1.

Two electrochemical cells, Zn^(2+)|Zn^(2+)||Cu^(2+)|Cu and Fe|Fe^(2+)||Cu^(2+)|Cu are connected in series. What will be the net e.m.f. of the cell at 25^(@)C ? Given : E^(@) of Zn^(2+)|Zn=-0.76" V", Cu^(2+)|Cu=+0.34" V",Fe^(2+)|Fe=-0.41" V"

Answer»

`+1.85" V"`
`-1.85" V"`
`+0.83" V"`
`-0.83" V"`

Solution :(a) emf of the cell,`Zn(s)|Zn^(2+)(aq)||CU^(2+)(aq)|Cu(s)` is :
`E_(Cu^(2+)//Cu)^(@)-E_(Zn^(2+)//Zn)^(@)=+0.34-(-0.76)=1.10" V"`
emf of the cell, `Fe(s)|Fe^(2+)(aq)||Cu^(2+)(aq)|Cu(s)` is :
`E_(Cu^(2+)//Cu)^(@)-E_(Fe^(2+)//Fe)^(@)=+0.34-(0.41)=0.75" V"`
Since the two electrodes are connected in SERIES, the net emf of cell is :
`=1.10+0.75=+1.85" V"`


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