• # question_answer 2) The ${{E}^{o}}$ for $C{{l}_{2}}/C{{l}^{-}}$is + 1.36, for iodine ${{I}_{2}}/{{I}^{-}}$is +0.53, for $A{{g}^{+}}/Ag$is $+0.79$,$N{{a}^{+}}/Na$ is$-2.71$and for$L{{i}^{+}}/Li$ is$-\text{ }3.04$. Arrange the following species in order of reducing strength: ${{I}^{-}},Ag,C{{l}^{-}},Li,Na$

Answer:

Reducing nature depends on the tendency of losing electron or electrons. More the negative reduction potential, more is the tendency to lose electron or electrons. Thus, reducing strength decreases from top to bottom in the electrochemical series. Thus, the order is: $Li>Na>{{I}^{-}}>Ag>C{{l}^{-}}$

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