Oxidation and Reduction

Report
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


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)

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
Agent
Electron exchange
Half-reaction
The atom is
Reducing Agent
Loses electrons
Oxidation
Oxidized
Oxidizing Agent
Gains electrons
Reduction
Reduced
Oxidation half-reaction
Zn(s) -> Zn+2 +
2eCu2+ +
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

If there is a difference in the number of
electrons in the equations, you must use
stoichiometrics!

Fe(s)
->
Au3+(aq) + 3e- ->

3 Fe(s) + 2 Au3+(aq) -> 3 Fe2+(aq) + 2 Au(s)

Fe2+ (aq)
Au(s)
+
2e-
x3
x2

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 oxidationreduction?

c) Find the oxidizing agent and the reducing agent

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)

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)




c) The oxidizing agent is the H+
gains electrons
The reducing agent is the Mg
loses electons
(s)
(aq)
because it
because it
Remember LEO the lion says GER

Do # 1, 2, 4, 5 on the worksheet

We will go over it next class

Have Fun!
Oxidation 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



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


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




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

Find the oxidation # for the following atoms
Atom
Charge
Oxidation #
Na
1+
+1
Sr
2+
+2
Ra
2+
+2
K
1+
+1
Li
1+
+1
Mg
2+
+2
Rb
1+
+1
Substances
Charge
Oxidation #
Elements in ground state (Li, Mg, Al, Fe, etc.)
0
0
Molecules of elements (H2, O2, Cl2, N2, S8, etc.)
0
0
Ions of alkali metals (Li+, Na+, K+, etc.)
1+
+1
Ions of alkaline earth metals (Ca2+, Mg2+, Be2+,
etc.)
2+
+2


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


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




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...

What are the oxidation numbers for the
atoms in an ammonia (NH3) molecule?


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


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


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


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!

Mn
+
4O
=
-1

X
+
4 (-2)
=
-1

X
+
-8
=
-1

X
=
7

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)


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



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
Reaction
Stronger Reducing Agent
Weaker Reducing Agent
Spontaneous
Solid
Aqueous solution
None
Aqueous solution
Solid




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?


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

An electrochemical cell is a device that can
spontaneously generate an electrical current
◦ i.e. A battery



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




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

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




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


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


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

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...


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



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

To find the cell potential, we must reverse the
reduction. In this case, the Mg reaction will be
the reduction

(Ag+ + e- -> Ag

Mg (s) -> Mg 2+ + 2 e- Reduction EO=2.37V
(s)
) x2
Oxidation EO=0.80V

Cell potential is calculate using the following
equation

EO
cell
= EO

EO
cell
= 0.80V + 2.37V

EO
cell
= 3.17V
oxidation
+ EO
reduction



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




A fuel cell can be represented in this
simplified way
Oxidation
Mg|Mg2+
||
Reduction
Ag+|Ag
According to convention, the oxidation
reaction is shown on the left

similar documents