1 Unit 5 - Electrochemistry Chemistry 12 Intro to Electrochemistry The branch of chemistry concerned with the conversion of chemical energy to electrical energy, and vice versa. Many of the concepts which we first learned in Acid-Base theory will apply to Electrochemistry. Consider the following reaction: 2 AgNO 3(aq) + Cu (s) → 2 Ag (s) + Cu(NO 3 ) 2(aq) As a net ionic equation: 2 Ag + (aq) + Cu (s) → 2 Ag (s) + Cu 2+ (aq) The previous reaction can be broken down into two halves: Reduction Reaction: A half-reaction in which a species GAINS electrons . Ag + (aq) + e - → Ag (s) (Ag + (aq) is being reduced) Oxidation Reaction: A half-reaction in which a species LOSES electrons. Cu (s) → Cu 2+ (aq) + 2 e - (Cu (s) is being oxidized) The overall balanced, net ionic equation is called a REDOX equation (REDuction + OXidation): Cu (s) → Cu 2+ (aq) + 2 e - 2 Ag + (aq) + 2 e - → 2 Ag (s) 2 Ag + (aq) + Cu (s) → 2 Ag (s) + Cu 2+ (aq) REDUCTION: OXIDATION: REDOX:
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Unit 5 - ElectrochemistryChemistry 12
Intro to Electrochemistry
� The branch of chemistry concerned with the conversion of chemical energy to electrical energy, and vice versa.
� Many of the concepts which we first learned in Acid-Base theory will apply to Electrochemistry.
� Consider the following reaction:
2 AgNO3(aq) + Cu(s) → 2 Ag(s) + Cu(NO3)2(aq)
� As a net ionic equation:
2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+
(aq)
� The previous reaction can be broken down into two halves:
� Reduction Reaction:
� A half-reaction in which a species GAINS electrons .
Ag+(aq) + e- → Ag(s)
(Ag+(aq) is being reduced)
� Oxidation Reaction:
� A half-reaction in which a species LOSES electrons.
Cu(s) → Cu2+(aq) + 2 e-
(Cu(s) is being oxidized)
� The overall balanced, net ionic equation is called a REDOX equation (REDuction + OXidation):
Cu(s) → Cu2+(aq) + 2 e-
2 Ag+(aq) + 2 e- → 2 Ag(s)
2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+
(aq)
REDUCTION:
OXIDATION:
REDOX:
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REMEMBER:
The number of electrons lost in one half reaction must equal
the number of electrons
gained in the other.
Law of Conservation of Charge
Here are two memory aids you can use:
� “LEO” the lion says “GER”
� LEO = Lose Electrons Oxidation
� GER = Gain Electrons Reduction
� “OIL RIG”
� Oxidation Involves Loss of electrons
� Reduction Involves Gain of electrons
� Going back to the previous example:
2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+
(aq)
What causes Cu(s) to become oxidized?
The Ag+(aq)!!!
� In this case, we call the Ag+(aq) the OXIDIZING AGENT.
� The oxidizing agent causes the other species to become oxidized, but is reduced itself.
� Every redox reaction will also have a REDUCING AGENT.
� The reducing agent causes the other species to become reduced, but is oxidized itself.
� For Example:
Zn(s) + I2(s) → Zn2+(aq) + 2 I-(aq)
� What are the half reactions?
� What is the reducing agent and the oxidizing agent?
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Homework:
� Read:
� Pages 192 - 200
� Do:
� #1, 2
Oxidation Numbers
� The charge that an atom would possess if the species containing the atom were made up of ions.
� Somewhat fictitious, but will allow us to determine if a species was reduced or oxidized.
Rules:
� The alkali metals (Li, Na, K, Rb, Cs) are ALWAYS +1.
� The alkaline earth metals (Be, Mg, Ca, Sr, Ba) are ALWAYS+2.
� The halogens (F, Cl, Br, I) are NORMALLY -1 (there are many exceptions).
