Chemistry 12
UNIT VI – ELECTROCHEMISTRY6.3
Put a strip of copper into a concentrated solution of nitric acid and it will quickly begin to
bubble, turning the solution green and eating away the metal. Put an identical strip of copper
into a concentrated solution of hydrochloric acid, and nothing happens.
The reason is based on ELECTROCHEMISTRY
ELECTROCHEMISTRY
ELECTROCHEMICAL
REACTIONS
Branch of chemistry concerned with the conversion of
chemical energy to electrical energy (and vice versa)
Reactions that involve the transfer (loss and gain) of electrons
between substances.
Often setup as a cell – a system that requires or produces
electrical energy
G1. OXIDATION AND REDUCTION PROCESSES
Consider the following reaction:
2 Al(s) + 3 CuCl2(aq)  2AlCl3(aq) + 3 Cu(s)
ionic: 2 Al(s) + 3 Cu2+(aq) + 2Cl-(aq)  2Al3+(aq) + 2Cl-(aq) + 3 Cu(s)
net: 2 Al(s) + 3 Cu2+(aq)  2Al3+(aq) + 3 Cu(s)
This reaction can be re-written as two separate HALF-REACTIONS.
OXIDATION
A half-reaction in which a species LOSES ELECTRONS
(e- are a product)
OXIDATION HALF-REACTION
Al(s) Al3+(aq) + 3e-
REDUCTION
A half-reaction in which a species GAINS ELECTRONS
(e- are a reactant)
REDUCTION HALF-REACTION
Cu2+(aq) + 2e-  Cu(s)
LEO the lion says GER
LEO - Loss of Electrons is Oxidation
GER
- Gain of Electrons is Reduction
2
REDUCING AGENT
Causes a substance to be OXIDIZED:
It LOSES electrons (becoming more positive/less negative) and donates those
electrons to another species, so that other speicies becomes reduced. An oxidation
causes a reduction.
Ex.
Zn  Zn2+ +2eU3+  U4+ + e2Cl-  Cl2 + 2e-
OXIDIZING AGENT
When a substance becomes REDUCED:
It GAINS electrons (becoming more negative/less positive) and accepts electrons from
another species which is being oxidized. A reduction causes oxidation. A substance
being reduced is referred to as the OXIDIZING AGENT
Ex.
Cu2+ +2e- Cu
V3+ + e-  V2+
F2 + 2e- 2F-
For every reduction, there must be an accompanying oxidation. A substance
can’t accept electrons if another substance won’t give them off in the first place.
REDUCTION-OXIDATION REACTION
Consist of 2 half-reactions. (reduction and one oxidation)
Often called REDOX
Ex. Consider the following reaction:
2 Al + 3 Cu2+  2 Al3+ + 3 Cu
Al:
Al  Al3+ + 3e-
Cu2+: Cu2+ + 2e-  Cu
Al is being oxidized, causing Cu2+ to be reduced.
Al is the REDUCTION AGENT
Cu2+ is being reduced, causing Al to be oxidized.
Cu2+ is the OXIDIZING AGENT
3
Ex. Are the following atoms being oxidized or reduced (and would that make them the
oxidizing agent or the reducing agent?)
a. I-  I3+ + 4eb. Au3+ + 3e-  Au
c.
Cu+  Cu2+ + e-
d.
F2 + 2e-  2 F-
Ex. For the following reactions indicate which substance is:
a. oxidized, b reduced, c. the oxidizing agent, d. the reducing agent.
Zn2+ + Mg  Zn + Mg2+
2 Na + Br2  2 Na+ + 2 Br-
Cu(NO3)2 + 2Ag  2AgNO3 + Cu
OXIDATION NUMBERS
OXIDATION
NUMBER
The charge that an atom would possess if the species containing
the atom were made up of ions. (Recall combining capacity)
ANY TIME THAT YOU SEE THAT AN ATOM OR ION HAS CHANGED ITS CHARGE
DURING A REACTION, YOU ARE DEALING WITH A REDOX REACTION.
You will need to be able to calculate the oxidation numbers of various atoms.
4
RULES FOR DETERMINING OXIDATION NUMBER
1.
The sum of the positive charges and the negative charges must equal the overall
charge on the species.
(In elemental form = charge is zero/neutral)
2.
Oxidation number is related to an atoms position on the periodic table (has to do with the
number of e- available for bonding)
i. Group 1 ions (Li, Na, K… Alkali metals) ALWAYS have an oxidation of 1+
when combined with other elements.
ii. Group 2 ions (Ca, Ba, Mg… Alkaline Earth Metals) ALWAYS have an oxidation
number of 2+ when bonded in a mlc.
iii. Group 17 (Cl, Br, F… Halogens) TEND to have an oxidation number of 1when bonded in a compound. (Many exceptions)
3.
