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For the reaction, Ag_((aq.))^(+) + Cl_((aq.))^(-) hArr AgCl_((s)) the Deltag^(@) Values for Ag_(aq.)^(@), Cl_(aq.) and AgCl_(s) are +77, -129 and -109 kj mol^(-1), Write the cell representation of above reaction and calculate E^(@) at 298 K, also calculate thelog_(10) K_(sp) of AgCl at 298K. (b) If 6.539 xx 10^(-2)g of metallic zinc is added to 100 mL saturated solution of AgCl, find the value of log_(10).([Zn^(2+)])/([Ag^(+)]^(2)). How many moles of Ag will be precipitated in this reaction ? Given, E_(Zn^(2+)//Zn)^(@) = -0.76 V. |
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Answer» Solution :`Ag_((s)) + (1)/(2)Cl_(2_(g))rarr AgCl_((s)), DeltaG^(@) = -109kJ` ....(1) `Ag_((s)) rarr Ag_(kJ)^(+) + e, DeltaG^(@) = +77 kJ` ....(2) `(1)/(2)Cl_(2(g))^(-) + e rarr Cl_((aq.))^(-), DeltaG^(@) = -129 kJ` ....(3) By eq. `(1)` - `(2)` - `(3)` `Ag_((aq.))^(+) + Cl_((aq.))^(-) rarr AgCl_((s)), DeltaG^(@) = -109 - 77 + 129 = -57 kJ` `:' -DeltaG^(@) = nE^(@)F` `57 xx 10^(3) = 1 xx E^(@)F` `:. E^(@) = 0.59 V` The CELL is `Ag|AgCl_((s))|Cl_((aq.))^(-)||Ag_((aq.))^(+)|Ag` `E_(cell) = E_(OP_(Ag//AgCl//Cl^(-))) -0.059log.(1)/([Cl^(-)])+E_(RP_(Ag^(+)//Ag)) + 0.059log [Ag^(+)]` at equilibrium `E_(cell) = 0`, thus `0 = E_(Ag//AgCl//Cl^(-))^(@) + E_(RP_(Ag//AgCl//Cl^(-)))^(@) + 0.059 log [Ag^(+)][Cl^(-)]` or `0 = E_(cell)^(@) + 0.059 log K_(SP)` or `E_(cell)^(@) = -0.059 log K_(SP)` or `logK_(SP) = -(E_(cell)^(@))/(0.059) = - (0.59)/(0.059) = -10` `:. K_(SP)AgCl = 1 xx 10^(-10) M^(2)` (b) Let solubility of `AgCl` be `S`, then `S = sqrt(K_(SP)) = sqrt(10^(-10)) = 10^(-5) M` Total milli-mole of `AgCl` in its SATURATED solution of `10^(-5)M` in `100 ML = 10^(-5) xx 10^(2) = 10^(-3)` `:.` Mole of `AgCl` in `100 mL` solution `= 10^(-3) xx 10^(-3) = 10^(-6)` Also mole of `Zn` added `= (6.539 xx 10^(-2))/(65.39) = 10^(-3)` `DeltaG^(@)` for `Ag rarr Ag^(+) + e` is `77 kJ` `DeltaG^(@) = nE^(@)F` `:. E_(Ag//Ag^(+))^(@) = (-77 xx 10^(3))/(1 xx 96500) = - 0.80 V` For the redox change on ADDITION of `Zn` `{:(ZnrarrZn^(2+)+2e,E^(@)=+0.76),(2Ag^(+)+2erarr2Ag,E^(@)=+0.80):}/(Zn+2Ag^(+)rarrZn^(2+)+2Ag,E_(cell)^(@)=0.76+0.80=1.56V)` `E_(cell) = E_(OP_(Zn))^(@) + E_(RP_(Ag))^(@) + (0.059)/(2)log.([Ag^(+)]^(2))/([Zn^(2+)])` At equilibrium, `E_(cell) = 0` `:.= E_(cell)^(@)+(0.059)/(2)log.([Ag^(+)]^(2))/([Zn^(2+)])` `:. log.([Zn^(2+)])/([Ag^(+)]^(@)) = (E_(cell)^(@) xx 2)/(0.059) = (1.56 xx 2)/(0.059) = 52.88` `K_(c)=([Zn^(2+)])/([Ag^(+)]^(2)) = 7.6 xx 10^(52)` Since `K_(c)` is appreciably HIGH and thus nearly whole of the `Ag^(+)` is converted to`Ag`. Thus moles of `Ag` formed `=` moles of `Ag^(+)` in `100 mL` saturated solution `= (10^(-3))/(10^(3)) = 10^(-6)` |
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