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

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For a cell involving two electrons changes, `E_(cell)^(@)=0.3" V"` at `25^(@)C`. The equilibrium constant for the reaction is :
A. `10^(10)`
B. `3xx10^(-2)`
C. 10
D. `10^(10)`
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Correct Answer - D
`logK_(C )=(nE_(cell)^(@))/(0.0591)=((2xx0.3" V"))/((0.0591" V"))~~=10`.
`K_(C )="Antilog "10=10^(10)`
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