Engineering Chemistry S Datta

254 ENGINEERING CHEMISTRY
E° cell
= ∈° Right
– ∈° Left
= 0.8 – (– 0.76) = 1.56 V.
Example 7. Two copper rods are placed in copper sulphate solution of concentration
0.1 M and 0.01 M separately in the form of a cell. Write the scheme of the cell and calculate its
e.m.f. at 298 K.
Sol.
Cell = Cu/Cu +2 (0.01 M) || Cu +2 (0.1 M)/Cu
data,
E cell
= 0.0592
n
E cell
= 0.0592
2
log C 2
, here C
C
2
= 0.1 M and C 1
= 0.01 M
1
01 . 0.0296
log = = 0.0296 V.
00110
. log
Example 8. Calculate the equilibrium constant for Daniel cell at 25°C from the following
R = 8.316 J. F = 96,500 C. S.E.P. (Ox) for Zn electrode.
= 0.765 V and S.E.P. (Ox) for Cu electrode = – 0.337
Sol. ∈° +
Zn/Zn = 0.765 V i.e., ∈° 2 +2 = – 0.765 V
Zn / Zn
and ∈° + = – 0.337 V
Cu/Cu 2 i.e., ∈° Cu
+2 /Cu
= 0.337 V
Cell = Zn/Zn +2 || Cu +2 /Cu
Half cell reactions Cu +2 + 2e Cu .....∈° = 0.337 V
Zn – 2e Zn +2 .....∈° = 0.765 V
Cell reaction: Zn + Cu +2 Zn +2 + Cu (E° cell
= ∈° Red
+ ∈° OX
) = 1.102 V.
∴ K =
+ 2
[Zn ] [Cu]
+ 2
[Zn] [Cu ]
+ 2
∈ + 2 = ∈° + 2
Zn / Zn Zn / Zn
+ 0 . 0592 log [ Zn ]
2 [ Zn]
∈° + 2 = ∈° + 2
Cu / Cu Cu / Cu
+ 0 . 059 log [ Cu ]
2 [ Cu]
At equilibrium, these two electrode potentials will be equal. Therefore,
∈° + + 0.0592
Zn
2 / Zn
2
∈° − ∈°
log [Zn ]
[Zn]
+ 2 + 2
Cu /Cu Z /Zn
+ 2
+ 2
=∈° +
+ 2
Cu /Cu
n
= 0.0592 log
2
or 0.337 – (– 0.765) = 0 . 0592
2
1.102 = 0. 0296 log
+
0.0592
2
[Zn ]
[Zn]
log [Zn
[Cu
log [Cu ]
[Cu]
+ 2 + 2
+ 2
+ 2
+ 2
−
]/[Zn]
]/[Cu]
[Zn ]/[Cu]
+ 2
[Cu ]/[Zn]
0.0592
2
log
[Cu ]
[Cu]