� Oxygen is normally -2.
� Hydrogen is normally +1.
� The rule which allows us to calculate oxidation numbers is simply:
The sum of the positive
charges and the
negative charges
must equal the overall
charge on the
species.
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For Example:
� What is the oxidation number of P in the molecule H4P2O7?
� What is the oxidation number of P in P4?
� What is the oxidation number of Cr in Cr3+?
� What is the oxidation number of Mn in MnO4-?
� We need to compare the oxidation number before and after the reaction has happened to see which species have been oxidized/reduced
� For Example:
Consider the following unbalanced half-reaction:
H2SeO3 → Se
Is Se oxidized or reduced?
H2SeO3 → Se
ON = +4 ON = 0
∆ON = 0 – 4 = -4
Since the ∆ON is negative, the species has gained electrons and is
therefore REDUCED.
To summarize:
� If the ON of a species increases(+ ∆ON), then the species has been oxidized (loses e-).
� If the ON of a species decreases(- ∆ON), then the species has been reduced (gains e-).
Examples:
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Predicting Spontaneity
� Using the “Standard Reduction Potentials of Half-Cells” table on page 8 of the data booklet, we can predict the spontaneity of a redox reaction.
� Know the trends on the table!!!
� For an isolated half-reaction, use a double arrow to indicate that the reaction may proceed in either the forward or backward direction:
Au3+(aq) + 3 e- Au(s)
� When a half-reaction goes through oxidation or reduction as a result of being part of a redox reaction, use a one-way arrow.
Au3+(aq) + 3 e- → Au(s)
For Example:
� Consider two half-cells:
� In one half-cell, there is a piece of Ag(s) in a solution of Ag+.
� The second half-cell has a piece of Cu(s) in a solution of Cu+.
� Each half reaction can be written as:
Ag+(aq) + e- Ag(s)
Cu+(aq) + e- Cu(s)
� Considering the two species, Ag+ is a stronger oxidizing agent and therefore has a greater tendency to reduce.
� The reduction reaction can be written as:
Ag+(aq) + e- → Ag(s)
� Since Cu+ is lower on the table, it is the stronger reducing agent and has a greater tendency to oxidize.
� Think of the reaction as going in reverse:
Cu+(aq) + e- ← Cu(s)
� When actually writing down the oxidation half-reaction, put Cu(s) on the reactants side.
Cu(s) → Cu+(aq) + e-
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� To find the overall redox reaction that will occur, add
together the two half reactions:
Cu(s) → Cu+(aq) + e-
Ag+(aq) + e- → Ag(s)REDUCTION
OXIDATION
REDOX Ag+(aq) + Cu(s) → Ag(s) + Cu+
(aq)
IF TWO HALF-CELLS ARE JOINED,
THE HIGHER HALF-REACTION ON
THE TABLE WILL UNDERGO
REDUCTION, AND THE LOWER
ONE WILL UNDERGO OXIDATION!
How do we do this?
1. Locate both the reactants on the table.
a) If both reactants only appear on the left, or both only
appear on the right, then…
THERE IS NO REACTION!
b) If one reactant appears on the left, and one reactant
appears on the right, there are two possible cases.
CASE 1:
� When the reactant that will be reduced (left side) is higher on the table than the reactant that will be oxidized (right side)…
A SPONTANEOUS REACTION WILL
OCCUR!
Case 2:
� When the reactant that will be reduced (left side) is lower on the table than the reactant that will be oxidized (right side)…
THERE IS NO REACTION!
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Examples:
NOTE:
� BE CAREFUL with half reactions that have H+ as a reactant.
� H+ must be treated as a reactant. These reactions can only occur in an ACIDICenvironment.
Homework:
� Read:
� Pages 193 - 210
� Do:
� #3 - 18
Balancing Half Reactions
� must balance for mass and charge.
