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Question
`"E"_"cell"^0` for the reaction, \[\ce{2H2O -> H3O+ + OH-}\] at 25°C is - 0.8277 V. The equilibrium constant for the reaction is ______.
Options
10-14
10-23
10-7
10-21
MCQ
Fill in the Blanks
Solution
`"E"_"cell"^0` for the reaction, \[\ce{2H2O -> H3O+ + OH-}\] at 25°C is - 0.8277 V. The equilibrium constant for the reaction is 10-14.
Explanation:
\[\ce{2H2O -> H3O+ + OH-}\]
`"E"_"cell"^0 = 0.0591/"n"`log K
log K = `("E"_"cell"^0 xx "n")/0.0591 = (- 0.8277 xx 1)/0.0591` = - 14
K = 10-14
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