UNIT 5 – ELECTROCHEMISTRY
- Oxidation and Reduction
o Oxidation is the loss of electrons, reduction is the gain of electrons
o If 1 atom or ion is oxidized in a reaction, another atom or ion must be reduced
 i.e. redox reactions
o Typically redox reactions are synthesis, decomposition, combustion, and single
displacement
 We take the redox reactions/ionic equations and write out all soluble ions
separately
 Any ion that appears on the reactant and product side are called
spectator ions and are not involved in the chemical reaction
o Ex. Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)
o Zn + Cu+2 + SO4-2 → Cu + Zn+2 + SO4-2
 Therefore omit SO4-2 (spectators)
 Net ionic equation:

o
o
o
Zn (s) + Cu+2 (aq) → Cu (s) + Zn+2 (aq)
 Cu+2 to Cu is a gain of 2e-, reduction
 Zn to Zn+2 is a loss of 2e-, oxidation
Zn can be considered an reducing agent because it donates
electrons and undergoes oxidation
Cu+2 can be considered an oxidizing agent because it accepts
electrons and undergoes reduction
Spontaneity of redox reactions:
 Write the net ionic equation
 Add arrows to indicate the gain and loss of electrons
 Using a data table identify which of the 2 metals is the stronger reducing agent
and which is the stronger oxidizing agent
 If stronger reducing agent is losing electrons and the stronger oxidizing agent is
gaining electrons
 The reaction will proceed spontaneously as written
 If the stronger reducing agent has gained electrons and stronger oxidizing agent
has lost electrons
 Reaction will not proceed spontaneously in the forward direction

Ex. Sn (s) + PtSO4 (aq) → SnSO4 (aq) + Pt (s)
 Sn + Pt+2 + SO4-2 → Sn+2 + SO4-2 + Pt
 Net ionic:
o Sn + Pt+2 → Sn+2 + Pt
o
o

-
Sn to Sn+2 loses electrons, oxidation; and Sn is the stronger
reducing agent
Pt+2 to Pt gains electrons, reduction, and Pt+2 is the stronger
oxidizing agent
Ex. Fe (s) + MgCO3 (aq) → Mg (s) + FeCO3 (aq)
 Fe + Mg+2 + CO3-2 → Mg + Fe+2 + CO3-2
 Net ionic:
o Fe + Mg+2 → Mg + Fe+2
o Fe to Fe+2 loses 2 electrons, oxidation; Fe+2 is the stronger
reducing agent
o Mg+2 to Mg gains 2 electrons, reduction; Mg is the stronger
oxidizing agents
Redox Reactions Involving Ionic Compounds
o You can balance redox reactions with a half-reaction which is an equation that describes
the changes in only the compound that is oxidized or only the compound that is reduced
o Balancing equations using half-rxn’s:
 Determine the lowest common multiple of the numbers of electrons in the
oxidation and reduction half reactions
 Multiply 1 or both half-rxn’s by the number that will bring the number of
electrons to the lowest common multiple
 Add the balanced half-rxn’s
 Cancel the electrons and any other identical molecules/ions present on both
sides of the equation
 If spectator ions were removed when forming half-rxn’s, add them back to the
equation
 Ex. Zn (s) + CuSO4 (aq) Cu (s) + ZnSO4 (aq)
o Oxidation half-rxn:
 Zn → Zn+2 + 2eo Reduction half-rxn:
 Cu+2 + 2e- → Cu
o Sum of half-rxn’s:
 Zn + Cu+2 + 2e- → Zn+2 + Cu + 2eo Net ionic equation:
 Zn + Cu+2 → Cu + Zn+2
o Ionic equation:
 Zn + Cu+2 +SO4-2 → Cu + Zn+2 +SO4-2
 Notice that atoms and charges are balanced

