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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

`1xx10^(-10)`

B

`29.5 xx10^(-2)`

C

10

D

`1xx10^(10)`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaG=-nFE_("cell")^(@) `
`DeltaG= -2.303RT logK , nFE_("cell")^(@) = 2.303RT logK`
`logK = (nFE_("cell")^@)/(2.303RT) = (2 xx96500xx0.295)/(2.303xx8.314xx298)`
`log K = 9.97 = K = 1xx 10^(10)`
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