Question

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"`
A. `+1.85" V"`
B. `-1.85" V"`
C. `+0.83" V"`
D. `-0.83" V"`

Answer 1

Correct Answer - A
(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"`