Slayt 1 -

Iron ores are minerals with a high iron content. One of them
hematite, Fe2O3 is chemically very similar to iron rust. The
reaction which iron metal is produced from hematite in a
blast furnace is described as:
Fe2O3(s) + 3CO(g)
2 Fe(l) + 3 CO2(g)
We can think of the CO(g) as taking O atoms away from
Fe2O3 to produce CO2(g) and the free element iron. A
commonly used term to describe a reaction in which a
substance loses O atoms, is « reduction» and gains O
atoms is «oxidation». An oxidation and reduction must
always occur together, and the reaction in which they do is
called an oxidation-reduction reaction.
Oxidation State (O.S) Changes
Fe2O3(s) + 3CO(g)
2 Fe(l) + 3 CO2(g)
The O.S of oxygen is -2 everywhere it appears. That of
iron(shwon in red) changes. It decreases from + 3 in
Fe2O3 to 0 in the free element, Fe. The O.S of carbon also
changes. It increases from +2 in CO to +4 in CO2 .
In terms of oxidation state changes,
in an oxidation process the O.S of some element
increases and
in reduction the O.S of an element decreases.
Identifying Oxidation- Reduction Reactions
Indicate whether each of the following is an oxidation-reduction
a) MnO2(k) + 4 H+(aq) + 2 Cl-(aq)  Mn2+(aq) + 2 H2O(s) + Cl2(g)
b) H2PO4 (aq) + OH (aq)  HPO42-(aq) + H2O(s)
a)The O.S of Mn in MnO2 decreases from +4 to +2 in Mn2+.
MnO2 is reduced to Mn2+. The O.S of O remains --2
throughout the reaction, and that of H at +1. The O.S of Cl
increases from -1 in Cl- to 0 in Cl2. Cl- is oxidized to Cl2 .The
reaction is an oxidation-reduction reaction.
b) The O.S of H is +1 on both sides of the equation. The O.S
of O remains -2 throughout the reaction. The O.S of
phosphorus is +5 in both H2PO4- and HPO42-. There are no
changes in O.S . This is not an oxidation-reduction reaction. It
is infact an acid-base reaction.
Oxidation and Reduction
Half reactions
The reaction illustrated below is an oxidation-reduction reaction
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
We can show this by evaluating changes in the O.S states. But we
may think of the reaction as involving two half-reactions occuring
at the same time. The overall or net reaction is the sum of these
two reactions:
Zn(k)  Zn2+(aq) + 2eCu2+(aq) + 2e-  Cu(k)
Zn(k) + Cu2+(aq)  Zn2+(aq) + Cu(k)
In the half-equation the O.S of Zn increases from 0 to +2,
corresponding to a loss of two electrons by each Zinc atom. In the
second half-equation the O.S of Copper decreases from +2 to 0
corresponding to the gain of 2e- by each Cu2+(aq) . To summarize;
Oxidation and Reduction Half
Oxidation, is a process in which the O.S of some
element increases, and in which electrons appear
on the right in a half-equation
Reduction, is a process in which the O.S of some
element decreases, and in which the electrons
appear on the left in a half equation.
Oxidation and reduction half-reactions must always
occur together, and the total number of electrons
associated with the oxidation must equal to the total
number associated with the reduction.
The same principles of equation balancing apply to oxidation-reduction (redox
equations are balance for numbers of atoms and balance for electric charge.Several
methods are possible but we emphasize the one described below:
The-Half reaction(Ion-Electron Method)
In this method of balancing a redox equation we:
 write and balance seperate half-equations for oxidation and reduction.
 adjust coefficients in the two-half equations so that the same number of electrons
appears in each half equation
 add together the two half-equations to obtain the balanced net equation.
Balancing the Equation for a Redox Reaction in
Acidic Solution
• Example: Write the balanced equation for the reaction:
SO32-(aq) + MnO4-(aq)  SO42-(aq) + Mn2+(aq)
• Step 1.Write the «skeleton» half-equations based on the species
undergoing oxidation and reduction. The O.S of sulfur increases from +4 in
SO32- to +6 in SO42-. The O.S. Of Mn decreases from +7 in MnO4- to +2 in
Mn2+. The skeleton half-equations are:
SO32-(aq)  SO42-(aq)
MnO4-(aq)  Mn2+(aq)
Step 2. Balance each half-equation atomically in this order
• Atoms other than H and O
• O atoms by adding H2O with the appropriate coefficient
• H atoms by adding H+ with the appropriate coefficient
• The other atoms( S and Mn) are already balanced in the skeleton halfequations. To balance O atoms we add one H2O molecule to the left of the
first half-equation and four to the right of the second.
