Chemistry 12
UNIT V – ELECTROCHEMISTRY v. 1.1
I. INTRODUCTION
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 Branch of chemistry concerned with the conversion of chemical energy
to electrical energy (and vice versa)
ELECTROCHEMICAL
REACTIONS
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
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.
A half-reaction in which a species LOSES ELECTRONS
OXIDATION
OXIDATION HALF-REACTION
Al(s)  Al3+(aq) + 3eREDUCTION
A half-reaction in which a species GAINS ELECTRONS
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
-2When a substance becomes 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.
A substance being oxidized is referred to as the REDUCING AGENT.
Ex.
Zn  Zn2+ + 2eU3+  U4+ + e2Cl-  Cl2 + 2e-
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 an 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
Consists of 2 half-reactions. One reduction and one oxidation.
Called a REDOX REACTION.
Ex. Consider the following reaction:
2 Al + 3 Cu2+  2 Al3+ + 3 Cu
Al  Al3+ + 3e-
Al is being oxidized, causing Cu2+ to be reduced.
Al is the REDUCTION AGENT
Cu2+: Cu2+ + 2e-  Cu
Cu2+ is being reduced, causing Al to be oxidized.
Cu2+ is the OXIDIZING AGENT
Al:
THE REDUCING AGENT IS OXIDIZED IN A REACTION
THE OXIDIZING AGENT IS REDUCED IN A REACTION
ANY TIME THAT YOU SEE THAT AN ATOM OR ION HAS CHANGED ITS CHARGE
DURING A REACTION, YOU ARE DEALING WITH A REDOX REACTION.
-3Ex. 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+ + ed.
F2 + 2e-  2 F-
Ex. For the following reactions indicate which substance is:
a. being oxidized, b. being reduced, c. the oxidizing agent, d. the reducing agent.
Zn2+ + Mg  Zn + Mg2+
2 Na + Br2  2 Na+ + 2 Br-
II. 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)
You will need to be able to calculate the oxidation numbers of various atoms.
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 I ions (Li, Na, K… Alkali metals) ALWAYS have an oxidation of 1+ when
combined with other elements.
-4ii. Group II ions (Ca, Ba, Mg… Alkaline Earth Metals) ALWAYS have an oxidation number
of 2+ when bonded in a mlc.
iii. Group XVII (Cl, Br, F… Halogens) TEND to have an oxidation number of 1- when
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.
6.
ELEMENTS IN THEIR ELEMENTAL FORM (NOT COMBINED WITH ANYTHING)
ALWAYS HAVE AN OXIDATION NUMBER OF 0 UNLESS OTHERWISE STATED.
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+
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
-5Consider the following
ClO3-  ClO4-
(Cl5+  Cl7+ = oxidation)
H2O2  H2O
Cr3+  CrO42NO2  N2O3
TIP – if the ratio of attached oxygen atoms increases in a reaction  oxidation
If the ratio of attached oxygen atoms decreases in a reaction  reduction
III. PREDICTING THE SPONTANEITY OF A REDOX REACTION
There is a table called the Table of Standard Reduction Potentials, which is partially shown below.
F2 + 2 e- ↔ 2 F
-
+
Ag + e- ↔ Ag
Cu
2+
+ 2 e- ↔ Cu
Zn
2+
+ 2 e- ↔ Zn
+
Li + e- ↔ Li
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
(Exceptions – Cu, Ag, Hg, Au are all clustered together on the table)
2.
In general, the halogens and oxyanions (anions containing oxygen) are found in the top half
of the table on the right.
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.
-6-
+
Copper
Cu + e- ↔ Cu(s)
Cu2+ + 2e- ↔ Cu(s)
Cu2+ + e- ↔ Cu+
H2O2 is found at the top left side (+1.78 V) and lower down on the right side (+0.70 V)
Watch for exceptions.
Oxidizing agents are found on the left of the table.
Reducing agents are found on the right.
ALL reactions on the table are written as REDUCTIONS () they gain e- in the forward rxn
The Table of Standard Reduction Potentials is used in a similar way to the Table of Relative Strengths
of Acids and Bases.
