Which has maximum potential for the half-cell reaction : `2H^(+)2e^(-) rarr H_(2)(g)`

Options

A. `1.0M` HCl
B. `1.0` M NaOH
C. Pure water
D. A solution with pH=4

Answer

A

Explanation

For `2H^(+)+2e^(-)toH_(2)`,
`E=E^(@)-(0.0591)/(2)"log"(1)/([H^(+)]^(2))=0+(0.0591)/(2)log[H^(+)]^(2)`
`=0.591log[H^(+)]=-0.0591pH`
(a) `[H^(+)]=1M,E=0.0591log1=0`
(b) `[OH^(-)]=10^(0),[H^(+)]=10^(-14)=-0.827V`
`E=-0.0591xx14=-0.827V`
(c) `[H^(+)]=10^(-7),pH=7,E=-0.0591xx7`
`=0.413`V
(d) pH=4,E=-0.0591`xx4=0.236V`.
Thus, potential E is maximum for 1 M HCl.