Balancing the Equation for a Redox Reaction
in Acidic Solution
SO32-(aq) + H2O(l)  SO42-(aq)
MnO4-(aq)  Mn2+(aq) + 4 H2O(l)
To balance H atoms we add two H+ to the right of the first half-equation and
eight to the left of the second
SO32-(aq) + H2O(l)  SO42-(aq) + 2H+(aq)
MnO4-(aq) + 8 H+(aq)  Mn2+(aq) + 4 H2O(l)
Step 3. Balance each half-equation electrically. Add the number of electrons
necessary to get the same electric charge on both sides of each half
equation.The half-equation in which the electrons appear on the right side is
the oxidation half reaction. The other half equation with electrons on the left
is the reduction half-equation.
SO32-(aq) + H2O(l)  SO42-(aq) + 2H+(aq) + 2e(net charge on each side, -2)
MnO4-(aq) + 8 H+(aq) + 5e-  Mn2+(aq) + 4 H2O(l)
(net charge on each side, +2)
Balancing the Equation for a Redox Reaction in
Acidic Solution
Step 4. Obtain the net redox reaction by combining the half-equations. Multiply
through the oxidation half-equation by 5 and through the reduction half-equation
by 2. This results in 10 e- on each side of the net equation:
5 SO32-(aq) + 5 H2O(l) 5 SO42-(aq) + 10 H+(aq) + 10e2MnO4-(aq) + 16 H+(aq) + 10e-  2 Mn2+(aq) + 8 H2O(l)
5 SO32-(aq) + 2 MnO4-(aq) + 5 H2O(l) + 16 H+(aq) 
5 SO42-(aq) + 2 Mn2+(aq) + 8 H2O(l) + 10 H+(aq)
Step 5. Simplify. The net equation should not contain the same species on both
sides. Subtract 5 H2O from each side of the equation in step 4. This leaves 3
H2O on the right. Also subtract 10 H+ from each side, leaving 6 on the left.
5 SO32-(aq) + 2 MnO4-(aq) + 6 H+(aq)  5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(s)
Step 6. Verify. Check the final net equation to ensure that is balanced both
atomically and electrically.e.g the net charge on each side of the equation is:
(5x2-) +(2 x 1-) + (6 x 1+) = (5 x 2-) + (2 x 2+) = -6.
Balancing the Equation for a Redox
Reaction in Basic Solution
Balance the equation for the reaction in which chromate ion
oxidizes sulfide ion in basic solution to produce free sulfur
and chromium(III) hydroxide:
CrO42- (aq) + S2-(aq) + OHCr(OH)3(s) + S(s) +H2O
• Solution: Initially, we treat the half-reactions as if they
occur in acidic solution, and then we adjust them for a
basis solution:
• STEP 1. write the two skeleton half-equations and balance
them for Cr and S atoms
• CrO42- (aq)
• S2-(aq)
• STEP 2. Balance each half-equation for H and O atoms.
Note that the second half-equation has no H or O atoms
• CrO42- (aq) + 5 H+(aq)
Cr(OH)3(s) + H2O
• S2-(aq)
Balancing the Equation for a Redox
Reaction in Basic Solution
STEP 3. Balance the half-equations for electric charge by adding
the appropriate numbers of electrons
Reduction: CrO42- (aq) + 5 H+(aq) + 3 eCr(OH)3(s) + H2O
Oxidation: S2-(aq)
S(s) + 2 eSTEP 4. Change from an acidic to a basic medium by adding OHions and eliminating H+.
The oxidation half equation is unaffected because it has no H+
ions. Add 5 OH- ions to each side of the reductionhalf equation;
combine H+ and OH- into H2O; eliminate H2O from the right side
of the half-equation.