The speices in the upper left corner have a tendency to proceed in the forward direction, while the
species in the lower right corner have a tendency to proceed in ther reverse direction. Each of the
reactions on the standard reduction table could proceed in either direction, providing the correct
substances are present.
Ex. The half reaction for Zn and 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
Zn  Zn2+ + 2e-
(written as a reduction)
(written as an oxidation)
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.
-7Try: State if the following ions could undergo reduction, oxidation, or both:
a. Ni2+
c. NO
e. Fe2+
b. Cl-
d. Cu+
f. Al3+
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) the higher
half reaction on the table will undergo reduction and the lower will undergo oxidation.
(Tip – look for the strong OA and the stronger RA present)
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-, take the following half
reactions and determine the overall redox reaction:
Br2 + 2e- ↔ 2BrI2 + 2e
-
-
↔ 2I
If you are only given two species (or ions) rather than the four needed for two complete half-reactions
as given above, you will need to be able to predict if that particular redox reaction will occur.
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-. Referring to the table, you see the half reaction
Cl2+2e- ↔ 2ClJust because it is on the table doesn’t mean it will happen. You have to have the reactants
present in order for a reaction to occur.
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)
-9There 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.
Both are only found as RA. No rxn can occur
Ex. Assume the only reactants are Br- and Cl-.
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.
Cu2+ is an OA, and is higher than Zn which is the RA.
The reaction will spontaneously occur.
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 IS ABOVE
THE REDUCING AGENT.
-10-
+
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
COMMENT ON H+
Some half-reactions require H+ to occur.
Ex.
H2O2 + 2H+ +2e- ↔ 2H2O
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 half-reaction where
it is the only substance involved.
2H+ + 2e- ↔ H2
-11IV. BALANCING HALF-REACTIONS
A half-reaction must be balanced, just as other chemical reactions must be balanced, however halfreactions 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.
2.
Balance the MAJOR ATOMS (atoms other than Oxygen or Hydrogen)
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
Major atoms
HYDROXIDE
O H - (charge)
-12Ex. Balance the half-reaction:
Ex. Balance the half-reaction:
RuO2 ↔ Ru
Cr2O72- ↔ 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 ↔ HPbO21. Balance the major atoms
2. Balance the oxygen atoms
3. Balance the hydrogen atoms
4. Balance the charge.
-13Now, CONVERT THE EQUATION TO BASIC CONDITIONS which is done by:
5. 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
V. BALANCING REDOX EQUATIONS USING HALF-REACTIONS
Note: There are two common methods for balancing redox equations: using half reaction and using
oxidation numbers. You are not required to know both methods (as they end up with the
same results) but it may be to your advantage to be familiar with both of them. Balancing
using half-reactions is easier for more complicated redox equations, the oxidation number is
easier for simpler redox equations.
STEPS IN BALANCING EQUATIONS USING HALF REACTIONS
Ex. Balance ClO4- + I2  Cl- + IO3- in acidic solution.
1.
Break into the two half reactions (look for similar species)
-14-
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.
-15DISPROPORTIONATION
REACTION
A redox reaction in which the SAME SPECIES IS BOTH
REDUCED AND OXIDIZED.
Ex. Balance ClO2-  ClO3- + Cl- in basic solution.
STEPS IN BALANCING REDOX EQUATIONS USING OXIDATION NUMBERS
This method is somewhat of a shortcut, based on that fact that since the total number of electrons lost
in an oxidation half-reaction must equal the total number of electrons gained in a reduction halfreaction, so the following two statements are true:
An increase in oxidation number in one species must be balanced by a decrease in oxidation
number of a second species.
In any redox reaction, the overall change in oxidation number must equal zero.
Ex. Balance the following redox reaction
1.
2.
ClO4- + I2  Cl- + IO3-
Balance the MAJOR ATOMS
Assign OXIDATION NUMBER TO ALL STOMS INVOLVED IN A CHANGE (Δ ON)
and determine the TOTAL CHANGE (this includes the number of atoms that changed also)
-163. Balance the change in oxidation number by multiplying to get the lowest common multiple for
each half reaction.
