Electrochemistry
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Electrochemical reactions are oxidation-reduction
reactions.
The two parts of the reaction are physically separated.
– The oxidation reaction occurs in one cell.
– The reduction reaction occurs in the other cell.
• There are two kinds electrochemical cells.
1. Electrochemical cells containing nonspontaneous
chemical reactions are called electrolytic cells.
2. Electrochemical cells containing spontaneous chemical
reactions are called voltaic or galvanic cells.
Electrical Conduction
• Metals conduct electric currents well in a
process called metallic conduction.
• In metallic conduction there is electron flow
with no atomic motion.
• In ionic or electrolytic conduction ionic
motion transports the electrons.
– Positively charged ions, cations, move toward the
negative electrode, cathode.
– Negatively charged ions, anions, move toward the
positive electrode, anode.
Electrodes
• The following convention for electrodes is correct
for either electrolytic or voltaic cells:
• The cathode is the electrode at which reduction
occurs.
• The cathode is negative in electrolytic cells and positive in
voltaic cells.
• The anode is the electrode at which oxidation
occurs.
• The anode is positive in electrolytic cells and negative in
voltaic cells.
The Half-Reaction Method
Half reaction method rules:
1. Write the unbalanced reaction.
2. Break the reaction into 2 half reactions:
― One oxidation half-reaction and
― One reduction half-reaction
Each reaction must have complete formulas for molecules and ions.
3. Mass balance each half reaction by adding appropriate stoichiometric
coefficients. To balance H and O we can add:
― H+ or H2O in acidic solutions.
― OH- or H2O in basic solutions.
4. Charge balance the half reactions by adding appropriate numbers of
electrons.
 Electrons will be products in the oxidation half-reaction.
 Electrons will be reactants in the reduction half-reaction.
5. Multiply each half reaction by a number to make the number of electrons in
the oxidation half-reaction equal to the number of electrons reduction halfreaction.
6. Add the two half reactions.
7. Eliminate any common terms and reduce coefficients to smallest whole
numbers.
The Half-Reaction Method
• Tin (II) ions are oxidized to tin (IV) by bromine.
Use the half reaction method to write and
balance the net ionic equation.
The Half-Reaction Method
• Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to
chromium (III) ions in acidic solution. Write and balance the net ionic equation
for the reaction.
The Half-Reaction Method
• In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH)4-, to
chromate ions, CrO42-. The hydrogen peroxide is reduced to hydroxide ions.
Write and balance the net ionic equation for this reaction.
The Half-Reaction Method
• When chlorine is bubbled into basic solution, it forms hypochlorite ions and
chloride ions. Write and balance the net ionic equation.
• This is a disproportionation redox reaction. The same species, in this case
Cl2, is both reduced and oxidized.
Stoichiometry of Redox Reactions
• Just as we have done stoichiometry with acid-base reactions, it can also be
done with redox reactions.
• What volume of 0.200 M KMnO4 is required to oxidize 35.0 mL of 0.150 M
HCl? The balanced reaction is:
Stoichiometry of Redox Reactions
• A volume of 40.0 mL of iron (II) sulfate is oxidized to iron (III) by 20.0 mL of
0.100 M potassium dichromate solution. What is the concentration of the iron
(II) sulfate solution? The balanced equation is:
Voltaic or Galvanic Cells
• Electrochemical cells in which a spontaneous chemical
reaction produces electrical energy.
• Cell halves are physically separated so that electrons (from
redox reaction) are forced to travel through wires and
creating a potential difference.
• Examples of voltaic cells include:
Auto batteries
Flashlight batteries
Computer and calculator batteries
The Construction of Simple Voltaic Cells
• Voltaic cells consist of two half-cells which
contain the oxidized and reduced forms of an
element (or other chemical species) in contact with
each other.
• A simple half-cell consists of:
– A piece of metal immersed in a solution of its ions.
– A wire to connect the two half-cells.