� Don’t be sloppy about writing the charges on ions, or you will make many mistakes!!!
� You will be given a “skeleton equation” which contains the major atoms involved.
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� We have to complete the balancing by supplying other species as follows:
� Balance the MAJOR atoms by inspection. (The “major
atoms” are any atoms other than O and H).
� Balance the OXYGEN atoms by adding H2O molecules. (The
reactions will generally occur in water).
� Balance the HYDROGEN atoms by adding H+. (The reactions
will initially be considered to occur in acidic solution).
� Balance the overall CHARGE by adding electrons.
NEVERNEVERNEVERNEVER vary the order
in which these steps
are carried out!!!
� We can summarize the procedure with another memory aid:
The officer’s name was: MAJOR HYDROXIDE.
“MAJOR HYDROXIDE” translates as “MAJOR OH-”
1. Balance the MAJOR species. (MAJOR)
2. Balance the O atoms. (O)
3. Balance the H atoms. (H)
4. Balance the charge, using electrons. (-)
For Example:
NH3 N2H4 (acidic)
CrO42- CrO2
- (acidic)
NO3- NO (acidic)
� If the solution is basic, we have to add another step:
� Using the self-ionization of water:
H2O(l) H+(aq) + OH-
(aq)
we can cancel out all the H+‘s.
� Since the self-ionization equation can be written in either
directions, write it in a way that will cancel out the H+’s in the
half-reaction.
� For Example:
PbO2 PbO (basic)
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Balancing Redox Reactions Using Half
Reactions
� Assume that the equation can be broken into separate reduction and oxidation half-reactions.
� After each is balanced, add the two reactions together to obtain the balanced redox equation.
For Example:
P4 → H2PO2- + PH3 (acidic)
Homework:
� Read:
� Pages 210 - 212
� Do:
� #19 – 25
� Study for your quiz!!!
REDOX TITRATIONS
� Redox titration calculations are the same as any other titration.
� They must be spontaneous reactions (higher on the left, lower on the right).
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Oxidizing Agents
� A very useful oxidizing agent is acidified KMnO4.
� The reaction:
MnO4-(aq) + 8 H+
(aq) + 5 e- → Mn2+(aq) + 4 H2O(l)
has a strong tendency to reduce and is therefore able to
oxidize many other substances (the K+ in KMnO4 is a spectator ion).
� Unlike an acid/base titration, the above reaction does not require the use of an indicator.
� The endpoint of this redox titration is the change from a clearsolution to a faint purple one.
� Consider the titration of acidified MnO4- with Fe2+(common
application):
MnO4-(aq) + 8 H+
(aq) + 5 Fe2+(aq) → Mn2+(aq) + 4 H2O(l) + 5 Fe3+
(aq)
� As we add MnO4- from a burette to a solution containing Fe2+, the
purple colouration of the MnO4- is continually destroyed.
� At the equivalence point, the last of the Fe2+ will have been used
up, so that the next drop of MnO4- will not react and a light
purple colour will remain in the solution.
Reducing Agents
� An example of a substance which is commonly used as a reducing agent is NaI or KI.
� Titrations involving this species usually occur in two steps:
1. The I- is oxidized to I2 by the substance being reduced.
2. Next, the I2 produced by the first step is reduced back to I-
by a second reducing agent.
� An example of a reaction involving I- is the reduction of laundry bleach (NaOCl):
2 I- → I2 + 2 e-
2 e- + 2 H+ + OCl- → Cl- + H2O
2 H+ + OCl- + 2 I- → Cl- + H2O + I2
NOTE:
All the OCl- must react, therefore, an excess of I- is added.
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� The second reaction involves reducing the I2 formed in the first reaction back into I- by S2O3
2- (usually Sodium thiosulphate, a strong reducing agent).