o
Balancing half-rxn’s occurring in acidic or basic solutions:
 Write unbalanced half-rxn’s that show the formulas for the given reactants and
products
 Balance any atoms other than oxygen and hydrogen first
 Balance any oxygen atoms by adding water
 Balance any H atoms by adding H ions
 If reaction is acidic go last step
 If reaction is basic continue with the rest of the steps
 Add to both sides the same number of hydroxide ions as the same number of H
ions already present
 For any H and OH ions on the same side of an equation combine to form water
 Cancel any water present on both sides
 Balance the charges by adding electrons
 Ex. S (s) + HNO3 (aq) → SO2 (g) + NO (g) +H2O (l)

o
Ex. 2K (s) + Cl2 (g) → KCl (aq)
o Oxidation half-rxn:
 K → K+ + eo Reduction half-rxn:
 Cl2 (g) + 2e- → 2Cl Need to multiply oxidation reaction by 2 to
balance electrons
o 2K → 2K+ + 2eo Sum of half-rxn’s:
 2K + Cl2 + 2e- → 2K+ + 2Cl- + 2eo Net ionic equation:
 2K + Cl2 → 2K+ + 2Cl Or:
o 2K + Cl2 → 2KCl
Ex. CN- (aq) + MnO4- (aq) → CNO- (aq) + MnO2 (s)
Some atoms of an element can undergo oxidation and other atoms of the same element
can undergo reduction in a single reaction called a disproportionation reaction
 Ex. Cu2O (aq) + H2SO4 (aq) → Cu (s) + CuSO4 (aq) + H2O (l)
 2Cu+ + O-2 + 2H+ + SO4-2 → Cu + Cu+2 + SO4-2 + H2O
 Net ionic:
o 2Cu+ + O-2 + 2H+ → Cu + Cu+ + H2O
o
-
Cu+ + Cu+ → Cu + Cu+2
 Oxidation half-rxn:
 Cu+ → Cu+2 + e Reduction half-rxn:
 Cu+ + e- → Cu
Redox and Molecular Compounds
o An oxidation number is a number equal to the charge that an atom would have if no
electrons were shared but instead were possessed by the atom with the greatest EN
 Ex. In water, oxygen is more EN than H, therefore, the oxygen atoms attract the
shared electrons more strongly than H
o If electrons are equally shared between atoms, i.e. neither gain or lose electrons, each
atom will have an oxidation number of zero, ex. Cl2 (g)
o Rules:
 A pure element has an oxidation number of zero
 Ex. Na (s), Br2 (l), P4 (s)
 Oxidation number of an element in a monatomic ion equals the charge on the
ion
 Ex. Al+3 – oxidation is +3, Se-2 – oxidation is -2
 Oxidation number of H in compounds is +1, but in metal hydrides it is -1
 Oxidation number of O in compounds is -2
 Exceptions with peroxides, H2O2, and superoxides, OF2
 In molecular compounds that don’t have H or O, the more EN element is
assigned a number equal to the negative charge it usually has
 Ex. PCl3 is -1, CS2 is -2
 The sum of the oxidation numbers of all the atoms in a neutral compound is
zero
 Ex. CF4 – fluorine is -1 for each atom which means C will be +4;
therefore +4-4=0
 The sum of the oxidation numbers of polyatomic ions equals the charge on the
ion
 Ex. NO2- - oxygen is -2 = -4; therefore N is +3 to end up with a -1 overall
charge
o
Ex. HClO4
o
Ex. Cr2O7-2

Possible ot have non-integer oxidation numbers
o Ex. Fe3O4-2
o
o
o
C3H6O
If oxidation numbers increase oxidation occurs, if oxidation number decreases reduction
occurs
 Ex. 2H2 (g) + O2 (g) → 2H2O (l)
 H and O have oxidation numbers of zero, H in water has an oxidation of
+1, and O in water has an oxidation of -2
o Therefore H gas is the reducing agent and O gas is the oxidizing
agent

Ex. CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)

