निम्नलिखित सेल अभिक्रिया के लिए `DeltaG^(@)` की गणना कीजिए
`Zn(s)+Ag_(2)O (s)+H_(2)O(I)`
`rarr Zn^(2+)(aq)+2Ag (s)+2OH^(-1)(aq)`
`E_(Ag^(+)//Ag)^(0)=+0.80 v ` और `E_(Zn^(+2)//Zn)^(0)=`
`-0.76 V, F =96500`
( दिया गया है AgOH का `K_(sp) = 2xx 10^(-8))`

Options

A. `-305` kJ /mol
B. 212 kJ/ mol
C. 305 kJ / mol
D. 301 kJ/ mol

Answer

B

Explanation

`E_((OH^(-)|Ag_(2)O|Ag))^(-)` के लिए गणना अपचयन `[2Ag^(+)+2e^(-) rarr 2Ag(s), DeltaG_(1)^(@)=-2`
`xx F xxE_(Ag+|Ag)^(@)`
आयनीकरण
`[Ag_(2)O+H_(2)O(1) rarr 2Ag^(+)+2OH^(-)`
`DeltaG_(2)^(@)=-2.303 ` RT log `k_(sp^(2))`
कुल `Ag_(2)O+H_(2)O(1) + 2e^(-) rarr`
`2 Ag(s)+ 2O H^(-), Delta G_(3)^(@)= -2 xxF`
`xx E_(OH^(-)|Ag_(2)O|Ag)^(@)`
`DeltaG_(3)^(@)= DeltaG_(1)^(@)+Delta_(2)^(@)`
`+2xx F xxE_(OH^(-)|Ag_(2)O|Ag)^(@)=+2xx F`
`xx E_(Ag^(+)|Ag)^(@)+(2.303 RT log k_(sp^(2)))/(2F)`
`E_(OH^(-)|Ag_(2)O|Ag)^(@)=E_(Ag^(+)|Ag)^(@)+(0.06)/(2)`
`xx 2 log k_(sp)`
`=0.8+0.06 log (2xx10^(-8))`
`= 0.8+0.06 [-8+0.3]`
`=0.8+0.06 xx(-7.7)`
`=0.8-0.462`
`=0.338` वोल्ट ` ~~0.34 V Delta G^(@)=- F E^(@)`
`=-nF(E_(c)^(@)-E_(A))_("अपचयन ")`
`=-2xx(E_(OH^(-)|Ag_(2)O|Ag)^(@)-E_(Zn^(+2)|Zn))`
`=-2 xx96500 xx (0.34 +0.76)`
`=- 2xx 96500 xx1.1` वोल्ट
`= 212300 J`
= 212.3 kj