In which case `(E_("cell")-E_("cell")^(@))` is zero

Options

A. `Cu|Cu^(2+)(0.01 M)||Ag^(+)(0.1 M)|Ag`
B. `Pt(H_(2))|pH=1||Zn^(2+)(0.01 M)|Zn`
C. `Pt(H_(2))|pH=1||Zn^(2+)(1 M)|Zn`
D. `Pt(H_(2))|H^(+)=0.01 M)||Zn^(2+)(0.01 M)|Zn`

Answer

A, B

Explanation

(a,b) are correct.
(a) `E_(cell)=E_(cell)^(@)-(2.303" RT")/(2F)"log"([Cu^(2+)])/([Ag^(+)]^(2))`
`E_(cell)-E_(cell)^(@)=(2.303" RT")/(2F)"log"(0.01)/((0.1)^(2))`
`E_(cell)-E_(cell)^(@)=0`
(b) `E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"([H^(+)]^(2))/([Zn^(2+)])`
`E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"((1xx10^(-1))^(2))/((0.01))=0`
(c ) `E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"([H^(+)]^(2))/([Zn^(2+)])`
`=-(2.3030" RT")/(2F)"log"((1xx10^(-1))^(2))/(1)ne 0`
(d) `E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"([H^(+)]^(2))/([Zn^(2+)])`
`=-(2.303" RT")/(2F)"log"((0.1)^(2))/((0.01))ne0`