The standard reduction potentials for `Zn^(2+)//Zn, Ni^(2+)//Ni, and Fe^(2+)//Fe` are -0.76, -0.23 and -0.44 V respectively. The reaction `X + y^(2+) rarr + Y` will be spontaneous when :

Options

A. `X= Ni, Y= Fe`
B. X=Ni,Y=Zn
C. X = Fe, Y = Zn
D. X= Zn, Y= Ni

Explanation

A cell is Sponteneous , if` DeltaG^(@)lt0`
Since,`DeltaG^(@)=-nFE_(cell)^(@)`
Thus `E_(cell)^(@)gt0,E_(cell)^(@)=E_("ox")^(@)+E_(red)^(@)`
(a)` X=Ni, Y=Fe`
`Ni+Fe^(2+)to X^(2+)+Fe`
`E_(Ni//Ni^(2+))^(@)=+0.23V`
`E_(Fe^(2+)//Fe)^(@)=-0.44V`
Thus, `E_(cell)^(@)=E_(Ni//Ni^(2+))^(@)+E_(Fe^(2+)//Fe)^(@)=-0.21V`
i.e, reaction is non-spontaneous .
(b)` X=Ni, Y=zn`
`Ni+zn^(2+)toNi^(2+)+zn`
`E_(Ni//Ni^(2+))^(@)=0.23V`
`E_(zn^(2+)//zn)^(@)=-0.76V`
`therefore E_(cell)^(@)=0.23+(-0.76)=-0.53V`
i.e is non -sponteneous .
(c)X=Fe. Y=zn
`Fe+zn^(2+)toFe^(2+)+zn`
`E_(cell)^(@)=E_(Fe//Fe^(2+))^(@)+E_(zn^(2+)//zn)^(@)=0.44-0.76=-0.32V therefore E_(cell)^(@)lt0`
i.e, reaction is non-sponteneous.
X=zn, Y=Ni
`Zn+Ni^(2+)toZn^(2+)+Ni`
`E_(cell)^(@)=E_(zn//zn^(2+))^(@)+E_(Ni^(2+)//Ni)^(@)`
`-0.76-0.23=-therefore E_(cell)^(@)gt0`
i.e, reaction is sponteneous.