1.

Consider the reaction : H_(2)(g),Pt|H^(+)(aq) , E=0.1" V" The pH of the solution is

Answer»

Solution :`H_(2)(g)+2e^(-) to2H^(+)(aq)`
`E=E^(@)-(0.0591)/(2)"LOG"[H^(+)]^(2)`
`0.1=0-(0.0591)/(2)2LOG[H^(+)]`
`0.1=0-0.0591" log " [H^(+)]`
`-log[H^(+)]=(0.1)/(0.0591)=1.69`
`pH=-log[H^(+)]=1.69=2`


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