CrO42- (aq) + 5 H+(aq) + 5 OH- (aq) + 3 eCr(OH)3(s) + H2O
+ 5 OH- (aq)
CrO42- (aq) + 5 H2O + 3 eCr(OH)3(s) + H2O + 5 OH- (aq)
CrO42- (aq) + 4 H2O + 3 eCr(OH)3(s) + 5 OH- (aq)
Balancing the Equation for a Redox
Reaction in Basic Solution
STEP 5. Combine the half-equations to obtain the net
redox equation. (multiply the reduction half-equation by 2
and the oxidation half reaction by 3)
2 CrO42- (aq) + 8 H2O + 6 eOH- (aq)
3 S2-(aq)
3 S(s) + 6 e2 CrO42- (aq) + 8 H2O + 3 S2-(aq)
OH- (aq) + 3 S(s)
2 Cr(OH)3(s) + 10
2 Cr(OH)3(s) + 10
In a redox reaction, the substance that makes it possible for
some other substances to be oxidized is called oxidizing agent
or oxidant. In doing so, the oxidizing agent is itself reduced.
Similarly, the substance that causes some other substances to
be reduced is called the reducing agent or reductant. In the
reaction the reducing agent itself is oxidized.
• An oxidizing agent(oxidant):
• contains an element whose O.S. decreases in a redox
• «gains» electrons(electrons are found on the left side of the
• A reducing agent(reductant):
• contains an element whose O.S. increases in a redox
• «loses» electrons(electrons are found on the right side of
the half-equation
Example: Hydrogen peroxide, H2O2 is a versatile chemical. Its
uses include bleaching wood pulp and fabrics and substituting
for chlorine in water purification. One reason for its versatility is
that it can be either an oxidizing or reducing agent. For the
following reactions, identify whether hydrogenperoxide is an
oxidizing or reducing agent.
a) H2O2 + 2 Fe2+(aq) + 2 H+(aq)
2 H2O + 2 Fe3+
b) 5 H2O2(aq) + 2 MnO4- (aq) + 6 H+
8 H2O +2
Mn2+(aq) + 5 O2(g)
Solution: a) Fe2+ is oxidized to Fe3+ and H2O2 makes it
possible; H2O2 is an oxidizing agent.
b) MnO4- is reduced to Mn2+ and H2O2 makes this possible; H2O2
is a reducing agent
Two kinds of interactions are possible between metal atoms on the electrode and
metal ions in solution
1. A metal ion Mn+ may collide with the electrode, gain n electrons and be
converted to a metal atom M, the ion is reduced
2. A metal atom M on the electrode may lose n electrons and enter the solution as
the ion Mn+. The metal atom is oxidized.
Electrode potential is a property proportional to the density of negative electric
charge. To measure a difference in potential, we need to connect two half-cells
in a special way.
The flow of electric current is in the form of migration of ions from the greater
negative charge to the other electrode.(here from Cu to Ag)
The net reaction that occurs as electric current flows through
the electrochemical cell is:
• The reading on the voltmeter is (0,463 V) significant. It is the potential
difference between two half cells. Because this potential difference is the «
driving force» for electrons, it is often called the electromotive force(emf) of
the cell or the cell potential .
• Why does copper not displace Zn2+ from solution?
• If we set up an electrochemical cell Zn(s) / Zn2+ half cell and Cu2+/Cu(s) half
cell, we find that electrons flow from the Zn to the Cu. The spontaneous net
reaction in the electrochemical cell is
• Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
• This is a spontaneous reaction and the reverse of this reaction does not occur
• A cell diagram shows the components of an
electrochemical cell in a symbolic way. We will use the
following conventions in writing cell diagrams:
• The anode, the electrode at which the oxidation occurs,
is placed at the left side of the diagram
• The cathode, the electrode at which reduction occurs,
is placed at the right side of the diagram.
• A boundary between different phases(e.g, an
electrode and a solution) is represented by a single
vertical line(/)
• The boundary between half-cell components, usually
a salt bridge, is represented by a double vertical line
• The electrochemical cells which produce electricity as a
result of chemical reactions are called voltaic or
galvanic cells
Representing a Redox reaction Through a Cell Diagram-Aluminium
metal displaces zinc(II) ion from aqueous solution
a) Write oxidation and reduction half equations and a net equation
for this redox reaction
b) Write a cell diagram for a voltaic cell in which this reaction
Solution: a) The term « displaces» means the Aluminium goes into
solution as Al3+ and Zn2+ comes out of the solution as zinc metal.