The oxidation number of oxygen is USUALLY 2-.
(The exception is in peroxides (H2O2) where the oxidation number is 1-.)
4.
The oxidation number of hydrogen is USUALLY 1+.
(Exception is metal hydrides (NaH, BaH2) where the oxidation number is 1-)
5.
All other oxidation numbers are variable. Each particular element has certain
tendencies and possibilities according to their position on the periodic table, but
should NEVER be assumed.
Ex. Determine the oxidation number of each atom for the following:
a. Na
c. N2O4
b. CCl4
d. PO43-
Try: Determine the oxidation number of each atom for the\ following:
a. P4
c. H4P2O7
b. PbSO4
d. NH4+
5
An atoms change in oxidation number indicates if it is being reduced or oxidized.
When it becomes more positive, it is being oxidized
When it becomes more negative, it is being reduced
ClO3-  ClO4-
(Cl5+  Cl7+ =
oxidation) H2O2  H2O
Cr3+  CrO42NO2  N2O3
G2. STRENGTHS OF REDUCING AND OXIDIZING AGENTS
There is a table called the Table of Standard Reduction Potentials, which is partially shown
below.
F2 + 2 e- ↔ 2 F
+
+ e- ↔ Ag
2+
+ 2 e- ↔ Cu
2+
+ 2 e- ↔ Zn
Ag
Cu
Zn
+
Li
-
-
+ e- ↔ Li
The table is written as reduction reactions (forward reaction)
Oxidizing Agents are found in the left hand column
Reducing Agents are found in the right hand column
Chart is organized in order of strength (reactions most likely to occur)
The following observations can be made from the Table of Standard Reduction
Potentials. (These will help you locate a half-reaction much more quickly)
1.
In general, metals are found in the bottom half of the table on the right
2.
In general, the halogens and oxyanions (anions containing oxygen) are found in the
top half of the table on the right.
6
3.
Some metals (Fe, Sn, Cr, Hg, Cu) have more than one common oxidation number.
In these cases there will be MORE THAN ONE POSSIBLE HALF REACTION
shown on the table involving these metals. A given ion may be found on either side
of the table.
Ex.
4.
Cu+ + e- ↔ Cu(s)
Cu2+ + 2e- ↔ Cu(s)
Cu2+ + e- ↔ Cu+
Copper
Water (and others) are found on both sides
YOU MUST PAY ATTENTION TO WHAT IS IN A SYSTEM (CONTAINER)
TO DETERMINE THE REACTION
Ex. The half reaction for Zn/Zn2+ is:
Zn2++2e- ↔ Zn
If there was a piece of Zn(s) in a solution of Zn2+(aq) you can write either:
Or
Zn2+ + 2e-  Zn (written as a reduction)
Zn  Zn2+ + 2e- (written as an
oxidation)
If only one species is given (i.e. the container has a solution of Zn2+) then only one of the
reactions can occur (the reduction in the above example)
Note – When referring to an ISOLATED half-reaction, use equilibrium arrows to show
that the reaction can go forward or backwards.
Ag+ + e- ↔ Ag
If the half-reaction is made to UNDERGO EITHER REDUCTION OR OXIDATION AS
A RESULT OF BEING PART OF A REDOX REACTION, then use a one-way
reaction arrow.
7
Try: State if the following ions could undergo reduction, oxidation, or both:
a. Ni2+
c. NO
e. Fe2+
b. Cl-
d. Cu+
f. Al3+
DETERMINING REDOX REACTIONS (STRONGEST WINS)
Become familiar with the idea that reduction equations are those on the table in the forward
direction and oxidation equations are those in the reverse direction.
Assume you have two different half-reactions in two different beakers. In one beaker there is
Zn(s) in a solution of Zn2+, and in the other is Cu(s) in a solution of Cu2+. The two possible
half-reactions are:
Zn2+ + 2e- ↔ Zn
Cu2+ + 2e- ↔ Cu
Of the two oxidizing agents (Cu2+ and Zn2+) , the Cu2+ is higher on the left side of
the table, and therefore has a greater tendency to become reduced. The reduction reaction
will be:
Cu2+ + 2e- ↔Cu
Of the two reducing agents ( Cu and Zn ) , the Zn is lower on the right side of the
table, and therefore has a greater tendency to be oxidized. The oxidation reaction will
be:
Zn2+ + 2e- ↔Zn or rewritten as
Zn
Zn2+ + 2e-
Recall: Reduction and oxidation must BOTH occur for a redox reaction to happen.
