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The e.m.f. of the cell Zn"|"Zn^(2+)(0....

The e.m.f. of the cell
`Zn"|"Zn^(2+)(0.1M)"||"Pb^(2+)(0.1M)"|"Pb` is ( standard electrode potentials for `Pb^(2+)"|"Pb "and " Zn^(2+)"|"Zn` electrode are - 0.126 V and - 0.763 V respectively.)

A

0.608 V

B

0.637 V

C

0.667 V

D

`-0.608` V

Text Solution

Verified by Experts

The correct Answer is:
B
`E^(@)=E(Pb^(2+)"|"Pb)-E^(@)(Zn^(2+)"|"Zn)`
`= - 0.126-(-0.763)=0.637 V`
`E=E^(@)- (0.059)/(2) log. ((0.1))/((0.1))`
`= 0.637-0=0.637V`
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