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For a cell reaction involving a two-ele...

For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be `0.295V` at `25^(@)C`. The equilibrium constant of the reaction at `25^(@)C` will be:

A

`29.5xx10^(-2)`

B

10

C

`1xx10^(10)`

D

`1xx10^(-10)`

Text Solution

Verified by Experts

The correct Answer is:
C
(c ) `"log "K=(nxxE_(cell)^(@))/(0.059)=(2xx0.295)/(0.059)=10`
`K="Antilog " 10=1xx10^(10)`.
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