2. Continue
• The changes in which electrical energy is produced
as a result of chemical change. The devices used to
produce electrical energy from chemical reactions
are called electrical cells, galvanic or voltic cells.
• In these cells, oxidation and reduction reaction
reactions occur in separate containers called half
cells and the red-ox reaction is spontaneous
3. Cell potential
• Cell potential, also known as electromotive force
(EMF), is a fundamental concept in
electrochemistry that represents the driving force
or electrical potential difference between the
electrodes of an electrochemical cell.
• It is a measure of the ability of a chemical reaction
to produce or consume electrical energy.
• In an electrochemical cell, chemical reactions occur
at the electrodes, which are typically made of
different materials.
4. Continue
• These reactions involve the transfer of electrons
between the electrodes and the surrounding
electrolyte solution.
• The cell potential is a measure of the energy
difference between the reactants and products of
these electrode reactions.
• The cell potential is represented by the symbol Ecell
and is usually measured in volts (V).
• It is determined by the difference in the standard
electrode potentials of the two half-cells involved
in the cell.
5. Continue
• The flow of electric current
in an electrochemical cell
indicates that a potential
difference exists between
two electrodes.
• Due to separation of
charges between the
electrode and the solution,
an electrical potential is set
up between metal
electrode and its solution
6. Continue
• The difference between the electrode potentials of
two half cells is called cell potential. It is known as
electromotive force (EMF) of the cell if no current is
drawn from the cell.
• E°cell = E°reduction - E°oxidation
• Ecell =Ecathode -Eanode.
• Since anode is put on left and cathode on right,
therefore it follows Ecell =ER -EL
7. More about cell potential
• The cell potential provides important information
about the feasibility and direction of the
electrochemical reaction.
• If the cell potential is positive (Ecell > 0), the
reaction is thermodynamically favorable, and
electrons will flow from the anode (the electrode
where oxidation occurs) to the cathode (the
electrode where reduction occurs).
• This generates an electric current in the external
circuit.
• On the other hand, if the cell potential is negative
(Ecell < 0), the reaction is not spontaneous, and an
external power source is required to drive the
reaction in the opposite direction.
8. Standard hydrogen electrode
• Since a half cell in an electrochemical cell can work
only in combination with the other half cell and
does not work independently, it is not possible to
determine the absolute electrode potential of an
electrode.
• We can, therefore, find only the relative electrode
potential.
• This difficulty can be solved by selecting one of the
electrodes as a reference electrode and arbitrarily
fixing the potential of this electrode as zero.
• For this purpose, reversible hydrogen electrode has
been universally accepted as a reference electrode.
It is called standard hydrogen electrode (S.H.E) or
9. Continue
• It consists of platinum wire
sealed in a glass tube and has a
platinum foil attached to it.
• The foil is coated with finely
divided platinum and acts as
platinum electrode.
• It is dipped into an acid
solution containing H+ ions in 1
M concentration (IM HCI).
• Pure hydrogen gas at 1
atmospheric pressure is
constantly bubbled into
solution at constant
10. SHE
• The standard electrode potential (E°) is a
measure of the tendency of a half-reaction to
occur under standard conditions (25°C, 1 atm
pressure, and 1 M concentration).
• It is expressed in volts. The standard electrode
potential values are tabulated and can be found
in reference tables
11. Factors affects EMF of the cell
• The EMF of the cell depends on
• The nature of the electrodes.
• Temperature.
• Concentration of the electrolyte solutions
12. Nernst equation
• The cell potential can be calculated using the
Nernst equation, which takes into account the
concentrations of the reactants and products in the
half-cells:
• Ecell = E°cell - (RT/nF)ln(Q) ,where
• Ecell is the cell potential
• E°cell is the standard cell potential
• R is the gas constant (8.314 J/(mol·K))
• T is the temperature in Kelvin
13. Continue
• n is the number of electrons transferred in the
balanced equation for the cell reaction
• F is Faraday's constant (96,485 C/mol)
• Q is the reaction quotient, which is the ratio of the
concentrations of the products to the
concentrations of the reactants raised to their
stoichiometric coefficients.