Therefore the two half-reactions must be connected or joined to allow the
transfer of electrons to occur (more later)
The overall reaction, which in this particular case will occur spontaneously, is found by
adding together the two half reactions – ONE OXIDATION AND ONE REDUCTION.
Cu2+ + 2e- ↔ Cu
Zn
↔ Zn2+ + 2e-
add the two ½ rexns (written correctly)
cross out what appears the same on
both sides
-8When two complete half-reactions (having all species present for each half-reaction)
are connected, the higher half reaction on the table will undergo reduction and the lower
will undergo oxidation.
Strongest Oxidizing Agent reacts and Strongest Reducing Agent reacts
Try: Which is the stronger oxidizing agent in each of the following pairs:
a. Ag+ or Cu2+
b. Co2+ or Au3+
Try: Which is the stronger reducing agent in each of the following pairs:
a. H2O or H2O2
Ex.
b. Sn2+ or Cu+
Assuming you have the following species present: Br2, Br-, I2, I-, what is the
overall reaction?
RULES FOR PREDICTING SPONTANEOUS REDOX REACTIONS
1. If only one species in a half reaction is present, don’t assume that the other species is also
present. You need to be explicitly told if the other species is present.
Ex. You are told a solution contains Cl-. Look for Cl- only (not Cl2)
2. If you are only given two potential reactants, rather than complete half-reactions, a
reaction may or may not occur. In order to determine if a reaction will occur, the first thing
to do is:
Locate each reactant on the table (on any side). Take note of the position – which is
higher or lower (stronger RA / OA)
There are three possibilities:
If BOTH species are OXIDIZING AGENTS or BOTH species are REDUCING
A.
AGENTS, then NO REACTION OCCURS. (Must have one reduction and one
oxidation)
Ex. Assume the only reactants are Zn and Cu.
B. If one is on the left of the table and one is on the right of the table two different
cases are possible:
i.
The oxidizing agent (LEFT column) is HIGHER than the reducing
agent (RIGHT column) =
THE REACTION WILL BE SPONTANEOUS
Ex. Assume the reactants in a vessel are Cu2+ and Zn.
ii.
The oxidizing agent (LEFT column) is LOWER than the reducing
agent (RIGHT column) = THE REACTION WILL NOT OCCUR
SPONTANEOUSLY (it is possible, but not spontaneous)
Ex. Assume the reactants are Zn2+ and Cu.
SUMMARY
A REACTION WILL ONLY BE SPONTANEOUS IF THE OXIDIZING AGENT (LEFT
COLUMN) IS ABOVE THE REDUCING AGENT (RIGHT COLUMN).
Try: Which of the following species will oxidize Cu+?
Ni2+
Ag+
Pb2+
I2
H2O
Try: Predict whether or not a reaction will occur when the following are mixed:
a. Cl2 and Brb. Sn and Mn
c. Ni2+ and Pb
d. Cl2 and K
MAKING A STANDARD REDUCTION POTENTIAL TABLE
Ex. Use the following information to make a SRP table with six half reactions.
1.
2.
3.
4.
5.
L + M2+ 2L+ + M
P- reduces D3+ to D2+
Element Q is the strongest oxidizing agent (Q + e-  Q-)
M + 2P  M2+ + 2PL+ oxidizes C to C2+
COMMENT ON H+
If H+ is present in a particular half-reaction, it must be treated like any other reactant. For
example, if you are asked whether the SO42- in a solution of Na2SO4 will reduce:
SO42- + 4H+ + 2e- ↔ H2SO4 + 2H2O
Your answer should be ‘there is no reaction unless H+ is could is present also” (or likewise
“only if the solution is acidic”), just as nothing would happen if SO42- wasn’t present – same
with H+. There will be no reaction if it were not present.
Note that H+ is necessary in many reduction half-reaction, but there is also one halfreaction where it is the only substance involved.
2H+ + 2e- ↔ H2
G3. BALANCING REDOX HALF-REACTIONS
A half-reaction must be balanced, just as other chemical reactions must be balanced,
however half- reactions are balanced for mass and CHARGE. Balancing half-reactions is
not overly complicated, but it is very easy to make mistakes if you are careless about
writing the charges on ions.
BALANCING HALF-REACTION STEPS
Usually when you are required to balance a half-reaction, you will be given a ‘skeleton
equation’ containing the major atoms. It is up to you to complete the balancing by
applying other species as follows:
1.
