JEE Advance - Chemistry (2001 - No. 3)
Pt | H2 (g) | HCl (aq) | AgCl (s) | Ag (s)
(i) Write the cell reaction.
(ii) Calculate $$\Delta H^o$$ and $$\Delta S^o$$m for the cell reaction by assuming that these quantities remain unchanged in the range 15oC to 35oC.
(iii) Calculate the solubility of AgCl in water at 25oC
Given : The standard reduction potential of the Ag+ (aq) / Ag (s) couple is 0.80 V at 25oC
Explanation
Given cell : $$Pt|{H_2}(g)|HCl(aq)|AgCl(s)|Ag(s)$$
(1) The half-cell reactions are as follows:
At anode : $${1 \over 2}{H_2}(g) \to {H^ + }(aq) + e$$
At cathode : $$AgCl(s) + e \to Ag(s) + C{l^ - }(aq)$$
Cell reaction : $${1 \over 2}{H_2}(g) + AgCl(s) \to {H^ + }(aq) + Ag(s) + C{l^ - }(aq)$$
(2) $$\Delta {S^0} = nF\left( {{{d{E^0}} \over {dT}}} \right)$$, where n = Number of electrons involved in the cell reaction, F = Faraday = 96500 C, dE0 = Difference of standard electrode potential at two different temperatures = (0.21 $$-$$ 0.23) = $$-$$ 0.02 V and dT = difference of two temperatures = (308 $$-$$ 288) K = 20 K
$$\therefore$$ $$\Delta {S^0} = 1 \times 96500 \times \left( {{{ - 0.02} \over {20}}} \right) = - 96.5$$ J K$$-$$1 mol$$-$$1
We know, $$\Delta {G^0} = - nF{E^0}$$
$$\therefore$$ $$\Delta G_{15^\circ \,C}^0 = - 1 \times 96500 \times 0.23$$ [$$\because$$ $$\Delta E_{15^\circ \,C}^0 = 0.23\,V$$]
= $$-$$ 22195 J . mol$$-$$1
$$\therefore$$ $$\Delta H_{}^0 = \Delta G_{}^0 - T\Delta S_{}^0$$
$$ = - 22195 - 288 \times ( - 96.5) = - 49987$$ J mol$$-$$1
(3) Given $$\Delta E_{(15^\circ \,C)}^0 = 0.23\,V$$ and $$\Delta E_{(35^\circ \,C)}^0 = 0.21\,V$$
$$\therefore$$ $${{\Delta {E^0}} \over {\Delta T}} = {{(0.21 - 0.23)} \over {20}} = - 0.01$$
$$\therefore$$ $$\Delta$$E0 for 10$$^\circ$$C = $$-$$0.01 $$\times$$ 10 = $$-$$0.1
$$\therefore$$ $$\Delta E_{(25^\circ \,C)}^0 = \Delta E_{(15^\circ \,C)}^0 + ( - 0.1) = 0.23 - 0.1 = 0.22\,V$$
$$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$$
$$0.22\,V = E_{C{l^ - }|AgCl|Ag}^0 - E_{2{H^ + }|{H_2}}^0 = E_{C{l^ - }|AgCl|Ag}^0 - 0$$
$$\therefore$$ $$E_{C{l^ - }|AgCl|Ag}^0$$ = 0.22 V and $$E_{A{g^ + }|Ag}^0$$ = 0.80 V (given)
$$E_{C{l^ - }|AgCl|Ag}^0 = E_{A{g^ + }|Ag}^0 - {{0.059} \over 1} \times \log {K_{sp}}(AgCl)$$
or, $$0.22\,V = 0.80\,V - {{0.059} \over 1}\log {K_{sp}}(AgCl)$$
$$\therefore$$ $${K_{sp}}(AgCl) = 1.47 \times {10^{ - 10}}$$
and $${K_{sp}}(AgCl) = [A{g^ + }] \times [C{l^ - }]$$
$$\therefore$$ $${[A{g^ + }]^2} = 1.47 \times {10^{ - 10}}$$
or, $$[A{g^ + }] = \sqrt {1.47 \times {{10}^{ - 10}}} = 1.21 \times {10^{ - 5}}$$
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