(3) To predict spontancity of redox rxn.:
For redox rxn. to be spotaneous, EMF of cell must be +ve.
Illustration: Cen Nickel vessel used to store CuSO
4 sol ?
= - 0.25 v
= 0.34 v
Ans: NI
2+/Ni || Cu
2+/Cu
E
0cell =
-
= 0.34 - (- 0.25) = 0.59 v
EMF comes out to be +ve. So, CuSO
4 will reacts with Ni. So, cannot store in nickel vessel.
Nerst's Equation:
M
n+ + ne
- M
Nerst eq. as follows
E = E
0 -
E
electrode potential at given conc. of M
n+ ions & Temp. T.
R
gas constant.
T
TEmp. in K.
F
1 Faraday.
n
no. of electrons involved in electrode rxn.
For pure solids or liquids or gases, at 1 atm, molar conc. taken as unity.
i.e. [M) = 1
So, E = E
0 -
Putting value of R, T & F R = 8.314, T = 298 K, F = 965000 coulom
......................................... (i)
At 298 K Temp.
Nerts eq. for EMF of cell:
Zn(s) + Cu
2+(ag)
Zn
2+(ag) + Cu(s)
For Zn(s) | Zn
2+ (ag.) electrode,
Zn
2+(ag) + 2e
- Zn(s)
=
+
[Zn
2+(ag.)]
For Cu/Cu
2+ electrode
Cu
2+ + 2e
- Cu(s)
=
+
ln[Cu
2+(ag.)]
Since oxidation takes place at Zn & reduction takes place at Cu electrode.
E
cell = E
cathode - E
anode
=
-
= {
+
ln[Cu
2+(ag.)]} - {
+
ln[Zn
2+(ag.)]}
= (
-
) +
n
= E
0cell -
ln
For cell rxn,
aA + bB
xX + yY
E
cell = E
0cell -
ln
At 296 K
Illustration: Calculate emf of cell
Cr|Cr
3+(0.1 M) || Fe
2+(0.01 M) | Fe
= - 0.75 v
= - 0.45 v
Ans: 2Cr + 3Fe
2+ 2Cr
2+ + 3Fe n = 6
E
cell = E
0cell -
= (
-
) -
= (- 0.45 + 0.75 v) -
E
cell = 0.2606 v
Eqm. constant form Nerst Eq.:
Zn + Cu
2+ Zn
2+ + Cu
eqn. const =
= KC
Eqn. occurs when E
cell = 0
0 = E
0cell -
log
E
0cell =
logK
c
At 296 k,
E0cell = logKc |
Illustration: Calculate eqm. constant for rxn.
Cu(s) + 2Ag
+ Cu
2+ + 2Ag
Ans: Here n = 2
E
0cell =
logK
c
-
=
logK
c
0.8 - 0.34 =
logK
c
K
c = 3.688 x 10
15
Dibb's Free Energy & Cell Potential:
-
G
0 = nFE
0cell
But for rxn. at eqm
E
0cell =
ln K
c
G
0 = - nF x
ln K
c
G0 = - 2.303 RT ln Kc |
Illustration: Calculate standard free energy change
Zn(s)|Zn
2+ || Cu
2+(1M)|CU(s)
= - 0.76 v
= + 0.34 v F = 96500 C mol
-1
Also calculate eqm constant for rxn.
Ans: mrxn. n = 2
G
0 = - nFE
0cell
E
0cell = 0.34 - (- 0.76) = 1.1 v
G
0 = - 2 x 96500 x 1.1 = - 212.3 KJ/mol
G
0 = - 2.303 RT logK
c
- 212300 = - 2.303 xn 8.314 x 298 x logK
c
K
c = 1.6 x 10
37