1.

The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. `0.51` voltB. 0 voltC. `-0.591` voltD. `0.059` volt

Answer» Correct Answer - C
`2H^(+)+2e^(-)rarr H_(2) pH = 10, [H^(+)]= 10^(-10)`
`E^(@)=0`
`E_("red") = (0.0591)/(2) "log" ([H^(+)]^(2))/(P_(H_(2))(g)) = (0.0591)/(2)xx2log[10^(-10)]`
`= (0.0591)/(2)xx2xx[-10]= -0.591 V`


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