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)

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