ElectrochemistryHard
Question
The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is
Options
A.0.5 mmol
B.1.0 mmol
C.2.0 mmol
D.2.5 mmol
Solution
AgCl + e- → Ag + Cl- Eo = 0.2 V
Ag → Ag+ + e- Eo = - 0.79 V
_________________________________
AgCl
Ag+ + Cl- Eo = - 0.59 V
Eo =
log K ⇒ - 0.59 =
log KSP ⇒ KSP = 10-10
Now solubility of AgCl in 0.1 M AgNO3
S (S + 0.1) = 10-10 ⇒ S = 10-9 mol/L
Hence 1 mole dissolves in 109 L solution
hence in 106 L amount that dissolves in 1 m mol.
Ag → Ag+ + e- Eo = - 0.79 V
_________________________________
AgCl
Ag+ + Cl- Eo = - 0.59 VEo =
log K ⇒ - 0.59 = log KSP ⇒ KSP = 10-10 Now solubility of AgCl in 0.1 M AgNO3
S (S + 0.1) = 10-10 ⇒ S = 10-9 mol/L
Hence 1 mole dissolves in 109 L solution
hence in 106 L amount that dissolves in 1 m mol.
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