Balance the MAJOR ATOMS (atoms other than Oxygen or Hydrogen)
2.
Balance the OXYGEN ATOMS by ADDING WATER
MOLECULES Most redox rxns occus in aqueous sol’n. Any
oxygen that is used as a reactant or given off as a product will be
in the form of water, NOT O or O2.
3.
Balance the HYDROGEN ATOMS BY ADDING H+. Redox reaction
are (initially) treated as if they occur in acidic conditions. (Basic will
require one extra step later)
Since water contains hydrogen, always balance the hydrogen after
you have added water to balance the oxygen.
4.
Balance the OVERALL CHARGE BY ADDING ELECTRONS. The
total charge (+ve, -ve, or neutral) must be the same on both sides.
Do this after the hydrogens, as they have a charge.
Note – NEVER vary the order of balancing, it will make it difficult to impossible if you don’t
follow the above order EXACTLY.
Memory aid:
MAJOR HYDROXIDE
Major atoms O H - (charge)
Ex. Balance the half-reaction:
RuO2 ↔ Ru
Ex. Balance the half-reaction: Cr2O7 2- ↔ Cr3+
Try: Balance the half-reaction: MnO4- ↔ Mn2+
BALANCING BASIC HALF-REACTIONS
All the above solutions were assumed to be in acidic solutions. Sometimes you will be
required to balance half-reactions in BASIC conditions.
First, balance as if it were in acidic conditions
Ex. Pb ↔ HPbO2-
Now, CONVERT THE EQUATION TO BASIC CONDITIONS which is done by:
Adding the water equilibrium equation in such a way as to CANCEL OUT ALL THE H+ in the
half reaction.
H+ + OH-↔ H2O
or
H2O ↔ H+ + OH-
Try: Balance the following:
a. HC2H3O2 ↔ C2H5OH in acidic conditions
b. MnO4- ↔ MnO2 in basic conditions
BALANCING REDOX EQUATIONS USING HALF-REACTIONS
1.
Break into the two half reactions (look for similar species)
Ex. Balance ClO4- + I2
Cl- + IO3- in acidic solution.
2. Balance the individual half reactions separately.
3. Multiply the half reactions by whole number so as to make the
TOTAL ELECTRONS LOST = TOTAL ELECTRONS GAINED (same number for both)
4. Add the two half reactions, canceling out any species common to both sides
Work THOUROUGHLY AND CAREFULLY. One small mistake can throw the whole
question off. Double check at the end. (atoms/e-/charrges balanced?)
Try: Balance MnO4- + C2O42-  MnO2 + CO2 in basic solution.
DISPROPORTIONATION
REACTION
A redox reaction in which the
SAME SPECIES IS BOTH
REDUCED AND OXIDIZED.
Ex. Balance ClO2-  ClO3- + Cl- in acidic solution.
Ex. Balance P4  H2PO2- + PH3 in basic solution
G4. REDOX TITRATIONS
Acid-base titrations are very useful as they allow an accurate determination of an unknown
concentration of an acid or a base. In a similar manner, there are many occasions where
you may need to know the concentration of a substance that is capable of undergoing an
oxidation or reduction reaction.
OXIDIZING AGENTS
One of the most useful oxidizing agents that you will encounter is acidic KMnO4. The half
reaction is:
MnO4- + 8H+ + 5e-  Mn2+ + 4H2O
It has such a strong tendency to reduce (note its position on the table – high on the top
left) that it is able to oxidize a large number of substances (the K+ in KMnO4 is left out as it
is a spectator ion).
Recall that acid base titrations use an indicator to help see the equivalence point of the
titration. The above redox titration also requires some way to identify the equivalence point.
Another reason the KMnO4 is so commonly used in redox titrations is that:
MnO4- + 8H+ + 5ePurple
Mn2+ + 4H2O
colourless
the species being reduced also acts
as The indicator. At the
equilvalence pont The colour
disappears.
Ex. A 100.0 mL sample containing FeCl2 is titrated with 0.100 M KMnO4 solution. If 29.15 mL
Of KMnO4 was required to reach the endpoint, what was the [Fe2+]?
Other commonly used chemicals in redox titrations are: dichromate and hydrogen peroxide.
Why?
H1. ELECTROCHEMICAL CELLS
Recall the half-reactions need to be somehow connected in order for both reactions to
occur (a donating and accepting of electrons needs to occur).
