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Question: \(Eº_{cell}\) for the reaction \(2H_{2}O \rightarrow H_{3}O^{+} + OH^{-}\) at \(25ºC\) is \(- 0.8277...

EºcellEº_{cell} for the reaction 2H2O→H3O++OH−2H_{2}O \rightarrow H_{3}O^{+} + OH^{-} at 25ºC25ºC is −0.8277V- 0.8277Vthe equilibrium constant for the reaction is

A

10−1410^{- 14}

B

10−2310^{- 23}

C

10−710^{- 7}

D

10−2110^{- 21}

Answer

10−1410^{- 14}

Explanation

Solution

2H2O→H3O++OH−2H_{2}O \rightarrow H_{3}O^{+} + OH^{-}

EºCell=0.0591nlog⁡KEº_{Cell} = \frac{0.0591}{n}\log K

log⁡K=Eºcell×n0.0591=−0.8277×10.0591=−14\log K = \frac{Eº_{cell} \times n}{0.0591} = \frac{- 0.8277 \times 1}{0.0591} = - 14

K=10−14K = 10^{- 14}