Redox Reactions
and
Voltaic Cells
Electrochemistry
• Electrochemistry is a branch of
chemistry that investigates the
interchanging between chemical
and electrical energy.
• In electrochemical reactions,
electrons are transferred from one
species to another.
Electrochemistry is described by two
processes, both involving
oxidation/reduction reactions.
1. The generation of an electrical
current from a spontaneous
chemical reaction.
2. The reverse process of using
current to produce a chemical
change.
Remember, in order to identify
redox reactions, you must identify a
change in the oxidation state of the
atoms in a chemical reaction.
Review of Oxidation Numbers
And
Balancing Redox Reactions
Examine the reaction that occurs when
potassium permanganate is added to a
solution of iron (II) chloride:
MnO4- + Fe2+ → Mn2+ + Fe3+
• Notice, this is an oxidation/reduction
reaction:
 Iron looses electrons
Fe2+ → Fe3+ + e Manganese gains electrons
MnO4- + e-→ Mn2+
Oxidation States
(Oxidation Numbers)
• A system of book-keeping for
electrons in molecules or ions
• The imaginary charges atoms
would have in molecules and
real charges in ionic compounds
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Rules of Assigning
Oxidation States
1.The oxidation state of an atom
in an element is zero
2.The oxidation state of a
monatomic ion is the same as
its charge
EXAMPLES
• SF6
• PbS
• CO2
• NO3• NH4+
• K2Cr2O7
Notice however, this reaction is
not balanced.
MnO4- + Fe2+ → Mn2+ + Fe3+
For a reaction to be balanced,
both the number of atoms and the
electrical charges must be equal on
both sides of the reaction.
3. Fluorine is always -1 in its
compounds
4. Oxygen is always -2 in its
compounds
5. Hydrogen is +1 in its covalent
compounds
6. Other oxidation states
calculated from algebraic
sum of known states
1. Identify the oxidation states for each
species in the potassium
permanganate / iron (II) chloride
reaction:
MnO4- + Fe2+ → Mn2+ + Fe3+
Now we see:
Fe2+ → Fe3+ + e- (Oxidation)
MnO4- + e- → Mn2+ (Reduction)
Before balancing a redox
reaction, you must decide whether it
occurs in an acid or basic solution.
Reactions in Acidic Solution
Step 1Write separate equations for the
oxidation and reduction half
reactions from the net ionic equation
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Step 2For each half-reaction:
 Balance all elements except H and O
 Balance O by adding H2O
 Balance H by adding H+
 Balance charge by adding electrons
Step 3Equalize the number of electrons in
the 2 half reactions by multiplying
one or both balanced half reactions
by an integer
2. Balance the potassium
permanganate / Iron (II) chloride
reaction in acidic solution.
8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
Step 3Cancel water molecules appearing
on both sides of the reaction.
Step 4Check that all ELEMENTS and
CHARGES balance
Step 4Add the half-reactions together and
cancel identical species from both sides.
Step 5Check that all ELEMENTS and
CHARGES balance
Reactions in Basic Solution
Step 1Completely balance the reaction as
described previously for reactions in an
acidic solution
Step 2 –
Add OH- ions to both sides equal to
the number of H+ ions present (the goal is
to add hydroxide to one side and make
waters on the other side)
Using the balanced chemical
reaction, it is now possible to see how
electrical energy is produced and the
resulting absorption of that energy
produces a chemical change.
8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
In this reaction, 5 electrons are
produced, and these 5 electrons change
permanganate to water and manganese
(II) in the presence of an acid.
What if we could harness this energy?
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To investigate how electrical
energy can be utilized from from a
chemical reaction, let us investigate
what happens when copper reduces
silver.
3. Write a balanced chemical reaction
for the reaction between copper
metal and silver nitrate.
So, how can we use the electrons
between the process for useful work?
By separating the components of
the reaction, we can create an
electrical gradient that will allow for
the transfer of energy from one
component to the other. It is this
transfer of electrical energy that is
harnessed for useful work.
Cu(s) + 2Ag+ (aq) → 2Ag(s) + Cu2+(aq)
Cu(s) → Cu2+(aq) + 2e(oxidized/ reducing agent)
2Ag+ (aq) + 2e- → 2Ag(s)
(reduced/ oxidizing agent)
1. Electrons are produced by the
oxidation of copper.
2. The electrons then reduce the silver.
DEMONSTRATION
We will separate each component
in the copper/silver reaction into their
respective redox half reactions.
In beaker 1: Cu(s) → Cu2+(aq) + 2eCopper(II) nitrate and solid copper
In beaker 2: 2Ag+ (aq) + 2e- → 2Ag(s)
Silver (I) nitrate and solid silver
DEMONSTRATION
DEMONSTRATION
If we read the balanced redox reaction
correctly:
Notice, this does not happen, as
would be signaled by the presence of a
voltage.
by connecting a wire between the
two metal strips, the copper should
oxidize and dissolve into aqueous
copper ions, sending electrons to the
silver where it can reduce the silver
ions in aqueous solution to solid silver
deposited on the silver strip.
What happened?
The charge did flow to the silver
strip where some silver was reduced,
however, because none of the positive
ions in the silver solution were
replaced, a build up in charge
occurred along the metal surface.
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What we have just built is called a
voltaic cell (Galvanic Cell)
DEMONSTRATION
This build-up in charge produces a
gradient favoring the reverse reaction (a
stone doesn’t roll up hill)
Voltaic cell:
A device in which electron transfer
takes place through an external pathway,
rather than directly between reactants.
The resulting
electron transfer can
be harnessed for
electrical work (i.e.
Lighting a light bulb,
or running a
calculator).
But, if we put in a
salt bridge, negative ions
can flow out of the
reduced compartment
and positive ions can flow
in to complete the circuit.
Hence an electrical
voltage flow, or current.
Electrodes
The metal strips act as electron
carriers and are called “electrodes”
The electrode where
oxidation occurs is called the
anode (-) “electronsproduced”
The electrode where
reduction occurs is called
the cathode (+) “electronsconsumed”
Electron flow is always from either
the source of highest concentration to
source of lowest concentration, or from
the source of greatest electrical
potential to the source of lowest
electrical potential.
ee-
K+(aq)
M1+(aq)
M1
X-(aq)
Anode
M1
M1+ + e-
NO3-(aq)
M2+(aq)
X-(aq)
M2
Cathode
M2+ + e-
We will discuss electrical potential
in our next lecture.
M2
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4. A reaction between iron(II) sulfate
and copper (II) sulfate could be used
to create a useful voltage. Write a
balanced chemical reaction and
draw a labeled diagram showing
how to create a galvanic cell.
Describe what is happening at both
the anode and cathode.
Quite often, chemist use a shorthand
notation to simplify cell descriptions
Voltaic cell line notations:
5. Write the line notation for both the
copper/silver and iron/copper
galvanic cells.
M1 M1+ M2+ M2
A single line is used to separate
phases in each compartment, while a
double line separates the anode
compartment from the cathode
department. The anode and cathode
metals are always listed on the outside.
6. The following reaction is spontaneous:
Cr2O72- + 14H+ + 6I- → 2Cr3+ + 3I2 + 7H2O
A solution of potassium dichromate and
sulfuric acid is place in one beaker with a
solution of potassium iodide in a second
with a joining salt bridge. A nonreacting metallic conductor (i.e. Pt) is
placed in both solutions connected by a
voltmeter. Draw the voltaic cell labeling
the anode, cathode, signs of electrodes,
electron and ion migration, and write the
reactions occurring at both electrodes.
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