Question

Knowing that $ {{K}_{sp}} $ for AgCl is $ 1.0\times {{10}^{-10}}. $ Calculate E for $ A{{g}^{+}}/Ag $ electrode immersed in M KCl at 25? C. $ (E_{A{{g}^{+}}/Ag}^{o}=0.79V) $

• A. 0.80V
• B. 0.208V
• C. 0.37V
• D. 0.18V

Answer 1

Answer: (B)

$ AgCl(s)A{{g}^{+}}(aq)+C{{l}^{-}}(aq) $ $ {{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}] $ $ A{{g}^{+}}=\frac{{{K}_{sp}}}{[C{{l}^{-}}]}=\frac{1\times {{10}^{-10}}}{1}=1\times {{10}^{-10}}M $ $ A{{g}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Ag $ $ E={{E}^{o}}-\frac{0.0591}{1}\log \frac{1}{1\times {{10}^{-10}}} $ $ =0.799-0.591=0.208V $