Consider the following data
E^o Cl₂/Cl^– = +1.36,  E^o I₂/I^– =+ 0.53,  E^o  Ag⁺ /Ag = +0.79,  E^o Na⁺/Na = –2.71 and  Li⁺ /Li is – 3.04.
Arrange the following ionic species in decreasing order of reducing strength: 

I^–, Ag, Cl^–, Li, Na.

1. Li > Na > I^– > Ag > Cl^–

2. Cl^– > Na > I^– > Ag > Li

3. Li < Na < I^– < Ag < Cl^–

4. Li > Na = I^– > Ag > Cl^–