1.

Zn(s)+Cl_(2)(1atm)toZn^(2+)+2Cl^(-). E_(cell)^(o) of the cell is 2.12 V. To increase E

Answer»

`[Zn^(2+)]` should be INCREASED
`[Zn^(2+)]` should be decreased
`[Cl^(-)]` should be decreased
`P_(Cl_(2))` should be decreased

Solution :According to Nernst's EQUATION
`E_(CELL)=E_(cell)^(o)(-0.059)/(2)"LOG"(["Anode"])/(["Cathode"])`
At anode`toZntoZ_(n)^(2+)+2e^(-)`
At cathode`toCl_(2)+2e^(-)to2Cl^(-)`
So, `E_(cell)=2.12(-0.059)/(2)"log"([Zn]^(++))/([Cl^(-)]^(2))`
E will be increase when CONCENTRATIONS of both `[Zn]^(++)` and `Cl^(-)` is decreased.


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