ELECTROCHEMICAL
CELL
A system of half reactions joined together in such a way as to
PRODUCE ELECTRICAL ENERGY. THE REACTION IS
SPONTANEOUS.
Consider the reaction:
Cu2+ + Zn  Cu + Zn2+
A spontaneous reaction will occur when zinc metal is placed into a solution of CuSO4.
However, The electrons would be transferred directly from the Zn to the copper,
being used up as the reaction proceeds.
The exact same reaction can be used to produce electricity if:
the half reactions are kept separate in such a way as to allow the reactions to still
occur, and forcing the electrons to flow through a wire.
ELECTRODE A conductor at which a half reaction occurs. (A general term)
ANODE
The electrode at which OXIDATION occurs.
It is the electrode receiving the electrons from a substance being
oxidized. All anions in the system travel towards the anode. Often
decreases in mass during a reaction.
CATHODE
The electrode at which REDUCTION occurs.
It is the electrode supplying electrons to a substance being
reduced. All cations travel towards the cathode. Often
increases in mass during a reaction.
Memory aid:
Oxidation occurs at the Anode
(vowels)
Reduction occurs at the Cathode (constonants)
EXAMPLE ELECTROCHEMICAL CELL DIAGRAM
1.The substances present are:
No appreciable reaction occurs before the cells are connected since there is nowhere for the
electrons to go.
2. After the half-cells are connected:
The Strongest OA:
The Strongest RA:
The reaction at the anode is:
What is happening at the anode is:
The reaction at the cathode is:
What is happening at the cathode is:
The electrons:
The salt bridge:
ELECTRONS AWAYS FLOW
ANODE CATHODE (A  C)
Ex. Assume two half-cells consisting of Pb(s) in a Pb(NO3)2 solution and Zn(s) in a ZnCl2
solution are connected to make an electrochemical cell. Draw and label the parts of the
cell, write the equations for the individual half reaction and overall reaction, state what
would be seen at each electrode as the reaction occurs, and indicate the direction in which
the ions and electrons move.
STANDARD REDUCTION POTENTIALS
VOLTAGE
The tendency of electrons to flow in an electrochemical cell is called the
VOLTAGE or ELECTRICAL POTENTIAL to do work.
The voltage is the WORK DONE PER ELECTRON TRANSFERED.
Since electrons cannot flow in an isolated half-cell, the voltage of an individual half-cell
cannot be determined. However:
The difference in electrical potential between two half-reaction can be measured
A ZERO-POINT is arbitrarily defined on the voltage scale. Specifically the voltage for the
HYDROGEN HALF-CELL:
Defined as
2H+(aq) + 2e- ↔ H2(g)
Eo = 0.00 V
Eo = the STANDARD REDUCTION POTENTIAL in volts. The ” o “ in Eo indicates it
is at standard state. (If it is not, it will just state E)
An electrochemical cell is said to be at STANDARD STATE if:
1.
2.
3.
4.
It is at 25oC
All gases are at 101.325 kPa (1 atm)
All elements are in their standard states (normal phase at 25oC)
All solutions involved in the half-cell (both reactants and products) have a
concentration of 1.0 M
ALL of these must be true
All voltages listed in the table of STANDARD REDUCTION POTENTIALS are determined at
standard state and are compared to the standard reduction potential of the hydrogen halfcell.
Ex. Cu2+ + 2e- ↔ Cu ……… Eo = + 0.34
This half-cell has a voltage 0.34 more than the hydrogen half-cell.
Ex. Zn2+ + 2e- ↔ Zn ……… Eo= -0.76
This half-cell has a voltage 0.76 less than the hydrogen half-cell.
Since the voltage is a measure of work done (so work is either done or is being done),
reversing a reduction reaction such as
Zn2+ + 2e- ↔ Zn ……… Eo= -0.76 V
produces:
an oxidation reaction with a CHANGED SIGN for Eo.
Zn  Zn2+ + 2e-
Eo = +0.76 V
IF A HALF REACTION IS REVERSED, THE SIGN OF ITS Eo VALUE IS
ALSO REVERSED.
When two half reactions are combined,
The voltage for the overall reaction is found to be the difference between the
voltages of the individual half cells (one reduction one oxidation)
Ex.
Hg2+ + 2e- ↔ Hg
Cu2+ + 2e- ↔ Cu
Eo = +0.85 V
Eo = +0.34 V
Since two half cells are added to give the redox reaction, the voltages can also be added.
Eo (cell) = Eo (reduction) + Eo (oxidation) ONCE you have adjusted the signs.
