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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

`-0.828 V`

B

0.828 V

C

0V

D

`-0.5 V`

Text Solution

Verified by Experts

The correct Answer is:
A
`H^(+) + e^(-) to 1/2 H_(2), E^(@) =0, DeltaG^(@) =0`
`H_(2)O `H_(2)O + e^(-) to 1/2 H_(2) + OH^(-), -1 xx E^(@) xx 96500 xx E^(@)` 298 ln `10^(-14)`
`E^(@) = -0.828` Volt.
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