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

• A. 0.5 mmol
• B. 1.0 mmol
• C. 2.0 mmol
• D. 2.5 mmol

Answer 1

Answer: (B)

1.0 mmol

Answer 2

$\text{AgCl + }\text{e}^{-}\overset{\quad\quad}{\rightarrow}\text{Ag + C}\text{l}^{-}\ E^{0}\text{ = 0.2 V}$

$\text{Ag }\overset{\quad\quad}{\rightarrow}\text{A}\text{g}^{+}\text{ + }\text{e}^{-}\ E^{0}\text{ = - 0.79 V}$

$E^{0}\frac{0.059}{n}\log{}K \Rightarrow - 0.59 = \frac{0.059}{1}\log K_{SP} \Rightarrow K_{SP} = 10^{- 10}$ Now solubility of AgCl in 0.1 M AgNO3

$\text{S }(\text{S + 0.1})\text{ = 1}\text{0}^{\text{-10}}\ \text{==>} \text{S = 1}\text{0}^{\text{-9}}\text{ mol/L}$

Hence 1 mole dissolves in 109 L solution hence in 106 L amount that dissolves in 1 m mol.