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E(Cu^(+2)//Cu)^(@)=0.34V, E(Zn//Zn^(+2))...

`E_(Cu^(+2)//Cu)^(@)=0.34V, E_(Zn//Zn^(+2))^(@)=0.76V`
A cell formed by the conbination of Cu and Zn
(a) when `CuSO_(4)` is added to `Cu^(+2)` compartment what is the effect on emf of cell
(b) when `NH_(3)` is added to `Cu^(+2)` compartment what is the effect on emf of cell
(c ) When `ZnSO_(4)` is added to `Zn^(+2)` compartment is the effect on emf of cell
(d) When `Zn^(+2)` is diluted what is the effect on emf of cell ?

Text Solution

Verified by Experts

(a) increases
(b) decreases
(c ) decreases
(d) increases
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If E_(Cu^(2+)|Cu)^(@) = 0.34V and E_(Cu^(2+)|Cu^(+))^(@)= 0.15 V then the value for disproportion for Cu^(+) is :

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

CuSO_(4)+Zn(s) to ZnSO_(4)+Cu

Consider the cell Zn|Zn^(2+) || Cu^(2+)|Cu. If the concentration of Zn and Cu ions are doubled, the emf of the cell.

Given: E_(Zn^(+2)//Zn)^(@) =- 0.76V E_(Cu^(+2)//Cu)^(@) = +0.34V K_(f)[Cu(NH_(3))_(4)]^(2+) = 4 xx 10^(11) (2.303R)/(F) = 2 xx 10^(-4) When 1 mole of NH_(3) added to cathode compartement than emf of cell is (at 298K) :

The cell reaction Zn(s) + Cu^(+2)rarr Zn^(+2) + Cu(s) is best represented as :

CuSO_(4)(aq.)+Zn(s) to ZnSO_(4)+Cu darr

CuSO_(4)(aq.)+Zn(s) to ZnSO_(4)+Cu uarr

The standard emf for the cell cell reaction Zn + Cu^(2+) rarr Zn^(2+) + Cu is 1.10 volt at 25^@ C . The emf for the cell reaction when 0.1 M Cu^(2+) and 0.1 M ZN^(2+) solutions are used at 25^@ =C is .