Question 13
Which of the following shows the increasing order of solubility?
a) KCl<pbs<agcl< li="" style="margin: 0px; padding: 0px;"></pbs<agcl<>
b) KCl<agcl<pbs< li="" style="margin: 0px; padding: 0px;"></agcl<pbs<>
c) PbS<agcl<kcl< li="" style="margin: 0px; padding: 0px;"></agcl<kcl<>
d) AgCl<pbs<kcl< li="" style="margin: 0px; padding: 0px;"></pbs<kcl<>
e) AgCl<kcl<pbs< li="" style="margin: 0px; padding: 0px;"></kcl<pbs<>
Correct Answer:
b) KCl<agcl<pbs< p="" style="margin: 0px; padding: 0px;"></agcl<pbs<>
Explanation:
KCl is highly soluble because its solubility is greater than 0.1M. AgCl is sparingly soluble because its
solubility is less than 0.01 M. PbS is least sparingly soluble becauseits solubility is very much less
than 0.01 M.
Question 14
Calculate the cell potential at 25
0
C for the following cell reaction using Nernst equation.E
o
ox
= -
3.402 V, E
0
red
=0.7996 V
Cu|Cu
2+
(0.024 M)||Ag
+
(0.0048 M)|Ag
1. 0.25 V
2. 0.30 V
3. 0.370 V
4. 0.5 V
5. 0.1V
Correct Answer:
c) 0.370 V
Explanation:
Oxidation: Cu→Cu
2+
+ 2 e
-
E
o
ox
= -(0.340 V) Reduction: Ag
+
+ e
-
→Ag E
o
red
= 0.799 V Overall cell
reaction is Cu(s) + 2 Ag
+
(aq)→ Cu
2+
(aq) + 2 Ag(s) E
0
cell
= E
0
red
+ E
0
ox
= 0.799 V + (-0.340 V) = 0.459
V Nernst equation, E
cell
= E
0
cell
– (0.0256/n) (In
ox
/In
red
) = 0.459
–(0.0256/2) * In [0.024 / (0.0048)
2
] =
0.459
– 0.0128 * In (1043) = 0.459 – 0.0128 * 6.95 E
cell
= 0.370 V
