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The cell, Zn//Zn^(2+)(1M)"//"Cu^(2+)(1M)...

The cell, `Zn//Zn^(2+)(1M)"//"Cu^(2+)(1M)//Cu(E_(cell)^(@)=1.10V)` , was allowed to be completely concentration of `Zn^(2+)` to `Cu^(2+)` is

A

`37.3`

B

`10^(37.3)`

C

`9.65xx10^(4)`

D

antilog `(24.04)`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(cell)=E_(cell)^(0)-(0.059)/2 "log" ([Zn^(2+)])/([Cu^(2+)]`
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