4.
Balance OXYGEN ATOMS by adding H2O and HYDROGENS by adding H+. DO NOT ADD
ELECTRONS (This has already been factored in by the previous step)
Ex. Balance P4  H2PO2- + PH3 in acidic solution
Try: Balance Zn + As2O3  AsH3 + Zn2+ in basic solution.
Try: Balance S
2-
-
-
-17-
+ ClO3  Cl + S (basic)
Try: Balance CN- + IO3-  I- + CNO- (acidic)
Try: Balance As4 + NaOCl + H2O  AsO43- + NaCl
VI. 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.
A. 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
-18It 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).
Ex. To find the [Fe2+] in a unknown solution, react it with acidic MnO4- as follows:
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+ + 5e-  Mn2+ + 4H2O
Purple
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+]?
1. Balance redox
2. Calculate moles MnO43. Use mol ratio to calculate Fe2+
4. Determine [ ]
B. REDUCING AGENTS
Two commonly used reducing agents are NaI and KI . A large number of substances can oxidize Ito I2 (as it is relatively low – about halfway down the table) according to the following half-reaction:
2I-  I2 + 2eTitrations involving I- generally involve two consecutive steps:
1. First the I- is oxidized to I2 by the substance being reduced.
2. Secondly, the I2 produced in step 1 is reduced back to I- by a second reducing agent such as
the thiosulphate ion (S2O32-)
An example of a reaction involving I- is the reduction of laundry bleach, NaOCl. The reaction
between I- and OCl- proceeds as follows:
-19-
No attempt is made to add exactly enough I- to react with the OCl-. Rather:
A deliberate excess of I- is added on order to ensure that all the OCl- has reacted. ([OCl-] is
what is being looked for). Any excess I- will not affect the results. The important part is that
every molecule of OCl- has reacted to form a molecule of I2 (or as much as possible)
The above reaction between OCl- and I- is the ‘initial’ reaction because the actual redox titration
involves a second reaction between the I2 produced, and another ion present in the titrating substance,
the reducing agent sodium thiosulphate, Na2S2O3. (which ionizes into sodium and thiosulphate)
When the addition of S2O32- has reacted most of the I2 present, the brown colour of the I2 almost
disappears (a diluted colour appearing pale yellow remains). Some starch solution is then added to the
titration, which produces a dark blue colour (this is caused by the reaction between starch and the
remaining I2 in solution). After adding the starch (which acts as a more noticeable and therefore more
precise indicator), the last of the S2O32- is added, causing the blue colour of the starch-I2 mixture to
fade – so that the last of the colour just disappears at the equivalence point between the I2 and the
S2O32-.
This whole process is done since there is no suitable indicator for the reaction between
OCl- and I-, so a substance is reacted that can then undergo another reaction that has a
suitable indicator.
Ex.
A 25.00 mL sample of bleach is reacted with excess KI according to the following equation:
2 H+ + OCl- + 2 I-  Cl- + H2O + I2
The I2 produced requires exactly 46.84 mL of 0.7500 M Na2SO3 to bring the titration to the
endpoint using starch solution as an indicator, according to the following equation:
2 S2O32- + I2  S4O62- + 2 IWhat is the [OCl-] in the bleach?
-20-
VII. 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
Consider the reaction:
A system of half reactions joined together in such a way as to
PRODUCE ELECTRICAL ENERGY
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.
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.
-21CATHODE
The electrode at which REDUCTION occurs.
It is the electrode supplying electrons to a substance being reduced.
All cations travel towards the cathode.
Memory aid:
Oxidation occurs at the Anode
Reduction occurs at the Cathode
(vowels)
(constonants)
EXAMPLE ELECTROCHEMICAL CELL DIAGRAM
Ag ↔ Ag+ + eCu ↔ Cu2+ + 2eBefore connecting, each cell has a small tendency to react (lose electrons and prduce positive ions),
such that any electrons given off remain on the metal. (Not to a great extent) Experimentally, Cu has
a greater tendency to oxidize.
