Oxidation and Reduction

Redox

Redox involves two simultaneous reactions◦ An oxidation and a reduction

Oxidation involves a loss of electrons

Reduction involves a gain of electrons

LEO the lion says GER

What exactly is a redox reaction?

Seeing how in each Redox reaction, there is said to be an oxidation half-reaction and a reduction half-reaction

One molecule will lose electrons (the oxidation) and the electrons will join another atom (reduction)

Redox

Oxidation half reaction

◦ Fe(s) -> Fe2+(aq) + 2e-

Iron is being oxidized to form the ferrous ion

Reduction half-reaction

◦ Au3+(aq) + 3e- -> Au(s)

The gold ion has been reduced to its ground state

Example

Agent Electron exchange

Half-reaction The atom is

Reducing Agent Loses electrons Oxidation Oxidized

Oxidizing Agent Gains electrons Reduction Reduced

To help simplify

Oxidation half-reactionZn(s) -> Zn+2 + 2e-

Cu2+ + 2e- -> Cu(s)

______________________________________________Cu2+ + Zn(s) -> Zn2+ + Cu(s)

By eliminating the common terms on either side, you are left with a simplified final equation

General equation for Redox

If there is a difference in the number of electrons in the equations, you must use stoichiometrics!

Fe(s) -> Fe2+ (aq) + 2e- x3

Au3+(aq) + 3e- -> Au(s)

x2

3 Fe(s) + 2 Au3+(aq) -> 3 Fe2+

(aq) + 2 Au(s)

General Equation continued

The reaction of a piece of magnesium (Mg) in hydrochloric acid (HCl) results in the formation of magnesium dichloride (MgCl2). The release of hydrogen (H2) can also be observed. During this reaction, metallic magnesium is oxidized into aqueous Mg2+ ions, while aqueous H+ ions of the acidic solution is reduced to hydrogen gas.

a) What are the half-reactions in this reaction?

b) What is the general equation for oxidation-reduction?

c) Find the oxidizing agent and the reducing agent

Example

a) Magnesium oxidized◦ Loses electrons

Oxidation reaction◦ Mg (s) -> Mg2+ (aq) + 2e-

Hydrogen reduced◦ Gains electrons

Reduction reaction◦ H+ (aq) + 1e- -> 1/2 H2 (g)

Solution

b) Add the two half reactions together

Mg (s) -> Mg2+ (aq) + 2e-

2x(H+ (aq) + 1e- -> ½ H2 (g)) 2H+ (aq) +2e- -> H2 (g)

______________________________________________ Mg (s) + 2H+ (aq) -> Mg2+ (aq) + H2 (g)

Solution (cont)

c) The oxidizing agent is the H+ (aq) because it gains electrons

The reducing agent is the Mg (s) because it loses electons

Remember LEO the lion says GER

Solution

Do # 1, 2, 4, 5 on the worksheet

We will go over it next class

Have Fun!

Problems

Oxidation NumberOxidation State

The oxidation number, also called the oxidation state, indicates the number of electrons an element has lost or gained, in relation to its ground state, during a redox reaction.

All elements in their ground state have an oxidation number of 0

They are considered to be atoms, due to them not having lost any electrons

Oxidation Number

When atoms are involved in redox reactions, their oxidation numbers vary

Oxidation numbers increase with an oxidation due to a loss in electrons

Oxidation numbers decrease with a reduction due to a gain in electrons

Oxidation number

To determine the oxidation number of an atom, we must determine whether it is part of an ionic or a covalent compound

Ionic compound◦ A bond between a metal and a non-metal which

share electrons

The oxidation number of each ion is equal to its charge

Oxidation Number

Example Calcium chloride (CaCl2) is composed of one

Ca2+ ion and two Cl- ions

To distinguish between an ion’s charge and its oxidation number, the convention is different

A charge is written as 2+ while an oxidation number is written as +2

Ionic compounds

Find the oxidation # for the following atoms

Exercises

Atom Charge Oxidation #Na 1+ +1Sr 2+ +2Ra 2+ +2K 1+ +1Li 1+ +1

Mg 2+ +2Rb 1+ +1

Substances Charge Oxidation #

Elements in ground state (Li, Mg, Al, Fe, etc.) 0 0Molecules of elements (H2, O2, Cl2, N2, S8, etc.) 0 0Ions of alkali metals (Li+, Na+, K+, etc.) 1+ +1Ions of alkaline earth metals (Ca2+, Mg2+, Be2+, etc.)

2+ +2

Oxidation Numbers for some Substances

When an atom is part of a molecule or polyatomic ion, convention determines its oxidation number by assigning each pair to the more electronegative atom in the bond, that is, the atom that is more likely to attract electrons to fill its outermost shell.

To determine the oxidation number of a molecule, the molecule can be represented with Lewis notation

Oxidation Number of Covalent Compounds

If you take a water molecule, you can see that the electrons are shared between the oxygen and the hydrogen

Since oxygen normally has 6 electrons◦ It gets 2 additional ones from hydrogen◦ Oxidation number of -2

Hydrogen atom has lost its electron◦ Oxidation number of +1

So what this means... In English...

In covalent molecules, the charge is not always the same as the oxidation number

There is no O2- or H+ in water

In covalent compounds, the electrons are not given to the other atom, but rather are shared between the two

Let’s try this again...

Oxidation Number of Covalent Compounds

What are the oxidation numbers for the atoms in an ammonia (NH3) molecule?