Al(s) is oxidized to Al3+ in the anode half-cell(left) and Zn2+ (aq)
is reduced to Zn(s) in the cathode half-cell(right)
To do a calculation of cell voltages, we choose a particular half-cell to
which we assign a potential of zero. We then compare other half cells
to this reference. The commonly accepted reference is the standard
hydrogen electrode
• Standard electrode potential, E˚, measures the tendency for a reduction
process to occur at an electrode.
• To determine the value of E˚ for an electrode, we compare it with a
standard hydrogen electrode(SHE). In the voltaic cell indicated below the
measured potential difference is 0,337 V, with electrons flowing from H2 to
the Cu electrode.
A standard cell potential, Ecell˚, is the potential difference
or voltage of a cell formed from two standard electrodes.
The net reaction that occurs in the voltaic cell is
The cell reaction indicates that Cu2+ (1M) is more easily
reduced than is H+(1M). The standard electrode potential
representing the reduction of Cu2+ (aq) to Cu(s) is
assigned a positive value.
When a standard hydrogen electrode is combined with a
standard zinc electrode, electrons flow in the opposite direction,
that is from the zinc to the hydrogen electrode.
In summary, the potential of the standard hydrogen electrode
is set at 0.
Any electrode at which a reduction half-reaction shows a
greater tendency to occur than does the reduction of H+(1M)
to H2(g) has a positive value for its reduction potential,E˚.
Any electrode at which a reduction half-reaction shows a
lesser tendency to occur than does the reaction of H+(1M) to
H2(g) has a negative value for its standard reduction
potantial, E˚.
The value of E˚ does not depend on the amount of
substances involved. E˚ values are unaffected by multiplying
half-equations by constant coefficients.
AT 25˚C
A new battery system currently under study for possible use in electric
vehicles is the zinc-chlorine battery. The net reaction producing electricity in
this cell is
What is the Ecell˚ of this voltaic cell?
Cadmium is found in small quantities wherever zinc is found. Unlike zinc, which
in trace amounts is an essential element, cadmium is an environmental poison.
To determine cadmium ion concentrations by electrical measurements the
standard reduction potential for the Cd2+/Cd electrode is needed. Th voltage of
the following voltaic cell is measured:
Cd(s) / Cd2+(1 M) // Cu2+(1 M) / Cu(s)
What is the standard reduction potential for the Cd2+/Cd electrode?
Determining a Free Energy Change from A Cell Potential. Determine ΔG˚
for the reaction
n mol e- = electric current(ampere) x
time(s) / 96485 C
For the reaction Cu2+(aq) + 2 eCu(s)
The mass of Cu accumulated on the
electrode= n/2 x MA(atomic mass of Cu)
Our main criterion for spontaneous change is that ΔG<0 .
However, according to the equation if ΔG< 0, then ΔG< 0
If Ecell>0 , a reaction occurs spontaneously in the forward
If Ecell<0, the reaction occurs spontaneously in the reverse
If Ecell=0, a reaction is at equilibrium.
If a cell reaction is reversed, Ecell changes sign
Making Qualitative Predictions with
Electrode Potential Data
Because the E˚value for the reduction of S2O82-(aq)
is larger than that for the reduction of Cr2O72-(aq) ,
S2O82-(aq) should be the better oxidizing agent.
R=8,314 Jmol-1K-1
at 25˚C
Example: What is the value of the equilibrium constant Keq for the reaction
between copper metal and iron(III) ions in aqueous solution at 25˚C?
Example : What is the value of Ecell for the voltaic cell pictured below?
Ecell= 0,011 V
Example: Will the cell reaction proceed spontaneously as written for the
following cell?
Any voltaic cell in which the net cell reaction involves only
a change in the concentration of some species( here H+)
is called a concentration cell.
A concentration cell consists of two half-cells with
identical electrodes but different ion concentrations.
