1.

Calculate the `EMF` of the following concentration cells at `30^(@)C` and predict whether the cells are exergonic or endergonic.

Answer» Note that at `30^(@)C` , the value of `2.303(RT)/(F)=0.06`.
`a. CH_(3)COONa` is a salt of `W_(A)//S_(B).` Thus
`pH_(a)=(1)/(2)(pK_(w)+pK_(a)=log c)`
`(1)/(2)(14+4.74=log10^(-2))`
`=(1)/(2)(14+4.74-2)=8.37`
`gt.Ph_(a)=8.37`
`NH_(4)NOH_(3)` is a salt of`W_(B)//S_(A)`. Thus,
`pH_(c)=(1)/(2)(pK_(w)-pK_(b)-log c )`
`=(1)/(2)(14-4.74-log 0.2)`
`=(1)/(2)[14-4.74-log 2xx10^(-1)]`
`=(1)/(2)(14-4.74-0.3+1)=4.98.`
Since `pH_(c)ltpH_(a)`, therefore,`EMF_(cell)=+ve` and the cell will be exergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(4.98-8.37)`
`=-0.06xx-3.39`
`=0.2034V`
`b. CH_(3)COONH_(4)` is salt of `W_(A)//W_(B)` and its `pH` is independent of the concentration.
`:. pH_(a)=(1)/(2)(pK_(w)+pK_(a)-pK_(b))`
`=(1)/(2)(14+4.74-4.74)=7`
`CH_(3)COONa` is a salt fo `W_(A)//S_(B).`
`:. pH_(c)=(1)/(2)(pK_(w)+pK_(a)+logc)`
`=(1)/(2)(14+4.74+log 10^(-3))=7.87`
Since `pH_(c)gtpH_(a)`
`:. EMF_(cell)=-ve` and the cell will be endergonic
`:. E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(7.87-7)`
`=-0.06xx0.87`
`=0.05522V`
`c. CH_(3)COH` is a weak acid `(W_(A)).`
Thus, `pH_(WA)=(1)/(2)(pK_(a)-log c)`
`=(1)/(2)(4.74- log 10^(-1))`
`=2.87`
`NH_(4)OH` is a weak base `(W_(B))`.
Thus, `pOH_(wB)=(pOH)_(c)=(1)/(2)(pK_(b)-logc)`
`=(1)/(2)(4.74-log 10^(-2))`
`=3.37`
Thus, `pOH_(wB)=pH_(c)=14-3.33=10.63`
Since `pH_(c)gtpH_(a)`, therefore, `EMF_(cell)=-ve`, and the cell will be endergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(10.63-2.87)`
`=-0.06xx7.67`
`-0.4656V`
`d.` Anode reaction`: cancer(Ag(s)) rarrAg^(o+)Ag^(o+)(0.1M)+cancel(e^(-))`
Cathode reaction `:`
`Ag^(o+)(1M)+cancel(e^(-)) rarr Ag(s)`
Cell reaction `:` `ulbar(Ag^(o+)(1M)rarrAg^(o+)(0.1M))`
Since `[Ag^(o+)]_(c)gt[Ag^(o+)]_(a),` therefore , `EMF_(cell)=+ve` and the cell will be exergonic.
`E_(cell)=E^(c-)._(cell)-(0.06)/(n_(cell))log.([Ag^(o+)])/([Ag^(o+)])`
`=0-(0.06)/(1)log.(0.1M)/(1M)`
`E_(cell)=-0.06[log 10^(-1)]=-0.06 xx -1=0.06V`
`e.` Anode reaction `:`
`2Cl^(c-)(10^(-3)M)rarr cancel(Cl_(2)(g)(1atm))+cancel(2e^(-)) `
Cathode reaction `:`
`cancel (2Cl(g)(1atm))+cancel(2e^(-))rarr2Cl^(c-)(10^(-2)M)`
`ulbar(2Cl^(c-)(10^(-3)M) rarr 2Cl^(c-)(10^(-2)M)`
Since `[Cl^(c-)]_(c)gt[Cl^(c-)]_(a)`
Therefore, `EMF_(cell)=-ve` and the cell will be endergonic.
`E_(cell)=E^(c-)._(cell)-(0.06)/(n_(cell))log.([Cl^(c-)]_(c)^(2))/([Cl_(a)^(2)])`
`=0-(0.06)/(2)log.((10^(-2))^(2))/((10^(-3))^(2))`
`=-(0.06)/(2)xx2 log 10=-0.06V`

`f.` Anode reaction `:`
`2Cl^(c-)(10^(-3)M) rarr Cl_(2)(g)(2atm)+cancel(2e^(-)) `
Cathode reaction `:` `Cl_(2)(g)(1 atm)+cancel(2e^(-)) rarr 2 Cl_(2)(g)(10^(-2)M)`
` Cell reaction `:` `ul bar(2Cl^(c-)(10^(-3)M)+Cl_(2)(g)(1atm)rarr2Cl^(c-)(10^(-2)M)+Cl_(2)(g)(2atm))`
`E_(cell)=E^(c-)._(cell)-(0.06)/(n_(cell))log.([Cl^(c-)]_(c)^(2)[P_(Cl_(2))]_(a))/([Cl^(c-)]_(a)^(2)[p_(Cl_(2))]_(c)).=0-(0.06)/(2)log.((10^(-2))^(2)xx2atm)/((10^(-3))^(2)xx1atm)`
`=-(0.06)/(2)[log 10^(2)xx2]`
`=-(0.06)/(2)[2 log 10+log 2 ]`
`=-0.03[2+0.3]=-0.03xx2.3=-0.069V`
Therefore, `EMF_(cell)` is negative and the cell will be endergonic.
`EMF_(cell)` is positive , if `[Cl^(c-)]_(a)gt[Cl^(c-)]_(c)` and `(p_(Cl_(2)))_(c)gt(p_(Cl_(2)))_(a)`


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