Ex. 3HNO2 (aq) → HNO3 (aq) + 2NO (g) + H2O (l)
Balancing equations using oxidation numbers
 Need to ensure that total increase in the oxidation number of the oxidized
element(s) equals the total decrease in oxidation number of the reduced
element (s)
 i.e. total electrons lost by reducing agent must be total number gained
by oxidizing agent
 Steps:
 Write an unbalanced equation
 Assign oxidation numbers
 Identify atoms undergoing a change in oxidation numbers
 Determine numerical values of the increase and decrease in oxidation
numbers
 Determine the lowest common multiple of the number of electrons lost
by the reducing agent and gained by the oxidizing agent
 Apply the numbers in the previous step to balance the increase in
oxidation number of 1 atom with the decrease in oxidation number of
another atom
 Balance all atoms


o If in acidic or basic solutions balance everything except H and O
For reactions in acidic solution balance any O by adding water and
balance any H by adding H+
For reactions in basic solution complete the previous step and then add
OH- ions to both sides to neutralize the H+
o Ex. Cr2O7-2 (aq) + C2H5OH (aq) → Cr+3 (aq) + CO2 (g)
o
-
Ex. ClO2 + SbO2- → ClO2- + Sb(OH)2-
Galvanic Cells
o A galvanic cell is a device that uses redox reactions to transform chemical energy into
electrical energy
 Ex. Zn (s) + Cu+2 (aq) → Zn+2 (aq) + Cu (s) + energy
o The key is to prevent direct contact between the reactants in the redox reaction
o Can then connect wires to allow electric current of flow out
o The metals are placed in aqueous solution which produces electrolytes; substances that
conduct electric current when dissolved in water
o The solutions are connected by a salt bridge
 An electrical connection between half cells that contains an electrolyte solution
 Allows current to flow and helps prevent contact between oxidizing
agent and reducing agent
o Ex. KCl
 The ends of the bridge are plugged with a porous material (cotton or glass wool)
 Ions diffuse through the plugs
o The plug prevents solutions from mixing
o A wire connects the 2 metals which act as electrodes (conductors)
 2 types:
 Anode – where oxidation occurs, negative
 Cathode – where reduction occurs, positive
o The salt bridge is selected so that is does not interfere in the reaction by forming a
precipitate
o The current will flow until the concentration of the solutions and changes in the
electrodes are so great that the process cannot continue

Net Ionic: Cu+2 (aq) + Zn (s) → Zn+2(aq) + Cu (s)

o
In order to form certain cells we must use inert electrodes (not solid metals, but
are still electrolytes or gases) which consist of neither the reactant nor product
but provides a surface on which redox reactions can occur
We can use cell notation to give a shorthand method for representing galvanic cells
 i.e. Zn (s) | Zn+2 (aq) || Cu+2 (aq) | Cu (s)
 Anode is usually on left side and cathode on the right
 Each single vertical line reps a phase boundary between the electrode
and the solution in the half-cell
o Double vertical lines rep that salt bridge between the half-cells
 Inert electrodes are also shown in the notation
 i.e.Pb (s) | Pb+2 (aq) || Fe+3 (aq), Fe+2 (aq) | Pt (s)
o We use commas to indicate they are in
the same phase