– And a salt bridge to complete the circuit, maintain
neutrality, and prevent solution mixing.
The Zinc-Copper Cell
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Cell components for the Zn-Cu cell are:
1. A metallic Cu strip immersed in 1.0 M copper (II) sulfate.
2. A metallic Zn strip immersed in 1.0 M zinc (II) sulfate.
3. A wire and a salt bridge to complete circuit
The cell’s initial voltage is 1.10 volts
The Zinc-Copper Cell
• In all voltaic cells, electrons flow spontaneously from the
negative electrode (anode) to the positive electrode
(cathode).
The Zinc-Copper Cell
• There is a commonly used short hand notation for
voltaic cells.
– The Zn-Cu cell provides a good example.
•
The Copper - Silver Cell
Cell components:
1.A Cu strip immersed in 1.0 M copper (II) sulfate.
2.A Ag strip immersed in 1.0 M silver (I) nitrate.
3.A wire and a salt bridge to complete the circuit.
•
The initial cell voltage is 0.46 volts.
The Copper - Silver Cell
• Compare the Zn-Cu cell to the Cu-Ag cell
– The Cu electrode is the cathode in the Zn-Cu cell.
– The Cu electrode is the anode in the Cu-Ag cell.
• Whether a particular electrode behaves as an anode or as a
cathode depends on what the other electrode of the cell is.
The Copper - Silver Cell
• These experimental facts demonstrate that Cu2+ is a stronger oxidizing
agent than Zn2+.
– In other words Cu2+ oxidizes metallic Zn to Zn2+.
• Similarly, Ag+ is is a stronger oxidizing agent than Cu2+.
– Because Ag+ oxidizes metallic Cu to Cu 2+.
• If we arrange these species in order of increasing strengths, we see
that:
Zn 2+ < Cu 2+ < Ag +
strength as oxidizing agent
Zn 0 > Cu 0 > Ag 0
strength as reducing agent
Easiest to reduce is
Strongest Oxidizing Agent
Ag +
Cu 2+
Zn 2+
(find order using left hand side of table)
Least likely to be reduced is
Strongest Reducing Agent
Zn
Cu
Ag
(find order using right hand side of table)
Standard Electrode Potential
• To measure relative electrode potentials, we must establish
an arbitrary standard.
• That standard is the Standard Hydrogen Electrode (SHE).
– The SHE is assigned an arbitrary voltage of 0.000000… V
The Zinc-SHE Cell
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•
For this cell the components are:
1. A Zn strip immersed in 1.0 M zinc (II) sulfate.
2. The other electrode is the Standard Hydrogen Electrode.
3. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.763 volts.
The Zinc-SHE Cell
• The cathode is the Standard Hydrogen Electrode.
– In other words Zn reduces H+ to H2.
• The anode is Zn metal.
– Zn metal is oxidized to Zn2+ ions.
The Copper-SHE Cell
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•
The cell components are:
1. A Cu strip immersed in 1.0 M copper (II) sulfate.
2. The other electrode is a Standard Hydrogen Electrode.
3. A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.337 volts.
The Copper-SHE Cell
• In this cell the SHE is the anode
– The Cu2+ ions oxidize H2 to H+.
• The Cu is the cathode.
– The Cu2+ ions are reduced to Cu metal.
Primary Voltaic Cells
• As a voltaic cell discharges, its chemicals are
consumed.
• Once the chemicals are consumed, further chemical
action is impossible.
• The electrodes and electrolytes cannot be regenerated
by reversing current flow through cell.
– These cells are not rechargable.
The Dry Cell
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•
•
One example of a dry cell is flashlight and radio batteries.
The cell’s container is made of zinc which acts as an electrode.
A graphite rod is in the center of the cell which acts as the
other electrode.
The space between the electrodes is filled with a mixture of:
1. ammonium chloride, NH4Cl
2. manganese (IV) oxide, MnO2
3. zinc chloride, ZnCl2
4. and a porous inactive solid.
