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:

Question

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:

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`

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
`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"`

4 views Answered 2023-02-05 15:29