Ex. Calculate the potential of the cell Ni2+ + Fe  Ni + Fe2+
If Eocell is positive for a redox reaction, the reaction is expected to be spontaneous
(positive voltage – produces electrical energy).
If Eocell is negative, then the reaction is non-spontaneous.
(negative voltage – requires electrical energy to occur)
Ex. Calculate the potential of the cell: Mn + Mg2+  Mn2+ + Mg
Ex. Calculate the potential of the cell: 3 Ag+ + Al  3Ag + Al3+
Try: Calculate the potential for the cell and state if it would be expected to be spontaneous:
2 H2O2  2 H2O + O2
Although Eo can be used to predict if a reaction is spontaneous, it has no correlation to the
rate of the reaction. Recall that the activation energy of a reaction determines the rate at
which it will proceed.
SURFACE AREA OF ELECTRODES
The surface area of electrodes has NO EFFECT on the cell potential. Increasing
the surface area will increase the rate of the reaction (number of electrons per
second – amps) but doesn’t change the voltage (work done by each). Increasing
the electrode surface area will also increase the length of time that the cell can
operate.
Also, since the electrodes are solids, changing the size of the electrodes will not
affect the cell (solids have constant concentration)
SELECTING PREFERRED REACTIONS
TO DETERMINE THE PREFERRED HALF-REACTIONS:
1. IDENTIFY ALL THE SPECIES PRESENT AND LOCATE ON CHART (look at both sides)
2. IDENTIFY THE STRONGEST RA (highest on left side) – IT WILL REACT FIRST
3. IDENTIFY THE STRONGEST OA (lowst on left side) – IT WILL REACT FIRST
Ex. A strip of iron metal is placed in a mixture of Br2(aq) and I2(aq). What is the preferred reaction
that will occur?
SPECTATOR IONS
Any ion capable of being reduced will be a spectator ion if:
There is another ion in the same solution that has a greater tendency to be
reduced.
Any ion capable of being oxidized will be a spectator ion if:
There is another ion in the same solution that has a greater tendency to be oxidized.
Some ions are more commonly used in making electrochemical cells because they have
such a low tendency to oxidize or reduce that they rarely come into play in a reaction.
Common Spectator Ions:
Na+
K+
Ca2+
Mg2+
SO42- (in neutral sol’n)
Cl-
HALF-REACTIONS NOT AT STANDARD STATE
If a half-cell is not at standard conditions, there will be a change in the potential.
Ex.
Cu2+ + 2e- ↔ Cu
Eo = +0.34 V
According to Le Chatalier’s principle, if the [Cu2+] so it is > 1.0 M, the equilibrium will shift
To the right, so the reduction potential increases.
Cu2+ + 2e- Cu
E = > 0.34 V
And if the [Cu2+] is decreased so it is < 1.0 M, the equilibrium will shift:
To the left, so the reduction potential decreases
Cu2+ + 2e-  Cu
E = < 0.34 V
Notice that the ‘o’ is dropped from the above equations as they are NOT at standard
state.
CELLS THAT REACH EQUILIBRIUM
Operating chemical cells ARE NOT AT EQUILIBRIUM. The reaction arrow is in one direction.
Ex.
2Ag+ + Cu  2Ag + Cu2+ Inutually, great tendency to form products. As rxn
proceeds, [ ]s change (rxt used, prod made)
THE REDUCTION REACTION (2Ag+ + 2e-  2Ag)
As the [Ag+] decreases as the rxn proceeds, the reduction potential decreases and
the half reaction lowers, ‘relative to the chart’. Due to the decrease in concentration,
the tendency to form products decreases as the cell operates.
THE OXIDATION REACTION (Cu2+ + 2e-  Cu)
As the [Cu2+] increases, the tendency for the reaction to undergo reduction
increases. Therefore the tendency to be oxidized is increasingly opposed by a
growing tendency to be reduced as the reaction proceeds.
Overall, the following occurs as cell go towards equilibrium:
INITALLY
AS RXN PROCEEDS
(TIME PASSES)
EVENTUALLY
.
The reduction potential of the reduction reaction decreases and the
reduction potential of the oxidation reaction increases as the reaction
proceeds until both half-reactions have the same reduction potential.
Eventually the difference reaches zero – at this point it is at equilibrium
H3. CORROSION OF METALS
CORROSION
Undesired oxidation of metals other than iron. (Getting ‘eaten
away’) (Undesired oxidation of iron = rusting)
When a drop of water rests on an iron surface, a spontaneous reaction occurs:
At the oxygen-poor region in the center of the drop, the iron oxidizes. Fe(s)  Fe2+ + 2eOnce the Fe2+ is formed, it tends to migrate away from the anode (random movement from
higher to lower concentrations). As they move away, more Fe(s) is exposed underneath
the drop.