The Dry Cell
• The anode reaction is:
• The graphite rod is the positive electrode (cathode).
• Ammonium ions from the NH4Cl are reduced at the cathode.
The Dry Cell
• The other components in the cell are included to remove the
byproducts of the reaction.
• MnO2 prevents H2 from collecting on graphite rod.
• At the anode, NH3 combines with Zn2+ to form a soluble
complex and removing the Zn2+ ions from the reaction.
The Dry Cell
• Alkaline dry cells are similar to ordinary dry cells except
that KOH, an alkaline substance, is added to the mixture.
• Half reactions for an alkaline cell are:
Secondary Voltaic Cells
• Secondary cells are reversible, rechargeable.
• The electrodes in a secondary cell can be regenerated by
the addition of electricity.
– These cells can be switched from voltaic to electrolytic cells.
• One example of a secondary voltaic cell is the lead
storage or car battery.
The Lead Storage Battery
• In the lead storage battery the electrodes are two sets of
lead alloy grids (plates).
• Holes in one of the grids are filled with lead (IV) oxide,
PbO2.
• The other holes are filled with spongy lead.
• The electrolyte is dilute sulfuric acid.
The Lead Storage Battery
• Diagram of the lead storage battery.
The Lead Storage Battery
• As the battery discharges, spongy lead is oxidized to lead
ions and the plate becomes negatively charged.
Pb 0s   Pb 2+  2 e This is the anode reaction.
The anode is the negative battery post during discharge.
• The Pb2+ ions that are formed combine with SO42- from
sulfuric acid to form solid lead sulfate on the Pb electrode.
Pb
2+

2SO4
 PbSO4 s
The Lead Storage Battery
• The net reaction at the anode during discharge is:
• As the cell discharges, the cathode reaction is:
• The cell reaction for a discharging lead storage battery is:
• As the cell discharges, the cathode reaction is:
The Lead Storage Battery
• What happens at each electrode during recharging?
• At the lead (IV) oxide, PbO2, electrode, lead ions are oxidized to lead
(IV) oxide.
• The concentration of the H2SO4 decreases as the cell discharges.
• Recharging the cell regenerates the H2SO4.
The Nickel-Cadmium (Nicad) Cell
• Nicad batteries are the rechargeable cells used in calculators,
cameras, watches, etc.
• As the battery discharges, the half-reactions are:
The Hydrogen-Oxygen Fuel Cell
• Fuel cells are batteries that must have their reactants continuously
supplied in the presence of appropriate catalysts.
• A hydrogen-oxygen fuel cell is used in the space shuttle
– The fuel cell is what exploded in Apollo 13.
• Hydrogen is oxidized at the anode.
• Oxygen is reduced at the cathode.
The Hydrogen-Oxygen Fuel Cell
• Notice that the overall reaction is the combination of
hydrogen and oxygen to form water.
– The cell provides a drinking water supply for the astronauts as
well as the electricity for the lights, computers, etc. on board.
• Fuel cells are very efficient.
– Energy conversion rates of 60-70% are common!
Uses of Standard Electrode Potentials
• Electrodes that force the SHE to act as an anode are assigned
positive standard reduction potentials.
• Electrodes that force the SHE to act as the cathode are assigned
negative standard reduction potentials.
• Standard electrode (reduction) potentials tell us the tendencies of
half-reactions to occur as written.
• For example, the half-reaction for the standard potassium
electrode is:
Uses of Standard Electrode Potentials
• Compare the potassium half-reaction to fluorine’s halfreaction:
Uses of Standard Electrode Potentials
• Use standard electrode potentials to predict whether an electrochemical
reaction at standard state conditions will occur spontaneously.
• Will silver ions, Ag+, oxidize metallic zinc to Zn2+ ions, or will Zn2+ ions
oxidize metallic Ag to Ag+ ions?
• Steps for obtaining the equation for the spontaneous reaction.
1.
2.
3.
