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.