1.

The potential of a hydrogen electrode at pH=10 is :

Answer»

`+0.59" V "`
`0.00" V"`
`-0.59" V "`
`-0.059" V "`

Solution :(C ) `H^(+)+e^(-) to 1//2 H_(2)`
`E=E^(@)-(0.059)/(N)"log"(1)/([H^(+)])=-(0.059)/(1)PH`
`=(0-0.059)/(1)xx10=0-0.59=-0.59" V"`


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