1.The possible half-reactions are:
2. After the half-cells are connected:
Cu has a greater tendency to oxidize, causing an excess of electrons to accumulate on the copper
electrode. This excess of electrons causes the electrons to flow from the Cu to the Ag electrode, but
by wires. As a result, the Cu loses electrons (oxidizes) and Cu becomes the ANODE.
THE ANODE IS WHERE OXIDATION OCCURS
3. As electrons are supplied to the Ag electrode: (via the wire)
The equilibrium Ag ↔ Ag+ + e- is upset, and according to Le Chataliers principle, the Ag+ will be
reduced to Ag. A reduction occurs at the Ag electrode, and it becomes the cathode.
THE CATHODE IS WHERE REDUCTION OCCURS
-224. Overall, the electron flow is:
from the Cu electrode (oxidation is producing electrons) to the Ag electrode (reduction is using the
electrons)
ELECTRONS AWAYS FLOW FROM ANODE CATHODE
(A  C)
5. The salt bridge:
Prevents free mixing of the solutions. Water and certain ions are able to pass through the salt bridge,
but are blocked somewhat. Ag+ ions are reduced to Ag(s) in one half-cell while being unable to
contact the Cu directly. (If it did, there would be direct transfer of electrons so they would not flow
through the wire and no electrical energy could be harnessed,) By separating the half cells, electrons
are forced to flow through the wire. After time, ions begin to move through the salt bridge to balance
out the charge difference.
A device such as a voltmeter is attached along the wire to measure the flow of electrons.
6. As Cu2+ ions are formed:
They accumulate around the anode. This excess positive charge is depleted by two simultaneous
migrations.
1. Cu2+ move away from the anode (random movement (greater probability it will
move away from the high [Cu2+] around the anode than approach it)
2. Negative ions (such as the SO42- ions in sol’n and the negative ions in the salt
bridge move towards the anode. Due to the positive ions produced at the anode,
they will attract the negative ions towards it.
7. As the [Ag+] is depleted around the cathode: (as it is being reduced to Ag(s))
The net amount of positive charge is decreased near the cathode. The deficiency of positive charge is
depleted by two simultaneous migrations:
1. Ag+ ions move towards the cathode (random movement – probability) as they
positive ions are depleted near it.
2. Negative ions (such as the NO3- in solution) move away from the cathode. The
greater amount of positive charge at the anode end of the cell will attract the
negative ions and they will begin to flow to it (into the salt bridge)
NOTE 1. NO electrons flow in sol’n, only ions. The electrons flow through the wire.
2. The number of electrons involved in the oxidation reaction must equal the number of
electrons involved in the reduction reaction.
-23Ex. 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, and indicate the direction in which
the ions and electrons move.
1. write possible ½ rxns
2. determine ox/red
3. write balanced rxn
4. draw cell and label
-24VIII. 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 half-cell.
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.
-25Since 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+
-26If Eocell is positive for a redox reaction, the reaction is expected to be spontaneous.
If Eocell is negative, then the reaction is non-spontaneous.
Ex. Calculate the potential of the cell: Mn + Mg2+  Mn2+ + Mg
Mn2+ + 2e- ↔ Mn
Mg2+ + 2e- ↔ Mg
Eo = -1.19 V
Eo = -2.37 V
Mg2+ + 2e- ↔ Mg
Mn
↔ Mn2+ + 2e-
Eo = -2.37 V
Eo = +1.19 V
Eo (cell) = -1.18 V
NOT SPONTANEOUS
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.
-27COMMENT ON WATER
Sometimes the reduction of neutral water is shown as:
2H+ (10-7M) + 2e- ↔ H2(g)
Eo= -0.41 V
Other times it is shown as:
2H2O + 2e- ↔ H2(g) + 2OH- (10-7 M)
Eo= -0.41 V
If a reaction occurs in a neutral solution, the reduction of neutral water (at – 0.41 V) may be a possible
reaction and must be considered along with any other reductions possible.
If a reaction occurs in an acidic solution, the reduction of H+ (at 0.00 V) may be a possible reaction
and must also be considered along with any other reductions possible.
