Q.10. Standard reduction potential for Cu
2+/Cu is +0.34 v. Calculate reduction potential at P
H = 14
K
sp of Cu(OH)
2 is 1.0 x 10
-19
Ans: For Cu(OH)
2, K
sp = [Cu
2+][on
-]
2
P
H = 14 So, [H
+) = 10
-14
K
w = [H
+)[OH
-)
[OH
-) =
= 1
So, [Cu
2+] =
= 1.0 x 10
-19 = 1 x 10
-19
E
cell = E
0cell +
log
10
= 0.34 +
log
10[Cu
2+]
= 0.34 +
log
10[1 x 10
-19]
E
cell = - 0.2205 v at P
H = 14
Medium Type
Q.1. In a fuel cell H
2 & O
2 react to produceelectricity. In process H
2 gas is oxidized at anode& O
2 at cathode. If 67.26 of H
2 atSTP reacts in 15 min. What is avg. current produced ? If entire current is used for electrodeposition of Cu from Cu
2+, how many gm of Cu are deposited ?
Cathode: O
2 + 2H
2O + 4e
- 4OH
-
Anode: H
2 + 2OH
- + 4e
- 2H
2O + 2e
-
Ans: Moles of H
2 reacting =
= 3
Eq. of H
2 used = Moles x n-factor
= 3 x 2 = 6
i = 643.33 A
Also, Eq. of H
2 = Eq. of Cu formed
w
Cu = 6 x
= 190.5 g
wt. of Cu deposited = 190.5 g
Q.2. 19 g fused SnCl
2 was electrolysed using inert electrodes. 0.119 g Sn was deposited at cathode. If nothing was given out during electrolysis, calculate ratio of wt. of SnCl
2 & SnCl
2 in fused state after electrolysis (at. et. Sn = 119).
Ans: electrolysis SnCl
2 yieldes:
Anode: 2Cl
- Cl
2 + 2e
-
Cathode: Sn
2+ + 2e
- Sn
Further Cl
2 formed at anode reacts with SnCl
2 to give SnCl
4
SnCl
2 + Cl
2 SnCl
4
Eq. of SnCl
2 lost during electrolysis = Eq. of Cl
2 formed during electrolysis = Eq. of Sn formed.
Eq. of Cl
2 formed =
= 2 x 10
-3
Eq. of SnCl
4 formed = 2 x 10
-3 (
whole Cl
2 react to from SnCl
4)
Now, total loss in Eq. of SnCl
4 during complete course
= Eq. of SnCl
2 lost during electrolysis + Eq. of SnCl
2 lost during reaction with Cl
2
= 2 x 10
-3 + 2 x 10
-3 = 4 x 10
-3
Initial eq. of SnCl
2 left in sol. = 2 x 10
-1 - 4 x 10
-3 = 0.196
Eq. of SnCl
4 formed = 2 x 10
-3 = 0.002
Q.3. Calculate emf of given cell rxn. and Pb(s) + Hg
2So
4 PbSO
4(s) + 2H
g(l)
Given
= 0.126
= - 0.789 v
K
SP of PbSO
4 = 2.43 x 10
-8 K
SP of Hg
2SO
4 = 1.46 x 10
-6.
Ans: Oxidation potential = - REduction Potential
Anode: Pb
Pb
2+ + 2e
2e-
Cathode: Hg
22+ + 2e
- HG
2
E
cell = E
0cell -
log
= (
-
) -
log
= 0.789 - (- 0.126) -
log
K
SP = [Pb
2+][SO
42-] = [Pb
2+]
2 [
[Pb
2+] = [SO
42-]]
[Pb
2+] =
Similarly,
E
cell = 0.789 + 0.126 -
log
= 0.941
Q.4. Two weak acid sol. HA
1 & HA
2 each with same conc. & having P
Ka values 3 & 5 are placed in contact with Hydrogen electrode (1atm, 25
0c) and are interconnected through salt bridge. Find emf of cell.
Ans: Cell is
Pt H
2(1 atm)|HA
2||HA
1|(H
2)(1 atm) P
t
At L.H.S.,
= 0 - 0.059 (P
H)
2
At R.H.S.,
= - 0.059 (P
H)
1
For acid HA
1,
HA
1 H
+ + A
1-
[H
+) = c
=
if
is small)
(P
H)
1 =
PK
a -
log
10c
Similarly,
(P
H)
2 =
PK
a -
logc
E
cell = - 0.059(P
H)
1 + 0.059(P
H)
2
= 0.059[
P
-
P
] =
[5 - 3]
= 0.059 v
Q.5. REduction potential diagram for Cu in acid solution is:
Calculate value of X.
Ans: Given, Cu
2+ + e
- Cu
+;
= 0.15 v ;
........................... (i)
Cu
+ + e
- Cu
+ ,
= 0.5 v ;
........................... (ii)
Cu
2+ + 2e
- Cu ;
= ? ;
........................... (iii)
= - n
F = - 1 x 0.15 F = - 0.15 F
= -n
F = - 1 x 0.5 x F = - 0.5 F
Adding
+
=
- 0.15 F + (- 0.5 F) =
= - 0.65 F
- n
F = - 0.65 F
=
= 0.325 volt.
X = 0.325 Volt