I2 + 2 e- → 2 I-
2 S2O32- → S4O6
2- + 2 e-
2 S2O32- + I2 → S4O6
2- + 2 I-
� When the S2O32- reacts with the I2 present, the brown colour
of the I2 will begin to become a pale yellow. At this point a starch solution is added.
� The starch will turn the solution a dark blue due to its reaction with the I2 in solution.
� The endpoint of the titration occurs when the blue colourhas just disappeared.
Homework:
� Read:
� Pages 215 - 232
� Do:
� #26 - 33
Electrochemical Cells
� An electrochemical cell is a device capable of deriving electrical energy from chemical reactions.
� Every electrochemical cell has two electrodes (a conductor at which a half-cell reaction occurs).
� Anode:
� The electrode at which OXIDATION occurs.
� The electrode toward which ANIONS travel.
� Cathode:
� The electrode at which REDUCTION occurs.
� The electrode toward which CATIONS travel.
For Example:
� Two half-cells consisting of Ag(s) in an 1.0 M AgNO3(aq)
solution and Cu(s) in a1.0 M CuSO4(aq) solution are connected to make an electrochemical cell.
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Standard Reduction Potentials
� Voltage:
� The tendency of electrons to flow in an electrochemical cell.
� Measured in VOLTS.
The voltage is always the work
done per electron transferred.
� Electrons cannot flow in an isolated half-cell, so we cannot determine individual half-cell voltages.
� We can measure the difference in electrical potentialsbetween two half-cells.
� To do this, we define the Hydrogen half-cell as a zero-point on the voltage scale:
2 H+(aq) + 2 e- H2(g) Eo = 0.00 V
Where:
Eo = the standard reduction potential, in Volts.
(The “o” in Eo implies that we are dealing with a standard state.)
� The cell is at 25oC.
� All gasses are at 101.3 kPa (1 atm).
� All elements are in their normal phase at 25oC.
� All solutions in the half-cell have a concentration of 1.0 M.
For Example:
Cu2+ + 2 e- Cu Eo = +0.34 V
� This reaction has a voltage which is 0.34 V more than that of the hydrogen half-cell.
Zn2+ + 2 e- Zn Eo = -0.76 V
� This reaction has a voltage which is 0.76 V less than that of the hydrogen half-cell.
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Something to remember:� If we reverse a reduction reaction and cause it to
become an oxidation reaction we find:
Zn2+ + 2 e- → Zn Eo = -0.76 V (as a reduction)
and
Zn → Zn2+ + 2 e- Eo = +0.76 V (as an oxidation)
If we reverse a half-cell, we reverse the sign of its Eo value.
� Since we can add two half-cells together to give a redox equation, the voltages associated with the half-cells can also be added.
� For Example:
� Imagine we connect these two half-cells together:
Fe3+(aq) + 3e- Fe(s) Eo = +0.77 V
Zn2+(aq) + 2e- Zn(s) Eo = -0.76 V
� After we decide which species will be oxidized and
reduced, we find the spontaneous reaction to be:
(Fe3+(aq) + 3e- → Fe(s)) x2 Eo
(RED) = +0.77 V
(Zn(s) → Zn2+(aq) + 2e-) x3 Eo
(OX) = +0.76 V
2 Fe3+(aq) + 3 Zn(s) → 2 Fe(s) + 3 Zn2+
(aq)Eo
(CELL) = + 1.53 V
� The value for EoCELL is the difference between the
voltages for the reduction and oxidation reactions as written on the table of “Standard Reduction Potentials of Half-Cells” in your data booklet.
� As easier way to calculate this is:
EoCELL = Eo
RED - EoOX
Use the values for Eo AS GIVEN on your table in your
calculations for EoCELL.
� Our value of Eo(CELL) also tells us whether the reaction will
be spontaneous or not.
� If EoCELL is positive for a redox reaction, the reaction is
expected to be SPONTANEOUS.
� If EoCELL is negative for a redox reaction, the reaction is
expected to be NON-SPONTANEOUS.
