Electrochemical cell
• Parts of a Voltaic Cell
• The electrochemical cell is actually
composed to two half cells. Each half
cell consists of one conducting
electrode ( a solid ) and a conducting
solution. Each electrochemical cell
consists of two solid conductors we call
electrodes
• Oxidation occurs in one half cell
this produces electrons.
• Reduction occurs in the other half cell
this accepts electrons.
• When the two half cells, consisting of a
metal electrode and a conducting solution
are connected with an external wire, the
strongest oxidizing agent will undergo a
reduction in one half cell and the strongest
reducing agent will undergo an oxidation in
the other half cell.
Key Points
1) Cathode: The electrode (solid) at which
reduction occurs
The charge on the cathode is positive
2) Anode: The electrode (solid) at which the
oxidation occurs
The charge on the anode is negative ,
because it producing electrons
3) Direction of current flow: The electrons
produced at the anode travel along the
external wire to the cathode.
4) Spontaneity: The reaction in an
electrochemical cell (also called a
voltaic cell) is always spontaneous.
5) Ion flow: In the operating
electrochemical cell, positive ions
(cations) in the conducting solutions
migrate toward the collection of
electrons on the positive electrode or
cathode, while negative ions migrates
(move) toward the negative electrode or
anode .
Voltaic
Cells
Anode:
negative
Cathode:
positive
Voltaic cells: Electrochemical cells in which a
spontaneous redox reaction can be harnessed
to produce an electric current.
Oxidation
occurs at the
anode
Reduction
occurs at the
cathode
The electrons are forced to flow through a wire by
separating the reducing agent from the oxidizing agent.
Calculating Voltages
a) Locate the two half cells reactions in the
table of standard reduction potentials.
b) The half reaction that has the higher
reduction potential will reduce and can
be written as you find it in the table.
c) The half reaction that has the lower
reduction potential must be reversed and
written as an oxidation.
d) The E0 value of the higher half reaction is
recorded unchanged from the table.
e) The sign of the E0 value of the lower half
reaction is changed and its value recorded.
f) The two half reactions are balanced for the
number of electrons exchanged but the value
of each E0 remains unchanged.
g) The two half reactions are then added
together and so are the E0 values. ( This
value will always be positive for an
electrochemical cell).
determine its voltage
Au (s) | Au 3+ (aq) ||Cu 2+ (aq) | Cu (s)
Gold
Copper
1) Gold is higher the table and is written
as found in table.
• Au 3+ (aq) + 3 e- ------> Au (s)
E0 = + 1.50 V
2) The copper is lower so it's reaction is
written as an oxidation (sign of E0
reversed ).
• Cu (s) ------> Cu 2+ (aq) + 2 eE0 = -0.34 V
3) Equation is balanced for the exchange
of electrons ( each needs six).
2 Au3+ (aq) + 6e- ----> 2 Au (s) +1.50 V 3
Cu (s) ---->3 Cu2+ (aq) + 6e- -0.34 V
• Note that the size of the values for
the E0 of the reactions are not
changed (even though the equations
are multiplied, E0 is not)
4) Add the two half reactions and the E0
values.
Cross off the electrons
2 Au3+ (aq) + 6e- ----> 2 Au (s)
3 Cu (s) ---->3 Cu2+ (aq) + 6e___________________________________
2 Au3+ (aq) + 3 Cu (s) --> 2 Au (s) + 3 Cu2+ (aq)
E0 = 1.16 v
Spontaneous because it is positive