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The standard potential of the reaction ...

The standard potential of the reaction
`H_(2)O + e^(-) rightarrow (1/2)H_(2) + OH^(-)` at 298 K by using `k_(w) (H_(2)O) = 10^(-14)`, is:

A

`-0.828V`

B

`0.828V`

C

`0V`

D

`-0.5V`

Text Solution

Verified by Experts

The correct Answer is:
A
`H^(+)+e^(-)to(1)/(2)H_(2),E^(@)=0,triangleG^(@)=0`
`H_(2)OhArrH^(+)+OH^(-),triangleG^(@)=-8.314xx298ln10^(-14)`
`H_(2)O+e^(-)to(1)/(2)H_(2)+OH^(-),-1xxE^(@)xx96500=-8.314xx298ln10^(-14)`
`E^(@)=-0.828"volt"`
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