Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Explanation

Cell potential of the cell formed by combining the hydrogen electrode in contact with solution having pH 10 with SHE is given by :
`pH=(E_("cell"))/(0.0591) `
`E_("cell") = 0.0591 pH = 0.0591 xx 10=0.591V`
`E_("cell")=E_("cathode")-E_("anode") = E_("SHE") -E_("el") = 0-E_("el")`
`0.591V = 0-E_("el") or E_("el")= -0.591V`
Alternatively, for hydrogen electrode,
`H^(+) +e^(-) to (1)/(2)H_(2)`
Applying Nernst equation,
`E_(H, (1)/(2)H_(2))^(+) =E_(H, (1)/(2)H_(2))^(Theta) -(0.0591)/(n)"log" (1)/([H^(+)])`
`=0-(0.0591)/(1)"log"(1)/(10^(-10)) [ :. [H^(+)] =10^(-pH) =10^(-10)M]`
`= -591 xx 10= -0.591 V`