4
5
Choose the appropriate half-reactions from a table of standard
reduction potentials.
Write the equation for the half-reaction with the more positive
E0 value first, along with its E0 value.
Write the equation for the other half-reaction as an oxidation
with its oxidation potential, i.e. reverse the tabulated reduction
half-reaction and change the sign of the tabulated E0.
Balance the electron transfer.
Add the reduction and oxidation half-reactions and their
potentials. This produces the equation for the reaction for
which E0cell is positive, which indicates that the forward
reaction is spontaneous.
Uses of Standard Electrode Potentials
Electrode Potentials for
Other Half-Reactions
• Will permanganate ions, MnO4-, oxidize iron (II) ions to iron (III) ions, or will iron
(III) ions oxidize manganese(II) ions to permanganate ions in acidic solution?
• Follow the steps outlined in the previous slides.
• Note that E0 values are not multiplied by any stoichiometric relationships in
this procedure.
Electrode Potentials for
Other Half-Reactions
• Will nitric acid, HNO3, oxidize arsenous acid, H3AsO3, in acidic solution? The
reduction product of HNO3 is NO in this reaction.
Effect of Concentrations (or Partial
Pressures) on Electrode Potentials
The Nernst Equation
• Standard electrode potentials, those compiled in
appendices, are determined at thermodynamic
standard conditions.
• Reminder of standard conditions.
The Nernst Equation
• The value of the cell potentials change if conditions are nonstandard.
• The Nernst equation describes the electrode potentials at nonstandard
conditions.
• The Nernst equation is:
• Substitution of the values of the constants into the Nernst equation at 25o C
gives:
• For this half-reaction:
• The corresponding Nernst equation is:
The Nernst Equation
• Substituting E0 into the above expression gives:
• If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard
conditions, then E = E0 because the concentration term
equals zero.
The Nernst Equation
• Calculate the potential for the Cu2+/ Cu+
electrode at 250C when the concentration of
Cu+ ions is three times that of Cu2+ ions.
The Nernst Equation
• Calculate the potential for the Cu2+/Cu+ electrode at 250C when the
Cu+ ion concentration is 1/3 of the Cu2+ ion concentration.
The Nernst Equation
• Calculate the electrode potential for a hydrogen electrode in which the [H+] is 1.0 x
10-3 M and the H2 pressure is 0.50 atmosphere.
The Nernst Equation
• The Nernst equation can also be used to calculate the potential for a cell that consists
of two nonstandard electrodes.
• Calculate the initial potential of a cell that consists of an Fe3+/Fe2+ electrode in
which [Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode in
which [Sn4+]=1.0 M and [Sn2+]=0.10 M . A wire and salt bridge complete the
circuit.
The Nernst Equation
• Calculate the E0 cell by the usual procedure.
Counting Electrons: Coulometry and Faraday’s
Law of Electrolysis
• A coulomb is the amount of charge that passes a given point
when a current of one ampere (A) flows for one second.
• 1 amp = 1 coulomb/second
1 faraday  6.022 10 e  96,487 coulombs
23
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Counting Electrons: Coulometry and Faraday’s
Law of Electrolysis
• Faraday’s Law states that during electrolysis, one faraday of
electricity (96,487 coulombs) reduces and oxidizes,
respectively, one equivalent of the oxidizing agent and the
reducing agent.
– This corresponds to the passage of one mole of electrons through
the electrolytic cell.
1 equivalent of oxidizing agent  gain of 6.022 1023 e1 equivalent of reducing agent  loss of 6.022 1023 e-
Counting Electrons: Coulometry and Faraday’s
Law of Electrolysis
• Calculate the mass of palladium produced by the reduction of
palladium (II) ions during the passage of 3.20 amperes of current
through a solution of palladium (II) sulfate for 30.0 minutes.
Counting Electrons: Coulometry and Faraday’s
Law of Electrolysis
• Calculate the volume of oxygen (measured at STP)
produced by the oxidation of water in the previous example.