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

Question

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

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)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`logK_(C )=(nE_(cell)^(@))/(0.0591)=((2xx0.3" V"))/((0.0591" V"))~~=10`.
`K_(C )="Antilog "10=10^(10)`

4 views Answered 2023-02-05 15:29