QJEEAdv05Ch3ea9: JEE Advanced 2005 Chemistry - Class 12 JEEAdv05Ch3ea9 Solution: (a). For the reaction Ag + (aq) + Cl - (aq) AgCl (s) Given: .tg …

Half-cell reactions are -

$A{g^ + }(aq) + e \to Ag(s)$

$Ag(s) + C{l^ - }(aq) \to AgCl(s) + e$

Cell reaction : $A{g^ + }(aq) + C{l^ - }(aq) \to AgCl(s)$

(1) The cell is : $Ag|AgCl(s)|C{l^ - }(aq)||A{g^ + }(aq)|Ag$

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

$\therefore$ $\Delta {G^0} = \Delta G_r^0(AgCl) - \Delta G_r^0(A{g^ + }) - \Delta G_r^0(C{l^ - })$

or, $\Delta {G^0} = [ - 109 - 77 - ( - 129)]$ kJ mol ^{$-$ 1}

or, $\Delta {G^0} = - 57$ kJ mol ^{$-$ 1}

But, $\Delta {G^0} = - nF{E^0}$ or, $- 57000 = - 1 \times 96500 \times {E^0}$

or, ${E^0} = {{57000} \over {96500}}$ or, ${E^0} = 0.59V$

The solubility equilibrium for AgCl is

$AgCl(s)$ $\rightleftharpoons$ $A{g^ + }(aq) + C{l^ - }(aq)$

For this reaction $E_{cell}^0 = - 0.59V$

$\therefore$ ${\log _{10}}{K_{sp}} = {{nF{E^0}} \over {RT}} = - {{0.59} \over {0.059}} = - 10$

(2) Amount of zinc added $= {{6.539 \times {{10}^{ - 2}}} \over {65.39}} = {10^{ - 3}}$ mol

Therefore, the following reactions will occur :

$2A{g^ + }(aq) + 2e \to 2Ag(s)$ ;

$Zn(s) \to Z{n^{2 + }}(aq) + 2e$ ;

$2A{g^ + }(aq) + Zn(s) \to Z{n^{2 + }}(aq) + 2Ag(s)$ ;

By Nernst equation, ${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$

At equilibrium, ${E_{cell}} = 0$ , $\therefore$ $1.58 = {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$

or, $\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}} = {{1.58 \times 2} \over {0.059}} = 53.47$

$\therefore$ Equilibrium constant, $K = {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$

$\therefore$ $\log K = 53.47$ or, $K = {10^{53.47}}$

Very high value of K indicates that the reaction goes to almost completion. Solubility of $AgCl = \sqrt {{K_{sp}}} = \sqrt {{{10}^{ - 10}}} = {10^{ - 5}}$ M

$\therefore$ $[A{g^ + }] = {10^{ - 5}}M$ . Hence, number of mole of Ag⁺ ions in 100 mL solution = 10 ^{$-$ 6}. Since, the reaction goes to almost completion, the amount of Ag formed = 10 ^{$-$ 6} mol.