Question

The oxidation potential of hydrogen half-cell will be negative if:
A. `p(H_(2))=1" atm and" [H^(+)]=1 M`
B. `p(H_(2))=1" atm and" [H^(+)]=2 M`
C. `p(H_(2))=0.2" atm and" [H^(+)]=1 M`
D. `p(H_(2))=0.2" atm and" [H^(+)]=2 M`

Answer 1

Correct Answer - B::C
(b c ) The Nernst equation for the oxidation reaction :
`H_(2) to 2H^(+)+2e^(-)` is :
`E_("ox")=E_("ox")^(@)-(0.0591)/(n)"log"([H^(+)]^(2))/(pH_(2))`
`E_("ox")=-(0.0591)/(2)"log"([H^(+)])/(pH_(2))`
n=2 , `E_(ox)^(@)`
Let us determine `E_("ox")` in different cases : (a)`p_((H_(2)))=1" atm", [H^(+)]=1" M"`
`E_("ox")=-(0.0591)/(2)"log"((1)^(2))/(1)=0`
(b)`p_((H_(2)))=1" atm", [H^(+)]=1" M"`
`E_("ox")=-(0.0591)/(2)"log"((2)^(2))/(1)`
`=-(0.0591)/(2)xx2 log2=-0.178" V"`
(c )`p_((H_(2)))=0.2" atm", [H^(+)]=1" M"`
`E_("ox")=-(0.0591)/(2)"log"((1)^(2))/(0.2)=-(0.0591)/(2)log5`
`=-0.027" V"`
(d)`p_((H_(2)))=0.2" atm", [H^(+)]=0.2" M"`
`E_("ox")=-(0.0591)/(2)"log"((0.2)^(2))/(0.2)=-(0.0591)/(2)log0.2`
`=-(0.0591)/(2)xx(-0.6989)=+0.0207" V"`