`E_"cell"^@` for the reaction , `2H_2O to H_3O^(+) + OH^(-)` at `25^@C` is -0.8277 V. The equilibrium constant for the reaction is

Question

`E_"cell"^@` for the reaction , `2H_2O to H_3O^(+) + OH^(-)` at `25^@C` is -0.8277 V. The equilibrium constant for the reaction is

`E_"cell"^@` for the reaction , `2H_2O to H_3O^(+) + OH^(-)` at `25^@C` is -0.8277 V. The equilibrium constant for the reaction is
A. `10^(-14)`
B. `10^(-23)`
C. `10^(-7)`
D. `10^(-21)`

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 - A
`2H_2O to H_3O^(+) + OH^(-)`
`E_"cell"^@=0.0591/n` log K
log K =`(E_"cell"^@xxn)/0.0591 = (-0.8277xx1)/0.0591= - 14 rArr K=10^(-14)`

4 views Answered 2023-02-05 15:29