Question

(I) For the reaction,

$A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq{)}_{\to }AgCl\right(s)$

Write the cell representation of above reaction and calculate $E^o$ at 298 K.

Given:

Species	$\Delta G_f\left(kJ/mol\right)$
$A{g}^{+}\left(aq\right)$	+77
$C{l}^{-}\left(aq\right)$	-129
$AgCl\left(s\right)$	-109

(II) Calculate the ${\log }_{10}{K}_{sp}\left(AgCl\right)$ at 298 K.

(III) If $6.539\times {10}^{-2}g$ of metallic zinc is added to 100 ml saturated solution of $AgCl$, find the value of ${\log }_{10}\frac{\left[Z{n}^{2+}\right]}{[A{g}^{+}{]}^2}$.

(IV)How many moles of Ag will be precipitated in the above reaction?

Given that-

$A{g}^{+}+e^{-}\to$ Ag; $E^o=0.80V$

$Z{n}^{2+}+2e^{-}\to$ Zn; $E^o=-0.76V$

(It was given that Atomic mass of Zn $=65.39$)

Answer 1

(I) Cell representation of the reaction is-
$Ag\left(s\right)|AgCl\left(s\right)|C{l}^{-}\left(aq\right)\parallel A{g}^{+}\left(aq\right)|Ag\left(s\right)$

$\Delta G_R^o=\Delta G_{AgCl}^o-(\Delta G_{A{g}^{+}}^o+\Delta G_{C{l}^{-}}^o)$
$=-109-77-(-129)$

$=-57kJmo{1}^{-1}$

$E_{ce11}^o=-\frac{\Delta G_R^o}{nF}=\frac{57\times {10}^3}{1\times 96500}=0.59V$

(II) The cell reaction is-

$A{g}^{+}\left(aq\right)+C{1}^{-}\left(aq{)}_{\overrightarrow{=}}AgCl\right(s)$

$\Delta G_R^o=2.303RT{\log }_{10}{K}_{sp}\left(AgCl\right)$

${\log }_{10}{K}_{sp}\left(AgCl\right)=-\frac{57\times {10}^3}{2.303\times 8.314\times 298}=-9.989\approx -10$

(III) $\cdot .K_{sp}\left(AgCl\right)={10}^{-10}{M}^2$

$[A{g}^{+}{]}_{sat}=\sqrt{K_{sp\left(AgCl\right)}}={10}^{-5}M$

Moles of Zn $=\frac{6.539\times {10}^{-2}}{65.39}={10}^{-3}$

$2A{g}^{+}\left(aq\right)+Zn$ ($s$) $\overrightarrow{=}2Ag\left(s\right)+Z{n}^{2+}\left(aq\right)$

Applying Nemst equation for the above reaction,

$E=E^o-\frac{0.059}{2}{\log }_{10}\frac{\left[Z{n}^{2+}\right]}{[A{g}^{+}{]}^2}$

At equilibrium, $E_{ce11}=0;E_{ce11}^o=1.56V$

$0=1.56-\frac{0.059}{2}{\log }_{10}\frac{\left[Z{n}^{2+}\right]}{[A{g}^{+}{]}^2}$

${\log }_{10}\frac{\left[Z{n}^{2+}\right]}{[A{g}^{+}{]}^2}=\frac{1.56\times 2}{0.059}=52.88$

Since the equilibrium constant $K_c$ for the reaction,

(IV) $2A{g}^{+}\left(aq\right)+Zn\left(s\right)\to 2Ag\left(s\right)+Z{n}^{2+}\left(aq\right)$ is very high i.e. ${10}^{52.88}$, so the reaction will almost go to completion. Therefore, the moles of Ag precipitated is ${10}^{-5}$ in 1000 ml solution. $\because K_{sp}={10}^{-10}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
Since, it is asked for 100 ml Solution so the no. of mole of $A{g}^{+}$ ppt. $=\frac{{10}^{-5}}{1000}\times 100={10}^{-6}moles$