4th Lecture on Electrochemistry Chemistry Part I12th Standard by Rizwana Mohammed

1. The Malegaon High School & Jr. College
Malegaon, (Nashik), 423203
4th Lecture on Electrochemistry
Chemistry Part I, 12th Science
By
Rizwana Mohammad

2. Quantitative aspects of electrolysis:
a. The mass of reactant consumed or the mass of product formed at
an electrode during electrolysis can be calculated by knowing
stoichiometry of the half reaction at the electrode.
i. Calculation of quantity of electricity passed:
Q(C) = I(A) X t(s)
ii. Calculation of moles of electrons passed:
Moles of electrons actually passed =
Q(C)
96500 (
C
mol e
)
iii. Calculation of moles of product formed:
To simplify further we introduce the entity mole ratio given by
Mole ratio =
moles of product formed in the half reaction
moles of electrons required in the half reaction
Therefore,
Mole of product formed = moles of electrons actually passed X mole ratio
=
Q(C)
96500 (
C
mol e
)
X mole ratio
=
I A X t(s)
96500 (
C
mol e
)
X mole ratio

3. iv. Calculation of mass of product:
Mass of product
W = moles of product X molar mass of product
=
I A X t(s)
96500 (
C
mol e
)
X mole ratio X molar mass of product
b. Suppose two cells containing different electrolytes are connected
in series. The same quantity of electricity is passed through them. The
masses of the substances liberated at the electrodes of the two cells
are related as given below:
The mass of the substance produced at the electrode of the first cell is
given by
W1 =
Q(C)
96500 (
C
mol e
)
X (mole ratio)1 X M1
Hence,
Q
96500
=
W1
molar ratio 1
X M1

4. Similarly mass of substance liberated at the electrode of second cell is
W2 in the equation,
Q
96500
=
W2
molar ratio 2
X M2
M1 and M2 are the molar masses of substance produced at the
electrodes of cell 1 and 2
W1
molar ratio 1
X M1
=
W2
molar ratio 2
X M2

5. Galvanic or voltaic cell:
In galvanic or voltaic cell, electricity is generated through the use of
spontaneous chemical reactions.
• A galvanic cell is made up of two half cells.
• Each half cell consists of a metal strip immersed in the solution of its own ions of
known concentration.
• The two half cells are placed in separate containers.
• The metal plates called electrodes are connected through voltmeter by a
conducting wire for transfer of electrons between them.
• To complete the circuit the two solutions are connected by a salt bridge.
Salt bridge:
• It is a U tube containing a saturated solute of an inert electrolyte such as KCl or
NH4NO3 and 5% agar solution.

6. • Salt bridge is prepared by filling U tube with hot saturated solution
of the salt and agar solution allowing it to cool.
• The cooled solution sets into a gel which does not come out on
inverting the tube.
• It is kept dipped in distilled water when not in use.
Functions of salt bridge:
• It provides an electrical contact between two solutions and thereby
completes the electrical circuit.
• It prevents mixing of two solutions.
• It maintains electrical neutrality in both the solutions by transfer of
ions.
Formulation or short notation of galvanic cells: A cell composed of
Mg (anode) and Ag (cathode) consists of two half cells, Mg2 (1M) |
Mg(s) and Ag (1M) | Ag(s) The cell is represented as
Mg(s) | Mg2 (1M) || Ag (1M) | Ag(s)
+
++
+

7. Writing The of cell reaction:
The cell reaction corresponding to the cell notation is written on the
assumption that the right hand side electrode is cathode (+) and left
hand side electrode is anode (-).
Consider the cell,
Ni(s) | Ni2 (1M) || Al3 (1M) | Al(s)
The oxidation at anode is
Ni(s) → Ni2 (1M)(aq) + 2e
The reduction half reaction is
Al3 (1M) + 3e → Al(s)
The net cell reaction is, thus,
3Ni(s) →3Ni2 (1M)(aq) + 6e (oxidation half reaction)
2Al3 (1M)(aq) + 6e → 2Al(s)(reduction half reaction)
3Ni(s) + 2Al3 (1M)(aq) → 3Ni2 (1M)(aq) + 2Al(s) overall cell reaction.
+ +
+ -
-+
+
+
-
-
+ +
e-