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E^(@) for the cell Zn(s)|Zn^(2+)(aq)|C...

`E^(@)` for the cell
`Zn(s)|Zn^(2+)(aq)|Cu^(2+)(aq)|Cu(s)` is 1.1V at `25^(@)C` the equilibrium constant for the cell reaction is about

A

`10^(-37)`

B

`10^(37)`

C

`10^(73)`

D

`10^(73)`

Text Solution

Verified by Experts

The correct Answer is:
B
At 298k,
`E^(@)=(0.0591)/(n)"log"k`
`1.1=(0.0591)/(2)"log"k` (Here n=2)
log `k=(1.1xx2)/(0.0591)=37.22`
`k="antilog" 37.22=1.65xx10^(37)`
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