Because the electrodes are identical, the standard halfcell potentials are numerically equal and opposite in sign.
This makes Ecell=0.
However, because the ion concentrations differ, there is a
potential difference between the two half-cells.
Spontaneous change in a concentration cell always
occurs in the direction that produces a more dilute
Constructing and using a hydrogen electrode is difficult. A better approach is
to replace the SHE by a different reference electrode and the other hydrogen
electrode by a glass electrode. A glass electrode is a thin glass membrane
enclosing HCl(aq) and a silver wire coated with AgCl(s)
The potential of the glass electrode depends on the hydrogen ion
concentration of the solution being tested. The difference in potential between
the glass electrode and the reference electrode is converted to a pH reading
by the meter
A device that stores chemical energy for later release
as electricity is called a battery
Primary batteries: The cell reaction is not reversible.
When the reactants have been mostly converted to
products, no more electricity is produced and the
battery is dead.
Secondary batteries: The cell reaction can be
reversed by passing electricity through the
battery(charging). This means that a battery can be
used several times by charging.
Flow batteries: Materials(Reactants, products,
electrolytes) pass through the battery, which is the
conversion of chemical energy into electrical energy
In this cell oxidation occurs at a zinc anode and
reduction at an inert carbon cathode. The electrolyte
is a moist paste of MnO2, ZnCl2, NH4Cl and carbon
Oxidation: Zn(s)
Zn2+(aq) +2 eReduction: 2 MnO2(s) +H2O + 2eMn2O3(s) +
2 OH-(aq)
An acid-base reaction occurs between NH4+(aq) + OH-(aq)
NH3(g) + H2O(l).
The Leclanche cell is a primary cell. It is cheap to make, but it
has some drawbacks. When current is rapidly drawn from the
cell, products build up on the electrodes(e.g NH3) and this
causes the voltage to drop. Also because the electrolyte medium
is acidic, zinc metal slowly dissolves. In order to overcome this
problem, an alkaline form of this battery can be produced . The
advantage of the alkaline battery are that zinc does not dissolve
in a base and the battery does a better job in maintaining the
The most common secondary battery is the automobile storage
battery (see below) . The electrodes are lead-antimony alloy. The
anodes are impregnated with lead metal and the cathodes with redbrown leaddioxide. The electrolyte is dilute sulfuric acid. When the
cell is allowed to discharge, following reactions occur:
Oxidation: Pb(s) + SO42-(aq)
PbSO4(s) + 2 eReduction: PbO2(s) + 4 H+(aq) + SO42-(aq)+ 2 ePbSO4(s) + 2
Net: Pb(s) + PbO2(s)+ 4 H+(aq)+SO42-(aq)
2 PbSO4(s)+2 H2O
To recharge it,electrons are forced in
the opposite direction by connecting
the battery to an external electric
source. The reverse of reactions
occurs when the battery is recharged
The cell-diagram of a silver-zinc cell is
Zn(s)/,ZnO(s)/ KOH(satd)/Ag2O(s),Ag(s)
Oxidation: Zn(s) + 2 OH-(aq)
ZnO(s) +H2O+ 2 eReduction: Ag2O(s)+H2O+ 2 e2 Ag(s)+ 2 OH-(aq)
Net: Zn(s) + Ag2O(s)
ZnO(s) + 2 Ag(s)
Because no solution species is involved in the net reaction, the quantity of the
electrode is very small, and the electrodes can be maintained very close
The storage capacity of a silverzinc cell is about six times as great
as a lead-acid cell of the same
These batteries are used in
watches, electronic calculators,
hearing aids and cameras
The essential process in a
fuel cell is fuel+oxygen
oxidation products.
Applied to methan, a fuel cell
can be obtained as the
picture to the next.
One of the simplest and most
successful fuel cells involves
the reaction between H2(g)
and O2(g) to produce water.
Oxidation: 2 H2(g) + 4 OH(aq)
4 H2O + 4 eReduction: O2(g)+ 2 H2O+4
e4 OH-(aq)
Net: 2 H2(g) + O2(g)
H O (l)
As long as fuel and O2 are available, the cell will produce electricity. It does
not have the limited capacity of a primary battery, but neither does have the
storage capacity of a secondary battery. Fuel cells have had their most
notable successes as energy sources in space vehicles.
