1.

For the half-cell raction, 2H_(2)O+2e^(-)toH_(2)+2OH^(-),E^(@)=-0.8277V at 298K. Autoprotolysis constant of water calculted from this value will be

Answer»

`1xx10^(-10)`
`1xx10^(-12)`
`1xx10^(-13)`
`1xx10^(-14)`

Solution :AUTOPROTOLYSIS constant of `H_(2)O=[H^(+)][OH^(-)]=K_(w)`
Redn. Half reaction:
`2H_(2)O+2e^(-)toH_(2)+2OH^(-),E^(@)=-0.8277V`
`underset("Ox. Half reaction:"H_(2)to2H^(+)+2e^(-),E^(@)=0.0)`
Cell reaction: `2H_(2)OhArr2H^(+)+2OH^(-),E^(@)=-0.8277V`
`E=E^(@)-(0.0591)/(2)"log"[H^(+)]^(2)[OH^(-)]^(2)`
At EQUILIBRIUM, E=0. Hence,
`E^(@)=(0.0591)/(2)log[H^(+)]^(2)[OH^(-)]^(2)`
`therefore-0.8277=0.0591log[H^(+)][OH^(-)]^(2)`
`=0.0591logK_(w)`
or `logK_(w)=14` or `K_(w)=10^(-14)`.


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