Question

Given the standard reduction potentials $ Z{{n}^{2+}}/Zn=-0.74\,V, $ $ C{{l}_{2}}/C{{l}^{-}}=1.36\,V, $ $ {{H}^{+}}/1/2{{H}_{2}}=0\,V $ and $ F{{e}^{2+}}/F{{e}^{3+}}=0.77\,V $
The order of increasing strength as reducing agent is:

• A. $ C{{l}^{-}},Zn,{{H}_{2}},F{{e}^{2+}} $
• B. $ {{H}_{2}},Zn,F{{e}^{2+}},Cl $
• C. $ C{{l}^{-}}F{{e}^{2+}},Zn,{{H}_{2}} $
• D. $ {{H}_{2}}F{{e}^{2+}},C{{l}^{-}},Zn $
• E. $ C{{l}^{-}},F{{e}^{2+}},{{H}_{2}},Zn $

Answer 1

Answer: (E)

The substances which have lower reduction potentials are stronger reducing agents. Hence the order of strength of reducing agent is increases in the following order:
$ \xrightarrow[strength\text{ }of \text{ }reducing\text{ }agent-increases]{Cl < F{{e}^{2+}} < {{H}_{2}} < Zn} $