• Nernst equation It relates the cell potential (Ecell) of
an electrochemical cell to the standard electrode
potential (E°cell) and the concentrations of the
reactants and products involved in the cell reaction.
14. Continue
• From standard Gibbs free energy change (ΔG°) for
an electrochemical cell reaction ΔG° = -
nFE°cell
• The relationship between Gibbs free energy change
and equilibrium constant (K) is given by:ΔG° = -
RTlnK
• By equating these two equations and rearranging,
we get -nFE°cell = -RTlnK
• Dividing both sides by -nF, we obtain
E°cell = (RT/nF)lnK
15. reaction quotient (Q)
• Now, we introduce the concept of reaction
quotient (Q), which is the ratio of the
concentrations of the products to the
concentrations of the reactants raised to their
stoichiometric coefficients:
• Q = [C]^c [D]^d / [A]^a [B]^b
• where:
• [A], [B], [C], [D] are the concentrations of species A,
B, C, and D, respectively
• a, b, c, d are the stoichiometric coefficients of A, B,
C, and D in the balanced equation for the cell
16. Continue
• Since at equilibrium, the reaction quotient equals
the equilibrium constant (Q = K), we can rewrite
the equation as:
• E°cell = (RT/nF)lnK = (RT/nF)lnQ
• Finally, we introduce the concept of the Nernstian
form of the equation by replacing E°cell with Ecell,
which represents the cell potential under non-
standard conditions:
• Ecell = E°cell - (RT/nF)lnQ
17. Task
• In pairs of threes think
about the application
of cell potential in our
every day life One
minute
18. Examples
• Problem 1:Consider a galvanic cell with the
following half-reactions
• Zn(s) Zn2+(aq)+2e- (E° = -0.76 V)
• Cu2+(aq) + 2e- Cu(s) (E° = 0.34V).Determine the cell
potential and the direction of electron flow in this
cell.
• 1.10 V ( Answer)
19. Answer
• The cell potential (Ecell) can be determined by
subtracting the reduction potential of the anode
from the reduction potential of the cathode:
• Ecell = E°cathode - E°anode
• Ecell = 0.34 V - (-0.76 V) = 1.10 V
• The positive cell potential indicates that the
reaction is spontaneous, and electrons flow from
the anode to the cathode
20. Problem 2:
• A concentration cell is constructed using two half-
cells with Cu2+ ions. One half-cell has Cu2+(aq)
with a concentration of 1 M, and the other half-cell
has Cu2+(aq) with a concentration of 0.1 M.
Determine the cell potential and the direction of
electron flow.
• Answer
21. Answer 2
• In a concentration cell, the difference in ion concentration
drives the electron flow. The cell potential (Ecell) can be
calculated using the Nernst equation:
• Ecell = E°cell - (RT/nF) * ln(Q)
• Since both half-cells contain the same species (Cu2+), E°cell
is 0. The Nernst equation simplifies to:
• Ecell = (RT/nF) * ln([Cu2+]1/[Cu2+]2)
• Using the given concentrations:
• Ecell = (0.059 V/n) * ln(1 M / 0.1 M) = 0.059 V/n * ln(10)
• The direction of electron flow depends on the higher
concentration of Cu2+. In this case, electrons will flow from
the 0.1 M Cu2+ solution to the 1 M Cu2+ solution
22. Problem 3
• A voltaic cell is constructed with the following half-
reactions:
• Fe3+(aq) + 3e- → Fe(s) (E° = -0.04 V)
• Mn2+(aq) + 2e- → Mn(s) (E° = -1.18 V)
• Determine the standard cell potential and the
oxidizing and reducing agents in the cell
23. Answer 3
• The standard cell potential (E°cell) can be
calculated by subtracting the reduction potential of
the anode from the reduction potential of the
cathode:
• E°cell = E°cathode - E°anode
• E°cell = -1.18 V - (-0.04 V) = -1.14 V
• Since the standard cell potential is negative, the
reaction is non-spontaneous under standard
conditions.
• In this cell, Fe3+(aq) acts as the oxidizing agent (it is
reduced) and Mn2+(aq) acts as the reducing agent