2H+ + 2e- ↔ H2(g)
Eo= 0.00 V
(you won’t be asked to deal with basic solutions)
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)
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
-28According 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 formed)
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
-29The 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.
IX. SELECTING PREFERRED REACTIONS
When a cell contains a mixture of substances, several different half-reactions (and therefore overall
reactions) may appear to be possible.
Consider the following cell:
The possible half-reactions are:
Ag+ + e- ↔ Ag(s)
Cu2+ + e- ↔ Cu(s)
Zn2+ + e- ↔ Zn(s)
Eo= +0.80 V
Eo= +0.34 V
Eo= -0.76 V
WHEN SEVERAL DIFFERENT REDUCTION HALF-REACTIONS CAN OCCUR
The half-reaction having the highest reduction potential will occur preferentially. (THE
STRONGEST RA – the highest one – WILL REACT FIRST)
Ag+ will reduce before the other two and form Ag(s)
WHEN SEVERAL DIFFERENT OXIDATION HALF-REACTIONS CAN OCCUR
The half-reaction haiving the lowest reduction potential will oxidize. (THE STRNGEST OA –
the lowest one – WILL REACT FIRST)
Zn will oxidize before the others and form Zn2+
TO DETERMINE THE PREFERRED HALF-REACTIONS:
1. IDENTIFY ALL THE SPECIE 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
-30Ex. 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-
X.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+ + 2e-
-31Once the Fe 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.
2+
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.
-32Ex. 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.
Ex.
Some buried gas and oil tanks made of steel have a thick braided wire connected to
them. The wire comes to the surface and is attached to a post in the ground that is
made of an easily oxidized metal such as magnesium or zinc. Because the post will
oxidize first, the buried tank is protected.
b. Change the conditions of the surroundings
When iron is placed in contact with water containing oxygen, the following halfreaction will oxidize the iron:
½O2 + 2H+(10-7 M) + 2e-  H2O
Eo = +0.82 V
If oxygen is removed from the system, the tendency for it to reduce is drastically
reduced. (Hydrogen will be reduced first as it is higher on the table).
2H+(10-7 M) + 2e- : H2(g)
Eo = -0.41 V
+
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.
XI. ELECTROLYSIS
ELECTROLYSIS The process of supplying electrical energy to a molten ionic compound or a
solution containing ions so as to produce a chemical change.
Supplying E to a non-spontaneous redox reaction, allowing them to occur.
Called an electrolytic cell.
-33ELECTROLYSIS OF A MOLTEN BINARY SALT
BINARY
SALT
A salt made up of only two different elements.
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.) Also note that there is no need for a salt bridge to keep the reactant separated as
no spontaneous reaction will occur between the reactants.
The only reactants present are Na+ and ClOA
RA
The anode reaction is:
2Cl-  Cl2 + 2e-
Eo = -1.36 V
The cathode reaction is:
Na+ + e-  Na
Eo = -2.71 V
The overall reaction is:
2Na+ +2Cl-  Cl2 + 2Na
Eo = -4.07 V
Recall: non-spontaneous reactions have a negative voltage (and the OA is below the RA on the
chart)
In order for the above cell to operate: AT LEAST + 4.07V must be added
Since the half cells are not at standard state, the reduction potentials will be different than those listed
on the table.
-34ELECTROLYSIS 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.
There are two possible oxidations:
There are two possible reductions:
So, in order to determine which reaction will occur, think back to the definition of electrolysis –
electrical energy is applied to produce a chemical change. It makes sense that the reaction that:
Requires the least voltage input will be preferred (will occur first). Just as before, the
strongest OA and the strongest RA will react first.
Looking at the above example. The preferred reaction (requiring the least voltage input) involves the
higher of the two possible reductions and the lower of the two possible oxidations. The same situation
applies to electrolysis as did for spontaneous reactions:
The half-reactions having the greatest tendency to reduce and greatest tendency to oxidize are
preferred.
