InterviewSolution
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InterviewSolution
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Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation. (b) For the cell reaction `Ni(s)|Ni^(2+)(aq)"||"Ag^(+)(aq)|Ag(s)` Calculate the equilibrium constant at `25^@C`. How maximum work would be obtained by operation of this cell ? `E_((Ni^(2+)//Ni))^(@)=-0.25V and E_((Ag^(+)//Ag))^(@)=0.80V` |
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Answer» Dissociation is a process in which molecules. Separate or split into smaller particles such as atoms. Degreed dissociation is the fraction of mole of reactant that underwent dissociation. `alpha=sqrt((K)/(C))` degree of dissociation of a weak electrolyte is propotional to the inverse square root of the concentration or the square root of the dilution. the concentration of any one ionic species is given by the root of the product of the dissociation constant and the concentration of the electrolyte. (b) `At Anode to Nito Ni^(2+)+2e^(-)` At Cathode `to [Ag^(+)+e^(-)to Ag]xx2` Cell reaction `-Ni+2Ag^(+) to Ni^(2+)+2Ag` `E_("Cell")^(@)=E_("Cathode")^(@)-E_("Anode ")^(@)=E_(Ag//Ag)^(@)-E_(Ni^(2+)//Ni^+)^(@) = 80V -(-0.250)` `E_("cell")^(@)=1.05V` `E_("cell")=E_("cell")^(@)-(0.059)/(n) log.([Ni^(2+)])/[Ag^+]^2` `n=2, E_("cell")^(@) =1.05V`, `[Ni^(2+)]=0.1M`, `[Ag^(+)]=1.0M` `^E.cell=1.06V-(0.059)/(2)log.((0.1))/((1^2))` `=1.05+0.0295V=1.0795V` `E_("cell")^(@)(0.0591)/(2)log KC` `1.05=(0.0591)/(2)logKC` `log KC = (1.05xx2)/(0.0591)=35.553` `KC=Anti (35.533) =3.412xx10^(35)` `DeltaG=-nFE^@=-2xx96500xx1.05=-202650J//Mol`. |
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