InterviewSolution
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InterviewSolution
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Calculate the `EMF` of the following concentration cells a `30^(@)C` and predict whether the cells are exergonic or endergonic. `[` Assume `Kw` does not chage at `30^(@)C]` `a. Pt|H_(2)(g)(1 atm)|H^(o+)(10^(-6)M)||H^(o+)(10^(-4)M)|H_(2)(g)(1atm)|Pt` `b. Pt||Hg_(2)(1atm)|NaOH(10^(-4)M)||H^(o+)(10^(-5)M)|H_(2)(g),(1 atm)|Pt` `c. Pt|H_(2)(g)(1 atm)|H_(2)SO_(4)(0.05M)||KOH(10^(-3)M)|H_(2)(g)(1atm)|Pt` `d. Pt|H_(2)(g)(1 atm)|CH_(2)COOH(10^(-2)M)||CsOH(10^(-3)M)|H_(2)(g)(1atm)|Pt(pK_(a) of CH_(3)COOH=4.74)` `e. Pt|H_(2)(g)(1atm)|H_(2)O||HCl(10^(-3)M)|H_(2)(g)(1 atm)|Pt` `f. |
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Answer» `a." "[H^(o+)]_(cathode )gt[H^(o+)]_(anode)or pH_(cathode)ltpH_(anode )` So `EMF` of the cell will be positive and the cell will be spontaneous or exergonix `(i.e., DeltaG=-ve).` `(pH)_(c)=4,(pH)_(a)=6` `:.E_(cell)=-0.060(pH_(c)-pH_(a))=-0.060(4-6)=0.12V` `b. (pH)_(c)=5,(pOH)_(a)=4`,or `(pH)_(a)=14-4=10` Since `(pH)_(c)lt(pH)_(a)`, so `EMF` of the cell will be positive and the cell will be spontaneous or exergonic. `:. E_(cell) =-0.06(5-10)=0.30V` `c.` Since`[H_(2)SO_(4)]=[H^(o+)]_(a)=0.05xx2(n` factor `)` `=0.1N=10^(-1)N` `pH_(a)=1,(pOH)_(c)=3,pH_(c)=14-3=11` Since `pH_(c)gtpH_(a)` so `EMF` of cell will be negative and the cell will not be feasible or non`-` spontaneous or endergonic `(i.e., Delta G=+ve)`. `E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(11-1)=-0.6V.` `d. CH_(3)COOH` is a weak acid and `CsOH` is a strong base. Thus, `pH_(wA)=(1)/(2)(pK_(a)-log c)` `=(1)/(2)(4.74-log 10^(-2))` `=(1)/(2)(4.74+2)=3.37=pH_(a)` `(pOH)_(c)=3,impliespH_(c)=14-3=11` Since `pH_(c)gtpH_(a),` so `EMF` of cell will be negative and is endergonic. `E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(11-3.37)` `=-0.06xx7.63` `=-0.457V` `e." "pH` of `H_(2)O=7=pHa` `pH_(c)=3 ` Since `pH_(c)ltpH_(a),EMF_(cell)=+ve,` cell is exergonic. `E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(3-7)=0.24V` `f.` `NH_(4)OH` is a weak base and `RbOH` is a strong base. Thus, `pOH_(wB)=(1)/(2)(pK_(b)-log C)=(1)/(2)(4.74-log10^(-2))=3.37` `pH_(a)=14-3.37=10.63` `(pOH)=3,impliespH_(c)=14-3=11` Since `pH_(c)gtpH_(a)`, therefore, `EMF_(cell)=-ve,` cell will be endergonic. `E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(11-10.63)` `=-0.022V` `g.` Mixture of `CH_(3)COOH(W_(A))` and salt of `W_(A)//S_(B)` `(CH_(3)COONa)` is an acidic buffer. Thus, `:. pH_(aci d b uffer )=pK_(a)+lo.([Sal t])/([Aci d])` `=4.74+log.(10^(-1)M)/(10^(-2)M)` `=4.74+1=5.74` Mixture of `NH_(4)OH(W_(B))` and salt of `W_(B)//S_(A)(NH_(4)Cl)` is a basic buffer. `:. pOH_(basic buffer)=pK_(a)+lo.([Sa l t])/([Base])` `=4.74+log `(0.4M)/(0.2M)` `=4.74+log 2 ` `=4.74+0.3 = 5.04` `:. pH_9c)=14-5.04=8.96` Since `pH_(c)gtpH_(a)` `:. EMF_(cell)=-ve,` cell is endergonic. `E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(8.96-5.4)` `=-0.06xx3.22` `=-0.1932V` |
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