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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

`10^(-10)`

B

`3xx10^(-2)`

C

10

D

`10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
D
`logK_(c )=(n E^(@))/(0.0591)`
`logK_(c )=(2xx0.3)/(0.0591)`
`logK_(c )=10.15`
`K_(c )="antilog" 10.15`
`K_(c ) = 1.419xx10^(10)`
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