1.

For galvanic cell, `Ag|AgCl(s),CsCl(0.1M)||CsBr(10^(-3)M),AgBr(s)|Ag.` Calculate the `EMF` generated and assign correct polarity to each electrode for spontaneous or exergonic process at `25^(@)C`. Given `:. K_(sp)` of `AgCl=3.0xx10^(-10),K_(sp)` of `AgBr=4.0xx10^(-13)`.

Answer» `LHS` electrode
`K_(sp)=[Ag^(o+)][Cl^(c-)]=[Ag^(o+)]_(a)(0.1)`
`3.0 xx 10^(-10)=[Ag^(o+)]_(a)(0.1)`
`:. [Ag^(o+)]_(a)=(3.0xx10^(-10))/(0.1M)=3.0xx10^(-9)M`
`RHS` electrode`:`
`K_(sp)=[Ag^(o+)]_(c)[Br^(c-)]`
`4.0xx10^(-13)=[Ag^(o+)]_(c)(10^(-3)M)`
It is a concentration cell, `:.E^(c-)._(cell)=0`

Half cell reactions are
Anode reaction `:` `Ag(s) hArr Ag^(O+)Ag^(o+)(aq)+e^(-)`
Cathode reaction `: Ag^(o+)(` cathode `)+e^(-) hArr Ag(s)`
Cell reaction `:` `ulbar(Ag^(o+)(cathode)hArrAg^(o+)(anode))`
Thus,
`E_(cell)=E^(c-)._(cell)-(0.059)/(1)log .([Ag^(o+)]_(a))/([Ag^(o+)]_(c))`
`=0-(0.059)/(1)log.(3.0xx10^(-9))/(4xx10^(-10))(` Take`0.059~~~0.06)`
`=-0.06[log 30-2 log 2]`
`=-0.06[1.48-2xx03]`
`=-0.0528V`
`E_(cell)=-ve,` suggest that cell is non`-` spontaneous or endergonic `(DeltaG=+ve)`. For spontaneous process, the polarity of electrodes has to be reversed, `i.e.,` change anode to cathode and cathode to anode.


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