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For a cell involving two electron change...

For a cell involving two electron changes, `E_("cell")^(@) =0.3V` at `25^(@)C`. The cell equilibrium constant of the reaction is

A

`2.95xx10^(2)`

B

10

C

`1xx10^(10)`

D

`1xx10^(-10)`

Text Solution

Verified by Experts

The correct Answer is:
C
`E^(@)=(0.059)/(n) log K`
or `log K=(n E^(@))/(0.059)=(2xx0.295)/(0.059)=10`
`:.K=1xx10^(10)`
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