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Knowing that ...

Question

Knowing that $K_{s p}$ for AgCl is $1.0 \times 10^{- 10}$ , Calculate E for a silver/silver chloride electrode immersed in $1.00 M K C l a t 25^{\circ} C$ . $E_{{A g}^{+} | A g}^{-} = 0.799 V$ .

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Solution

$A g + C l^{-} \to A g C l + e^{-}$
$E = E^{0} -$ $\frac{0.0591}{1}$ $log \frac{1}{[{C l}^{-}]}$
$= 0.799 - 0.0591 log (\frac{1}{105})$ $= 0.5035 V .$

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