1.

Pb|PbSO_(4)|H_(2)SO_(4)(aq)||PbCl_(2) saturated solution |Cl_(2)|Pt As the cell discharged:

Answer»

Conductivity of ANODE solution decreases
Concentration of `PB^(2+)` increases at anode solution and decreases at cathode
`E_("cell")^(@)` depends on `K_(sp)` of `PbSO_(4)` and `K_(sp)` of `PbCl_(2)`
`E_("cell")^(@)` is zero

Solution :Anode: `Pb(s)+SO_(4)^(2-)toPbSO_(4)(s)+2e^(-)`
Cathode: `Cl_(2)+2e^(-)to2Cl^(-)`
`2Cl^(-)+Pb^(2+)toPbCl_(2)(s)`
At anode, concentration of `SO_(4)^(2-)` FALLS, so conductivity of anode solution decreases.
As `[SO_(4)^(2)]` decreases `[Pb^(2+)]` increase at anode to satisfy `K_(sp)`. At cathode concentration of `Pb^(2+)` decreases.
`E_("cell")^(@)=E_(Pb^((2+))/(Pb))^(@)+(0.059)/(2)logK_(sp)(PbCl_(2))-`
`(E_((Pb^(2+))/(Pb))+(0.059)/(2)logK_(sp)(PbSO_(4)))`
`=(0.059)/(2)log(K_(sp)(PbCl_(2)))/(K_(sp)(PbSO_(4))): E_("cell")^(@)ne0`


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