(i) Arrange the following metals in the order in which they displace each other : Al, Cu, Fe, Mg, Zn.
(ii) Electrode potential of metals in their respective solutions are provided. Arrange the metals in their increasing order of reducing power.
`K^+//K=-2.93V,` `Ag^+//Ag=+0.80V,` `Hg^+//Hg=0.79V`
`Mg^(2+)//Mg=-2.36V,` `Cr^(3+)//Cr=-0.74V`

Explanation

(i) The given metals and the corresponding half-cells, and their standard potentials `(E^@)` are

A metal is able to displace other metals from their salt solutions, if the metal has lower (less positive or more negative) electrode potential than the others. The above metals arranged in the increasing order of their `E^@` values are given below:
`underset("Metal:")(E^@,V":")underset(Mg)(-2.36)underset(AI)(-1.66)underset(Zn)(-0.76)underset(Fe)(-0.44)underset(Cu)(+0.34)`
Thus, Mg is able to displace Al, Zn, Fe, and Cu from their salt solutions.
Al is able to displace Zn, Fe and Cu.
Zn is able to displace Fe and Cu.
Fe displaces only Cu.
Cu is not able to displace any of these metals.
(ii) The reducing power of a metal is determined by its electron losing capacity, i.e., by its capacity to get oxidised. So, the metal having the lowest electrode potential (most negative) should therefore be the strongest reducing agent. Therefore, the arrangement in the increasing order of their reducing power is,
`underset("Metal:")(E^@//"Volts:")underset(Ag)(+0.80)underset(Hg)(+0.79)underset(Cr)(-0.74)underset(Mg)(-2.36)underset(K)(-2.93)`
i.e., in this list is the strongest and Ag is the weakest reducing agent.