Engineering Chemistry S Datta

232 ENGINEERING CHEMISTRY
For general purpose, a
M + n = C M +n
So, then the expression is : ∈ = ∈° + 0. 059
n
log 10 C M
+n
[A form of Nernst equation]
When a = 1, then ∈ = ∈°
∈° is known as Standard electrode potential of metal M.
If we construct the following cell:
Zn (–) | Zn +2 || Cu +2 | (+) Cu, E cell
= 1.015 V
For this cell Zn/Zn +2 is anode and Cu/Cu +2 is cathode and at anode, Zn is ionised to Zn +2
i.e., oxidation takes place at anode and at the cathode Cu +2 is neutralised i.e., reduction takes
place at the cathode.
Total cell reaction is:
Zn + Cu +2
Zn +2 + Cu.
Highlights:
According to IUPAC convention,
• The single vertical lines denote electrical contact.
• The double vertical lines denote the junction of the two solutions and this liquid
junction potential is eliminated or minimised by the use of a salt bridge.
• A complete cell consists of two half cells. The right hand half cell is regarded as
the positive electrode and the left hand half cell is regarded as negative electrode.
• Electrons flow through the external wire from the negative electrode to the positive
electrode. So reduction takes at the right hand electrode and oxidation takes place
in the left hand electrode.
• E.m.f. of the cell E = ∈ right
– ∈ left
(when both are reduction potentials)
E = ∈ red
+ ∈ ox .
• Oxidation potential may be taken as reduction potential with a minus sign.
Determination of Standard Potential ∈° of a Metal Electrode
The electrode potentials cannot be determined straight. We are to assign a definite
value of electrode potential for a definite electrode. For example, to determine the electrode
potential, we take the ∈° value for hydrogen electrode to be ‘zero’ i.e., a solution of acid with
([H + ] = 1) and to which hydrogen gas is bubbled at one atmosphere and platinum foil is dipped.
We have a zero potential for this electrode. Now if we want to determine standard electrode
potential of zinc (∈° Zn
) we are to construct the following cell. (See page 227)
Zn|ZnSO 4( solution)
HSO( 2 4 solution)|H 2 (Pt)
( a + = 1)
( a + = 1)
Zn 2
H