation state is the hypothetical charge of an atom, if all of it’s bonds to other atoms were fully ionic.

Question 2

What is the oxidation state of all pure elements?

Answer 2

0, all pure elements are atoms and do not have a charge.

This is not the reason for it being 0, it’s just saying they’re atoms.

Question 3

What is the oxidation state of all ions?

Answer 3

Charge/Subscript
Example: Hg2^2+, Hg’s oxidation state is 1+

Single ion, not polyatomic ion.

Question 4

What is the oxidation state of fluorine in compounds?

Answer 4

-1

Question 5

What is the oxidation state of oxygen (oxide) in compounds?

Answer 5

-2 unless bonded to fluorine

Question 6

What is the oxidation state of oxygen (peroxide) in compounds?

Answer 6

-1

Question 7

What is the oxidation state of oxygen (superoxide) in compounds?

Answer 7

-0.5

Question 8

What is the oxidation state of hydrogen when bonded to nonmetals?

Answer 8

+1

Question 9

What is the oxidation state of hydrogen when bonded to metals?

Answer 9

-1

Question 10

How can you determine which element in a compound typically holds the negative oxidation state?

Answer 10

The one that’s more electronegative.

Question 11

What is the oxidation state of chlorine in compounds?

Answer 11

-1

Question 12

Given a compound, find one element’s oxidation state given the oxidation state of the rest.

Answer 12

Since compounds have an overall charge of 0, we can write a chemical equation, S= Subscript and E = Element
S1E1 + S2E2 + … = 0
Substitute each E for their oxidation state besides the unknown:
S1E1 + S2O2 + … = 0
Than solve from there:

Example:
MgCl2 with Cl’s charge = -1
Mg + 2Cl = 0
Mg + 2(-1) = 0
Mg + -2 = 0
Mg = +2

Typically, the element given has a higher electronegativity

Question 13

What is the typical charge of alkaline earth metals?

Answer 13

+2

Question 14

What is the oxidation state of aluminum in compounds?

Answer 14

+3

Question 15

Given a polyatomic ion, find one element’s oxidation state given the oxidation state of the rest.

Answer 15

S = Subscript, E = Element and O = oxidation state
S1E1 + S2E2 + … = Charge of ion
S1E1 + S2O2 + … = Charge of ion
Solve from there
Example:
SO4 with oxygen state = -2
S + 4O = -2
S + 4(-2) = -2
S + -8 = -2
S = 6

Typically, the element given has a higher electronegativity

Question 16

How can you determine which element takes the negative oxidation state in a compound?

Answer 16

The one with a higher electronegativity

Question 17

What does it mean if an element’s typical oxidation state is a decimal in compounds?

Answer 17

It means that this is the average oxidation state. Example: Iron, oxidation state is 2.67, but it has oxidation states of 2 and 3 in compounds.

Question 18

How do you calculate the oxidation state of an element in a compound that has multiple oxidation states in that compound, given the other oxidation states, the possible oxidation states, and the average oxidation state?

Answer 18

The total charge is 0. First add up the charges of the other elements. Since the total charge is 0, you know the charge of the element has to cancel it out, so you just find the configuration that works with the average.
Example: Fe3O4
O has an oxidation state/-2
Fe is 2.67 with 2 and 3
There is a charge of -8 from the O, so the 3 fe has to match that. +2, +3 and +3 is the only combination that works, so it’s the answer.

Question 19

What is the difference between being oxidized and reduced?

Answer 19

In a redox reaction, often times the oxidation numbers of elements either go up in the product or go down in the product. When it goes up, it’s oxidation , when to goes down, it’s reduced.

Example (Oxidation state in brackets):
Mg (0) + O2 (0) = Mg(2)O(-2)
Mg has been oxidized and O2 has been reduced.

Question 20

In a transfer of electrons, certain elements are oxidized and reduced. Which ones gain and lose electrons?

Of the oxidized and reduced

Answer 20

The oxidized lose electrons (lower electronegativity) and the reduced gain electrons (higher electronegativity)

Question 21

What are the reducing agents and oxidizing agents?

Answer 21

The reducing agent is the element in the reaction that loses electrons (oxidized), the one that gains electrons (reduced) is the oxidizing agent.

Question 22

What is a half reaction?

Answer 22

A half reaction is a separated part of an overall reaction that represents how an element gains or loses electrons. There are at least two half-reactions in a redox reaction. Example:
In Mg + O2 = MgO, Mg –> Mg^+2 + 2e^- and O2 + 2e^- –> O2^2- are the two half reactions.

Question 23

How can you determine if a reaction is most likely a redox reaction by looking at pure elements and compounds?

Answer 23

If there is a pure element on one side and that pure element is part of a compound on the other side, it’s most likely a redox reaction.

Question 24

What types ofreactions are redox reactions?

Answer 24

Single replacement reactions, Combustion reactions, Combination/Synthesis reactions, Decomposition reactions

Question 25

What is the half-reaction method? Describe how to use it.

Answer 25

The half reaction method is a method used to balance redox reactions. On both sides of the equation, charges must be balanced. First, we separate the reaction into half-reactions. Then, we take the LCM (Least Common Multiple) between the electron on the half reactions, and multiply each element by the half reactions required coefficient.

Example:
Al + Ni^2+ –> Al^3+ + Ni
Al –> Al^3+ + 3e^-
Ni^2+ + 2e^- –> Ni
LCM 3 and 2 is 6
2(Al –> Al^3+ + 3e^-)
3(Ni^2+ + 2e^- –> Ni)
2Al –> 2Al^3+ + 6e^-
3Ni^2+ + 6e^- –> 3Ni
The electrons cancel (6e^- on the left side of the nickel equation and the right side of the aluminum equation).
2Al + 3Ni^2 –> 2Al^3+ + 3Ni

Question 26

How to you write a half reaction where a polyatomic ion becomes an ion under acidic conditions?

Answer 26

The rule with half-reactions under acidic conditions is that you can add H^+ ions and H2O molecules where necessary. Let’s do an example BrO3^- –> Br^-
First, we need to add oxygen to the right side
BrO3^- –> Br^- + 3H2O
Balance the hydrogen
BrO3^- + 6H^+ –> Br^- + 3H2O
The left side has a net charge of 5+ and the right has a net charge of 1-, so we need to add 6 electrons to the left side
BrO3^- + 6H^+ + 6e^- —> Br^- + 3H2O

Question 27

How to you write a half reaction where a polyatomic ion becomes an ion under basic conditions conditions?

Answer 27

The rule with half-reactions under acidic conditions is that you can add OH^- ions and H2O molecules where necessary. Let’s do an example Al –> Al(OH)4^-
First, add hydroxide ions to the left side
Al + 4(OH)^- –> Al(OH)4^-
The charge on the left is 4- and the charge on the right is 1- so you add 3 electrons to the right
Al + 4(OH)^- –> Al(OH)4^- + 3e^-