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

Options

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

Answer

A

Explanation

The reaction `X + Y^(2+) to X^(2+) + Y` will be spontaneous is `E_(cell)^(@)` is +ve .
a) `Zn + Ni^(2+) to Zn^(2+) + Ni` ,
`E_(cell)^(@) = E_(Ni^(2+) | Ni)^(@) - E_(Zn^(2+)|Zn)^(@)`
= `-0.23-(-0.76 ) V = + 0.53` V
b) `Ni + Fe^(2+) to Ni^(2+) + Fe` ,
`E_(cell)^(@) = E_(Fe^(2+) | Fe)^(@) - E_(Ni^(2+) | Ni)^(@)`
`= -44- (-0.23) V = -0.21V`
c) `Ni + Zn to Ni^(2+) + Zn` ,
`E_(cell)^(@) = -0.53` V from (a)
d) `Fe + Zn^(2+) to Fe^(2+) + Zn` ,
`E_(cell)^(@) = E_(Zn^(2+) | Zn)^(@) - E_(Fe^(2+) | Fe)^(@)`
`= -0.76 - (-0.44V) = -0.32` V
Hence (a) is the correct option .