� For Example:
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A few final comments…
� The surface area of an electrode has no effect on the cell potential.
� Since solids have a constant concentration, there is no shift
in equilibrium by increasing or decreasing surface area.
� Omit the “o” symbol in Eo when cells are not at standard conditions.
� Le Chatelier shifts occur with cells not at standard conditions. Consider the following:
Cu2+(aq) + 2e- Cu(s) E
o = +0.34 V
Shift left, Eo �(less work done by e-)
Shift right, Eo �(more work done by e-)
� Operating electrochemical cells ARE NOT at equilibrium, but tend towards it as they die out. Homework:
� Read:
� Pages 226 - 236
� Do:
� #34 – 46
� Study for your quiz!!!
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Selecting Preferred Reactions
� Sometimes several different reactions appear to be possible simultaneously.
� When several different reduction half-reactions can occur, the half-reaction having the highest tendency to accept electrons (highest reduction potential) will occur preferentially.
� When several different oxidation half-reactions can occur, the half-reaction having the highest tendency to lose electrons (lowest reduction potential) will occur preferentially.
What do we do?
� List all the species present. Break up all ionic compounds into ions.
� Starting in the upper left of the table (reduction side), look down the table until you find the first match with a species on the list.
� Starting in the bottom right of the table (oxidation side), look up the table until you find the first match with a species on the list.
For Example:
Applied Electrochemistry
� Examples include: (see Hebden pages 228 – 233)
� The Breathalyzer
� Lead-Acid Storage Battery
� Zinc-Carbon Battery
� Alkaline Dry Cell
� Fuel Cells
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Corrosion of Metals� Oxidizing (rusting of metals) is a spontaneous
electrochemical reaction:
The oxygen rich environment promotes reduction of
water:
½ O2(g) + H2O(l) + 2e- → 2 OH-(aq)
The electrons for the reduction come from the
oxidation of the Iron metal:
Fe(s) → Fe2+(aq) + 2e-
The yellow solid
produced is a
precipitate of Fe(OH)2(s).
� Isolate the metal from it’s environment.
� Apply plastic or paint on the surface. Oxygen can’t reach
the surface.
� Apply a metal which is corrosion resistant on the surface of
the original metal.
� For Example:
Chrome plated bumpers.
Zinc coated metals (galvanized).
� Cathodic Protection
� The process in which a substance is protected from
unwanted oxidation by connecting it to a substance having
� Magnesium is a stronger reducing agent then Iron, and
will therefore oxidize preferentially, sparing the Iron.
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� Another example:
� Ocean vessels are even more prone to oxidation.
� The salt water acts as an electrolyte, increasing the rate of
oxidation.
� These vessels may be
protected by using Zinc
strips on the hull (same
effect as the underground
gas tank).
� The same effect can be
achieved using a DC power
source.
� Change the conditions in the chemical environment by removing O2.
� Storing the metal in oil or another Oxygen free environment.
Sodium is stored
in oil to prevent its
spontaneous
oxidation with
Oxygen.
Homework:
� Read:
� Pages 237 - 241
� Do:
� #47, 49 - 63
Electrolysis
� The process of supplying electrical energy to a molten ionic compound or to a solution containing ions so as to produce a chemical change.
� The process takes place in an ELECTROLYTIC CELL.
� Electrolysis reactions are non-spontaneous (EoCELL< 0). We
have to supply the energy so that the reaction can occur.
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The Electrolysis of Molten NaCl
� This simple electrolytic cell only requires a source of ions and a source of electrical energy.
� Recall that NaCl is an ionic solid; when melted, the ions are mobile.
� There is no need for a salt bridge to keep the reactants separated because there is no spontaneous reaction which occurs between the reactants.