Whenever a separation of charge exists a force acts on the charged particles
 Therefore they have potential energy
The electrical potential difference between 2 points is the amount of energy
that a unit charge would gain by moving from 1 point to the other
 In galvanic cells the 2 points are the electrodes
o A voltmeter measured the potential difference
 The electrical potential difference between the electrodes is called the
cell potential
o Cell potential depends on:
 Nature of oxidizing and reducing agent
 Concentration of salt solutions in the half-cell
 Temperature of solutions
 Atmospheric pressure
o
o
o
The standard cell potential, symbolized by E°cell, is the difference between electrodes
when the concentrations of the salt solution are 1.0 mol/L, the atmospheric pressure is
101.325 kPa (1.0 atm), the electrodes are pure metals, the temperature is 25°C, and
there is no electric current flowing
 When all these factors are constant, only the nature of the oxidizing and
reducing agent is a factor
You can define a half-cell potential only by choosing an arbitrary reference point
 Half reaction between H gas and H ions
o H2 (g) → 2H+ (aq) + 2e Platinum electrode is immersed in a 1.o mol/L solution of monoprotic
acid with H gas being bubbled past the electrode at 1.0 atm
 Half cell potential of H half cell is defined as zero
 Therefore half cell potential of all other cells is defined as the potential
difference between the H half cell and the chosen cell
 The H half cell can be the reducing or oxidizing half cell depending on
whether H is a stronger or weaker or reducing agent
o Ex. Zn (s) | Zn+2 (aq) || H+ (aq) | H2 (g) | Pt (s)
 Zinc half cell potential is 0.76 V
 Zinc is oxidized to give 0.76 V, however when in
a reaction with a different half cell the potential
difference value will switch signs if reduction is
now occurring
 i.e. Zn+2 (aq) + 2e- → Zn (s) E° = - 0.76 V
Cu+2 (aq) + 2e- → Cu (s) E° = + 0.34 V
To calculate standard cell potential
 E°cell = E°cathode – E°anode
 A positive cell potential means reaction will proceed spontaneously in direction
indicated
 Reaction will always proceed spontaneously if reducing agent on left
side is stronger than on the right
 If cell potential is negative, reaction will not proceed spontaneously in the
direction indicated
 The reaction actually proceeds in the reverse direction
o Ex. 2I- (aq) + Br2 (l) → I (s) + 2Br- (aq)
o
Non-spontaneous reactions
 A cell that uses an external source of electrical energy to drive a nonspontaneous redox reaction is called an electrolytic cell
 Converts electrical energy into chemical energy in a process called
electrolysis
 Similar setup to a galvanic cell, just works in an opposite way (uses a battery
instead of a voltmeter)
 If E cell is negative the reaction is not spontaneous
 Ex. Electrolysis of water
o Some water molecules are oxidized at the anode and other
water molecules are reduced at the cathode
 Oxidation:
 2H2O (l) → O2 (g) + 4H+ (aq) + 4e Reduction:
 2H2O (l) + 2e- → H2 (g) + 2OH- (aq)
o
Rusting is an example of corrosion which is a spontaneous redox reaction of materials
with substances in their environment
Many metals are easily oxidized by oxygen
Rust is hydrated iron (III) oxide, Fe2O3 · xH2O (where x can be any number of water
molecules)
The surface of iron behaves as though it consists of many galvanic cells in which
electrochemical reactions form rust
 Iron acts as the anode, and the cathode is inert and may be an impurity such as
soot
Water, in the form of rain, reacts with carbon dioxide in the air to form carbonic acid,
H2CO3 (aq)
 This weak acid partially dissociates into ions
 H2CO3 (aq) ↔ H+ (aq) + HCO3- (aq)
o Therefore carbonic acid is an electrolyte for the corrosion
process
 Oxidation half cell:
o Fe (s) → Fe+2 (aq) + 2e Reduction half cell:
o O2 (g) + H2O (l) + 4e- → 4OH- (aq)
 Overall:
o 2Fe (s) + O2 (g) + 2H2O (l) → 2Fe+2 (aq) + 4OH- (aq)
 Nothing stops the Fe+2 and OH- ions from mixing
 So; 2Fe (s) + O2 (g) + 2H2O (l) → 2Fe(OH)2 (s)
o Fe(OH)2 undergoes further oxidation with oxygen in the air
 4Fe(OH)2 + O2 + H2O → 4Fe(OH)3
o
o
o
o

o
o
Fe(OH)3 readily breaks down to form iron (III) oxide,
 Fe2O3 · xH2O
Other metals do not corrode to the same extent and their metal
oxide corrosion layer acts as a protective layer
 Ex. Al2O3, Cr2O3, MgO
Corrosion prevention:
 Try and keep the metal cool, dry and clean
 Paint an iron object, which helps prevent air and water from touching the metal
surface
 Can also use grease, oil, or plastic
 Can put a layer of another metal that is more resistant to corrosion
 Iron can form an alloy with a different metal
 Galvanizing is a process in which iron is covered by a protective layer of zinc
 Zinc is more easily oxidized, therefore it becomes the anode and will
completely react before iron is affected
o Iron also acts as the cathode during this reaction