Question

The standard reduction potential of two reactions are given.
$AgC{l}_{\left(s\right)}+e^{-}\to A{g}_{\left(s\right)}+C{l}_{\left(aq\right)}^{-}$; $E^{\ominus }=0.22$V (i)
$A{g}_{\left(aq\right)}^{+}+e^{-}\to A{g}_{\left(s\right)};E^{\ominus }=0.80$V (ii)
The solubility product of $AgCl$ under standard conditions of temperature is:

• A. $1.6\times {10}^{-5}$
• B. $1.5\times {10}^{-8}$
• C. $3.2\times {10}^{-10}$
• D. $1.5\times {10}^{-10}$

Answer 1

The correct option is D $1.5\times {10}^{-10}$

(cathode)

$AgC{l}_{\left(s\right)}+e^{-}\to A{g}_{\left(s\right)}+C{l}_{\left(aq\right)}^{-}$; $e^{\ominus }=0.22$V

$A{g}_{\left(aq\right)}^{+}+e^{-}\to A{g}_{\left(s\right)};E^{\ominus }=0.80$V canode

$E_r^o=E_c^o-E_a^o=-0.58V$

$E=Eo-\frac{0.0591}{n}lo{g}_{10}\frac{\left[A\right]}{\left[B\right]}$

here $E=0$ because at equilibrium $AG=0$

So $E=0$

So

$0=Eo-\frac{0.0591}{n}lo{g}_{10}ksp$ (put value)

$0.58=-\frac{0.0591}{1}lo{g}_{10}ksp$ $(n=1)$

$lo{g}_{10}ksp=-9.8138$

Take cmtilog

$ksp=1.5\times {10}^{-10}$