A fuel cell reaction is often rated in terms of the efficiency value,
ε= ΔG/ΔH. For the methane fuel cell ε= - 818/-890 = 0,92
Construction of a
Exchange Membrane
Fuel Cell
Another kind of flow battery, because it uses oxygen from air,
is known as air battery. The substance that is oxidized is
typically a metal.
One heavily studied battery system is the aluminium-air
battery. In this battery oxidation occurs at an aluminium anode
and reduction at a carbon-air cathode. The electrolyte
circulated through the battery is NaOH(aq).The half reactions
and the net reaction are:
Oxidation: 4 { Al(s) + 4 OH-(aq)
[Al(OH)4]-(aq) + 3 e-}
Reduction: 3{O2(g) + 2 H2O + 4 e4 OH-(aq)
4 Al(s) + 3 O2(g)+ 6 H2O + 4 OH-(aq)
Al-air batteries have one of the highest energy densities of the
all bateries, but they are not widely used because of previous
problems with cost,shelf life, start up time, by-product removal
which have restricted their use to mainly military applications
Corrosion is the wearing away of metals due to
a chemical reaction.
If an iron nail is embedded in a gel of agar(acidbase indicator Phenolphtalein and the
potassiumferricyanide) in water, it is observed
that at the head and tip of the nail a deep blue
precipitate forms. This shows us that oxidation
Oxidation : 2 Fe(s)
2 Fe2+(aq) +4 eReduction: O2+ 2 H2O+4 e4 OH- (aq)
Some metals such as Aluminium form corrosion
products that adhere tightly to the underlying
metal and protect it from further corrosion. Iron
oxide flakes off and constantly exposes fresh
surface. The difference in corrosion behaviour
explains why cans made of iron deteriorate
rapidly in the environment, whereas aluminium
cans have an almost unlimited lifetime. The
simplest method of protecting a metal from
corrosion is to cover it with a paint.
Rust, the most familiar
example of corrosion
Another method of protecting an iron surface is to plate it with a thin layer of a
second metal. iron can be plated with copper by electroplating or with tin by
dipping the iron into the molten metal. In either case the underlying metal is
protected only as long as the coating medium remains intact.
When iron is coated with zinc(galvanized iron) the situation is different. Zinc is
more active than iron . If a break occurs in the zink plating, the iron is still
protected.Zinc is oxidized instead of iron.(see below)
a sacrifical anode
protection in the hull of a
Cathodic protection is a technique is to
control the corrosion of a metal surface by
making the surface the cathode of an
electrochemical cell. This is a useful
protection method used with large iron
and steel objects in contact with
water,moist soils-ships,storage tanks,
pipelines, etc.
This involves connecting a chunk of
magnesium, aluminium, zinc or some
active metal to the object either directly
through a wire. Oxidation occurs at the
active metal. The iron behaves like a
cathode and supports a reduction halfreaction. As long as some active metal
remains, the iron is protected. The tive
metal is called sacrificial anode.
When the cell functions spontaneously, electrons flow from the zinc to the
copper and the net chemical change in the voltaic cell is
Zn(s) + Cu2+
Zn2+(aq) + Cu(s) Ecell= + 1,100 V
Now suppose we connect the same cell to an external energy source of
voltage greater than 1,100 V. That is the connection is made so that the
electrons are forced into the zinc electrode(cathode) and removed from the
copper electode(the anode)
Oxidation Cu(s)
Reduction Zn2+(aq) + 2 eNet
Cu(s) + Zn2+(aq)
- ECu2+/Cu˚= - 0,337 V
EZn2+/Zn = - 0,763 V
Cu2+(aq)+ Zn(s) Ecell= - 1,100 V
We conclude that when the cell reaction in a voltaic cell is reversed by
reversing the direction of the electron flow , the voltaic cell is changed to an
electrolytic cell.
Generally, we do not measure electric charge directly, we measure electric
current. One ampere(A) of electric current represents the passage of 1
coulomb of charge per second (s).
1 mol e-= 96485 C
number of mol e- = current(C/s) x time(s) x 1 mol e96485 C
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