Without losinf its concentration, `ZnCl_(2)` solution can't be kept in contact with

Options

A. `Al`
B. `Ag`
C. `Pb`
D. `Au`

Answer

A

Explanation

`ZnCl_(2)` solution can't be kept in contact with `Al` because `E_(Zn^(2+)//(Zn)^(@)`. Thus, `Zn^(2+)(aq.)` will vbe reduced by `Al` The correct order of reduction potentials is
`E_(Au^(3+)//Au)^(@) gt E_(Pb^(2+)//Pb)^(@) gt E_(Zn^(2+)//Zn)^(@) gt E_(Al^(3+)//Al)^(@)`