The standard reduction potential for `Zn^(2+)//Zn, Ni^(2+)//Ni` and `Fe^(2+)//Fe` are respectively - 0.76, -0.23 and -0.44 V. The reaction `X + Y^(2+) rarr X^(2+) + Y` will have more negative `Delta G` value when X and Y are

Options

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

Answer

D

Explanation

Given, `Zn^(2+)//Zn = -0.76 V`
`Ni^(2+)//Ni = -0.23 V`
`Fe^(2+)//Fe = -0.44 V`
and, relation
`X + Y^(2+) rarr X^(2+) + Y`, means
X is oxidised to `X^(2+)` (acts as anode)
`Y^(2+)` is reduced to Y (acts as cathode)
Also, `E_("cell")^(@) = E_(C)^(@) - E_(A)^(@)`
Now, for
(a) X = Ni and Y =Fe
`therefore E_("cell")^(@) = E_(Y)^(@) - E_(X)^(@)`
`E_("cell")^(@) = -0.44 + 0.23 = -021 V`
Thus, `Delta G^(@) = -nFE^(@)`
`= (-2 xx -0.21)F`
`= +0.42 F " "(n = 2)`
(b) X = Ni, Y = Zn
`therefore E_("cell")^(@) = -0.76 - (-0.23) = -0.53 V`
`therefore Delta G^(@) = -nFE^(@) = (-2 xx -0.53) F`
`Delta G^(@) = + 1.06 F`
(c ) X = Fe, Y = Zn
`E_("cell")^(@) = -0.76 - (-0.44) = -0.32 V`
`therefore Delta G^(@) = -n FE^(@) = (-2 xx -0.32)F = +0.64 F`
(d) X = Zn, Y = Ni
`E_(cell")^(@) = -0.23 - (-0.76) = +0.53`
`therefore Delta G^(@) = -nFE^(@) = (-2 xx 0.53)F`
`= -1.06 F`
(e) X = Fe, Y= Ni
`E_("cell")^(@) = -0.23 - (-0.44) = +0.21 V`
`therefore Delta G^(@) = -nFE^(@) = (-2 xx 0.21)F`
`Delta G^(@) = 0.42 F`
Hence, maximum -ve value of `Delta G^(@)` is possible in case (d).
In other words,
when `E^(@) = -v e, Delta G^(@) = + v e` and
when `E^(@) = + v e, Delta G^(@) = - v e`
Thus, options (a), (b) and (c ) give positive `Delta G^(@)` and options (d) and (e) give `Delta G^(@)` negative. But for option (d), `Delta G^(@)` is more -ve.