� From the above set up we have:
Na+(l) + e- → Na(s) Eo
RED = - 2.71 V
2 Cl-(l) → Cl2(g) + 2 e- EoOX = - 1.36 V
2 Na+(l) + 2 Cl-(l) → 2 Na(s) + Cl2(g) Eo
CELL = - 4.07 V
� In ALL electrolytic cells, the reduction half-reaction will be belowthe oxidation half-reaction on the table.
� To make this cell operate we must add at least +4.07 V.
The Electrolysis of Aqueous NaI
� Since we are talking about a salt in an aqueousenvironment, we must now consider the water present.
� There are two possible reduction reactions:
2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) (10-7 M)
AND…
Na+(aq) + e- → Na(s)
� There are two possible oxidation reactions:
H2O(l) → 1/2 O2(g) + 2 H+(aq)(10-7 M) + 2 e-
AND…
2 I-(aq) → I2(s) + 2 e-
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� Consider this:
� According to the above voltage differences, we can get a reaction to occur if we add at least +0.95 V.
How do we select the preferred half-reaction when
we are dealing with electrolysis reactions?
� Therefore:
The preferred reaction will be the one requiring the
least voltage input!!
� The preferred reaction (the one requiring the least voltage input)involves the higher of the two possible reductions and the lowerof the two possible oxidations.
� During electrolysis, ALWAYS consider the possibility of water being reduced and/or oxidized.
� We get:
2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq)(10-7 M) Eo
RED = - 0.41 V 2 I-(aq) → I2(s) + 2 e- Eo
OX = - 0.54 V
2 H2O(l) + 2 I-(aq) → H2(g) + I2(s) + 2 OH-(aq)(10-7 M) Eo
CELL = - 0.95 V
The Overpotential Effect
� The measured voltage we find when we connect two half-cells together will differ from the voltage expected from the table of Standard Reduction Potentials.
� This is a result of the activation energies required for the reactions to occur at the electrodes.
� For most substances, the “overpotential” is generally small or negligible (+0.02 V).
� Water has a substantial overpotential.
� On the table we see:
� The reduction of neutral water theoretically occurs at
- 0.41 V, but actually occurs at around - 0.80 V.
� The oxidation of neutral water theoretically occurs at
+0.82 V, but actually occurs a bit above +1.36 V.
� The following exceptions occur:
� OXIDATION REACTIONS:
� We will only consider Cl- and Br-.
� These ions will oxidize preferentially to the H2O.
� REDUCTION REACTIONS:
� We will only consider Zn2+ and Cr3+.
� These ions will reduce preferentially to the H2O.
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Electrolysis Example:
Homework:
� Read:
� Pages 243 - 246
� Do:
� #64 - 72
Applications of Electrolysis
� Electroplating:
� A process in which we cause a metal to be reduced or “plated” at a cathode.
� The cathode is made out of the material which will receive the metal plating.
� The electroplating solution will contain the ions of the metal we want to “plate” onto the cathode.
� In some instances, but not always, the anode may be made of the same metal we want to have plated out on our cathode.
For Example:
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� Electrorefining:
� The process of purifying a metal by electrolysis.
� Consider the following cell:
� At the anode…
� Apart from the small amounts of Pb and Zn, the Cu has the
greatest tendency to oxidize.
� The small amount of Zn or Pb present is
preferentially oxidized as it is exposed
at the surface.
� When any exposed Zn or Pb atoms
have reacted, only the Cu atoms are
available to be oxidized.
� Any Ag, Au, or Pt atoms present can’t be oxidized because the
anode is mostly copper, which is oxidized in preference to Au,
etc.
� These metal particles simply drop off the anode and
accumulate on the bottom of the electrolytic cell.
� At the cathode…
� The Cu2+ in solution is preferentially reduced at the cathode
and none of the Pb2+ and Zn2+ can be reduced since Cu2+
exists in larger concentrations and has a higher reduction
potential than Zn2+ and Pb2+.
� The result of the electrorefining is that the PURE metal is deposited on the CATHODE.
� Very pure metals (>99.99%) can be produced in this mannor.