-35The 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)
2I I2 + 2eEo = -0.54 V (anode)
-7
o
2H2O + 2I  H2(g) + 2OH (10 M) E cell = -0.95 V
Min 0.95 V must be input
Note – The concentrations of the materials in cells in not relevant – as long as there is sufficient
material in the cell you can assume that the reactions proceed as predicted. In NEUTRAL
AQUEOUS SOLUTIONS there are two equations involving water that must be considered:
Oxidation
½ O2(g) + 2H+ (10-7 M) + 2e-  H2O
Eo = 0.82 V
Reduction
2H2O + 2e-  H2(g) + 2OH-(10-7 M)
Eo = -0.41 V
In ACIDIC SOLUTIONS there are also two equations involving water that must be
considered:
Oxidation
½ O2(g) + 2H+ + 2e-  H2O
Eo = 1.23 V
Reduction
2H+ + 2e-  H2(g)
Eo = 0.00 V
Basic solutions will not be used with cells in Chemistry 12, as OH- can often precipitate metals
out of solution.
IN REALITY
It is most often found that a higher potential than calculated must be applied to cause electrolysis.
This is due to:
Factors such as internal resistance and others.
The difference between actual potentials required for electrolysis and the calculated potentials are
termed overpotentials. As a result of the overpotential effect, when dilute neutral solutions ( <1.0 M)
containing Cl- or Br- are electrolyzed:
-36Oxidation:
Cl2 + 2e-  2ClBr2 + 2e-  2Br½ O2 + 2H+(10-7M) + 2e-  H2O
Eo = -1.36 V
Eo = -1.09 V
Eo = -0.82 V
You would expect that O2 would be produced. In reality it is found that Cl2 or Br2 are actually
produced. This is indicated on the table – the lines that state ‘overpotential effect’
When dilute solution of certain metals are electrolyzed:
Reduction:
2H2O + 2e-  H2(g) + 2OH- (10-7M)
Fe2+ + 2e-  Fe
Cr3+ + 3e-  Cr
Zn2+ + 2e-  Zn
Eo = -0.41 V
Eo = -0.45 V
Eo = -0.74 V
Eo = -0.76 V
You would expect that H2 is produced, but in practice the metals are produced.
Electrolysis of aqueous solutions containing Cl- or Br- will produce Cl2 or Br2 at the anode
Electrolysis of aqueous solutions containing Fe2+, Cr3+, or Zn2+ will produce Fe, Cr, or Zn at
the cathode
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? Determine the minimum
voltage required.
1. detemine species present
2. determine strongest OA/RA
3. determine overall reaction
4. calculate answer
Ex.
-37What is the overall reaction which occurs when a 1.0 M solution of HCl(aq) is electrolyzed
using carbon electrodes?
XII. ELECTROPLATING
ELECTROPLATING
The Cathode is:
Electroplating is an electrolytic process in which a metal is reduced or
‘plated out’ at a cathode
the site where plating occurs, so it is the material that will receive the metal
plating.
The Electroplating Solution is:
The Anode is:
a solution containing ions of the metal which is to be ‘plated’
onto the cathode.
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.
-38Ex. 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
cathode – the medallion is to have Ni plated onto it, so it must be the cathode.
anode – can make it the same material to be plated, so often it is Ni to provide a supply of Ni2+.
ions – must have Ni2+ in sol’n to provide nickel to plate onto the cathode. NO3- is used as a spectator.
Ni] flow towards the –ve cathode, plating onto it.
CATIONS – FLOW TOWARDS CATHODE
ANIONS – TOWARDS ANODE
Electrons – flow A C
ELECTROREFINING
The process of purifying a metal by electrolysis
At the anode the small amounts of Zn or Pb present is preferentially oxidized as it is exposed at the
surface. When any exposed Pb/Zn atoms have oxidized and gone into solution as ions, only the Cu
atoms are available to be oxidized. Any Au, Ag, or Pt atoms present cannot be oxidized because the
anode is mostly copper which is oxidized in preference to Au, Ag, or Pt which simply drop off and
accumulate on the bottom of the cell. This ‘anode sludge’ can be purified to obtain the valuable
metals.
-39XIII. 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 is: Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2e- (or Pb  Pb2+ + 2e-)
The cathode reaction is: 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.)
-402. 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).