At the same time, the reaction:
½ O2 + H2O + 2e-  2OH-
Is occurring at the oxygen-rich outer surface of the drop. When the Fe2+ reaches this
region, it encounters the OH- and precipitates as Fe(OH)2(s).
The Fe(OH)2 is eventually oxidized to a complex mixture of Fe2O3 and H2O by the O2 in the
air. Rust is Fe2O3.XH2O where ‘X’ can change. Rust can have numerous different colours
(red, brown, yellow, black) since differing numbers of water molecules attached to the iron(III)
oxide will change the colour of the compound.
A metal can corrode if it touches a different type of metal in the presence of an electrolyte
solution exposed to oxygen. For example, if iron touches copper wire and the spot where
they touch gets wet, then:
Fe  Fe2+ + 2e- (Fe has a greater tendency to oxidize than Cu)
The copper conducts the electrons away from the Fe and makes them available to the
oxygen/water touching the wire.
½ O2 + H2O + 2e-  2OHPREVENTING CORROSION
There are several ways to stop or at least slow down corrosion, all of which fall into
two main categories.
1. ISOLATING THE METAL FROM ITS ENVIRONMENT
a. Apply a protective coating (paint or plastic) to the surface. If oxygen and water
can’t get to the metal, it won’t corrode.
b. Apply a metal which is corrosion resistant to the surface of the original metal.
Ex. Tin cans – steel is plated with tin (steel – strong, tin oxidizes and
forms a strong protective coating of tin oxide which adheres
strongly to the steel underneath, preventing further corrosion)
2. ELECTROCHEMICAL METHODS
a. Cathodic protection
The process of protecting a substance from unwanted oxidation by connecting
it to a substance having a higher tendency to oxidize.
Ex.
Both Mg and Zn are stronger reducing agents than iron, so if pieces of
magnesium or zinc are attached to the surface of iron, the Mg or Zn
will:
be preferentially reduced and act as the anode. That will force the Fe to
act as a cathode relative to the Mg (Fe remains in it’s reduced form)
Ex.
Strips of zinc are often bolted to the iron-hull of ships below the water line. The
zinc is ‘sacrificed’ (and eventually needs to be replaced) to keep the iron of the
ship from oxidizing and corroding. Some ships even pass a low voltage electric
current into the hull from an electric generator. This forces electrons into the
metal and prevents it from being oxidized.
Ex.
Galvanized iron simply has a zinc coating. The zinc reacts preferentially with
air and water, forming a zinc oxide layer which protects the iron as it adheres
strongly and prevents the exposure of the underlying metal to air and water.
b. Change the conditions of the surroundings
When iron is placed in contact with water containing oxygen, the following
half- reaction will oxidize the iron:
½O2 + 2H+(10-7 M) + 2e-  H2O
Eo = +0.82 V
If oxygen is removed from the system, the reaction cannot occur.
Another method is to lower the [H+] ions by adding OH- ions. A piece of iron
will not rust (or a very little amount until all the oxygen has reacted) in a basic
solution.
H4. ELECTROLYSIS
ELECTROLYSIS The process of supplying electrical energy to so as to produce a
chemical change. (Needs a battery or DC power supply)
Supplying E to a non-spontaneous redox reaction, allowing them to
occur. Called an electrolytic cell.
Only need 1 container (don’t hae to separate chemicals with barrier
or salt bridge as it is non-spontaneous)
ELECTROLYSIS OF A MOLTEN BINARY SALT
A salt made up of only two different elements.
BINARY
SALT
NaCl KBr
MgI2 AlF3
When such a salt is melted, the ions are free to move in the liquid form (molten NaCl is
NaCl(l), as only Na+ and Cl- ions are present. Do not confuse it with NaCl(aq) which is a
solution containing Na+, Cl-, and H2O.)
The only reactants present are Na+ and ClThe anode reaction is:
2Cl-  Cl2 + 2e-
Eo = -1.36V
The cathode reaction is:
Na+ + e-  Na
Eo = -2.71V
The overall reaction is:
2Na+ +2Cl-  Cl2 + 2Na
Eo = -4.07 V
In order for the above cell to operate: AT LEAST + 4.07V must be added
In reality, slightly more than the theoretical voltage is required due to factors such as
friction, internal resistance etc.