Calculate the Oxidation Number

1- Since there are no metallic atoms, we know that ammonia is a covalent compound

2- Using the periodic table, we can determine which molecule is more electronegative. In this case, it is Nitrogen. The electrons in the bond are assigned to it. Nitrogen needs 3 electrons to reach its ground state. The oxidation number of Nitrogen is -3

How you do it...

3- As for Hydrogen, each hydrogen ion has lost its one electron, which gives it an oxidation number of +1

Ans: The oxidation number of nitrogen is -3 and hydrogen is +1

Continued

The same element can have a different oxidation number depending in which molecule it is found in.

There are a few non-metals that almost always have the same oxidation number◦ Oxygen is always -2, except in peroxide (-1)

Hydrogen is also almost always the same, +1

Overview

What is the oxidation number for the manganese atom in the permanganate ion (MnO4

-), a very strong oxidizing agent? The permanganate ion’s negative charge indicates that the net charge is 1-. Furthermore, since the oxidation number of oxygen is -2 and there are 4 oxygen atoms in this molecule. How would you be able to figure out the oxidation number for the manganese?

Hint: math!

Example #2

Mn + 4 O = -1

X + 4 (-2) = -1

X + -8 = -1

X = 7

Solution

What are the oxidation numbers of all of the atoms in the following redox reaction

Fe2O3 (s) + 2 Al (s) -> Al2O3 (s) + 2 Fe (l)

Example #3

Seeing how Fe (s) and Al (s) are both solid, their oxidation numbers are 0

For Fe2O3 and Al2O3, oxygen has an oxidation number of -2◦ There are three of them, for a total of -6

The iron and aluminum need an oxidation number of +3 to balance everything out

Solution

Metals can be classified by their reduction power

Spontaneous reactions

For spontaneous reactions to occur, the stronger reducing agent must be in the solid state and the weaker one must be in its ionic form

Reducing Power of Metals

Reaction Stronger Reducing Agent

Weaker Reducing Agent

Spontaneous

Solid Aqueous solution

None Aqueous solution Solid

TABLE!

You have Aluminum (Al) and Copper (Cu)

Which would have to be solid and which would have to be aqueous?

Same situation, but with Gold (Au) and Copper (Cu)

Which would have to be solid and which would have to be aqueous?

Practice

For the Aluminum (Al) and Copper (Cu) reaction, the Al would need to be solid and the Cu in solution

For the Gold (Au) and Copper (Cu) reaction, the Cu would have to be solid and the Au in solution

Answers

An electrochemical cell is a device that can spontaneously generate an electrical current◦ i.e. A battery

Electrochemical Cell

The electrochemical cell is composed of two electrodes, also called half-cells.

They are called half-cells due to the half reactions taking place in the cell

The two electrodes are joined by a wire and are connected by a salt bridge◦ A tube of salt water plugged with a porous

membrane

Electrochemical Cells

The classic example of an electrochemical cell is a piece of zinc immersed in a solution of zinc sulfate (ZnSO4) and a piece of copper in a solution of copper sulfate (CuSO4)

Since zinc is a better reducing agent than copper, it gives up its electrons to the copper

Positive terminal is called the anode and the negative is the cathode

Zinc is oxidized and the copper is reduced

Example

The concentration of each half-cell’s ions varies◦ The Zn2+ ions increase◦ The Cu2+ ions decrease

The ions in the salt bridge move towards their respective poles to maintain the number of positive and negative ions◦ Positive towards the Cathode◦ Negative towards the Anode

As the Electrochemical Cell works

Electrochemical Cells

During a redox reaction, one of the metals has greater potential energy than the other because it has a greater reducing power.

Without this potential difference between two metals of different types, no chemical reaction would take place

To calculate the potential, we need a reference electrode

In this case, we use the hydrogen electrode

Reduction and Oxidation Potential

2 H+ + 2 e- H2

This reaction is considered to have a redox potential of 0.00 V

EO = 0.00V

Where EO is redox potential

Hydrogen electrode

To measure the EO, calculate the potential difference, create an electrochemical cell using the Hydrogen and the electrode which you want to test

The voltage which is read on the voltmeter will be the redox potential of the unknown

How we measure EO

Redox Potential

Redox Potential Table

With this table, we can compare the oxidation or reducing power of substances

And, we can calculate the cell potential

That’s about it for you guys...

What we can do with this...

The cell potential is equal to the sum of the oxidation potential and the reduction potential of the cell

Calculating the potential of an electrochemical cell made up of two different electrodes can be done without having to use a reference electrode

Cell potential

First, determine the stronger reducing agent

If we use Silver (Ag) and Magnesium (Mg) as an example, the magnesium is stronger since it is found below silver in the table of redox potentials

This makes it the electrode which will lose electrons and undergo oxidation

How to figure it out...

To find the cell potential, we must reverse the reduction. In this case, the Mg reaction will be the reduction

(Ag+ + e- -> Ag (s) ) x2 OxidationEO=0.80V

Mg (s) -> Mg 2+ + 2 e- ReductionEO=2.37V

Continued

Cell potential is calculate using the following equation

EO cell = EO oxidation + EO reduction

EO cell = 0.80V + 2.37V

EO cell = 3.17V

Cell Potential

The electrochemical cell can also be called a fuel cell

If the EO is above 0, then the reaction is said to occur spontaneously.

If the EO is below 0, then the fuel cell is said to not be spontaneous

What the results mean

A fuel cell can be represented in this simplified way

Oxidation Reduction Mg|Mg2+ || Ag+|Ag

According to convention, the oxidation reaction is shown on the left

Fuel Cell