ELECTROLYSIS OF AN AQUEOUS SOLUTION
Consider the electrolysis of aqueous sodium iodide, NaI(aq). This involves another
consideration – now water is also present in the system. Inert electrodes are used and the
cell is set up as follows:
During the electrolysis of aqueous solutions you must always consider the possibility
that H2O may either oxidize and/or reduce.
ALWAYS START BY LISTING ALL THE SPECIES PRESENT
The reaction that requires the least voltage input will be the preferred reaction and will
occur first (always the SOA and SRA)
The half-reactions and overall reaction for the electrolysis of NaI(aq) is:
2H2O + 2e-  H2(g) + 2OH-(10-7 M)
Eo = -0.41 V (cathode)
2IEo = -0.54 V (anode)
 I2 + 2e-7
o
2H2O + 2I  H2(g) + 2OH (10 M) E cell = -0.95 V
Min 0.95 V must be input
Ex.
What products are formed at the anode and cathode and what is the overall reaction
when a solution containing NiSO4(aq) is electrolyzed using inert electrodes? Draw and
label a cell diagram. Determine the minimum voltage required.
Ex.
What is the overall reaction which occurs when a 1.0 M solution of HCl(aq) is
electrolyzed using carbon electrodes?
H5. ELECTROLYTIC CONCEPTS
ELECTROPLATING
The Cathode is:
Electroplating is an electrolytic process in which a metal is reduced
or ‘plated out’ at a cathode
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the site where plating occurs, so it is the material that will
receive the metal plating.
The Electroplating Solution is:
be ‘plated’
a solution containing ions of the metal which is to
onto the cathode.
The Anode is:
made of either an inert material or made of the same metal
which is to be plated onto the cathode.
You will only be asked about electroplating in NEUTRAL solutions.
Ex. Design a cell to electroplate a copper medallion with nickel metal. Include in the
design:
a. the ions present in the solution
b. the direction of ion flow
c. the substance used for the anode and cathode
d. the direction of electron flow when the cell is connected to a DC power
source
ELECTROREFINING The process of purifying a metal by electrolysis
(similar to electroplating).
APPLIED ELECTROCHEMISTRY
A. THE BREATHALYSER
When alcohol is consumed, it is absorbed into the blood stream from the stomach. Some of it
passes through the cell walls of blood capillaries into the alveoli (air sacs that make up the
lungs), in a similar manner that CO2(g) and O2(g) enter in and out of the lungs. The process
of alcohol entering the bloodstream and entering the lungs is at equilibrium so that the greater
the concentration of alcohol in the blood, the greater the concentration of alcohol in the lungs.
When the lungs exhale, the ethanol in the lungs is expelled.
Ethanol undergoes oxidation by an acidic solution of dichromate ions as follows:
3 C2H5OH + 2 K2Cr2O7 + 8 H2SO4  3 CH3COOH + 2 Cr2(SO4)3 + 2 K2SO4 + 11 H2O
(Cr2O72- is orange/yellow)
(Cr3+ is dark green)
The expelled air is blown into a breathalyzer which can measure and record the amount of
green colourization. The more green the more alcohol in the breath. The machine is
calibrated to be able to accurately calculate the blood alcohol content.
B. BATTERIES
1. The Lead-Acid Storage Battery
A car battery is of this type, consists of alternating pairs of plates of Pb(s) and PbO2(s)
immersed in dilute sulphuric acid.
The anode reaction:
Pb(s)+ HSO4-(aq)  PbSO4(s) + H+(aq) + 2e- (or Pb  Pb2+ + 2e-)
The cathode reaction:
PbO2(s) + HSO4-(aq) +3H+ +2e-  PbSO4(s) + 2H2O (or Pb4+ + 2e-  Pb2+)
The overall reaction (Pb + Pb4+  Pb2+ + Pb2+) occurs when the battery is
discharging – spontaneously reacting to produce electrical energy. This forms an
insoluble PbSO4(s) layer on the plates of the battery. When an external charge is
applied, the reaction is driven backwards. (Over time the PbSO4 tends to flake off
the plates so less Pb and PbO2 can be formed, so eventually the battery needs to
be replaced.)
2. The alkaline battery
Gets its name from the alkaline (basic) electrolyte (KOH) that is used.
The cathode reaction: 2 MnO2(s) + H2O(l) +2e-  Mn2O3(s) + 2
OH-(aq) The anode reaction: Zn(s) + 2OH-(aq)  ZnO(s) + H2O(l)
+2e-
Cheap to make, but cannot be recharged (or reversed), has a relatively short shelf life
(reaction continually occurs slowly whether the battery is being used or not).