The $E^{Θ}$ for $C l_{2} / C l^{-}$ is $+ 1.36$ for $I_{2} / I^{Θ}$ is +0.53 for $A g^{+} / A g$ is 0.79 $N a^{+} / N a$ is -2.71 and for $L i^{+} / L i$ is -3.04. Arrange the following ionic species in decreasing order of reducing strength : $I^{-}, A g, C l^{-}, L i, N a$

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$I^{-} > N a > L i > A g > C l^{-}$

$L i > N a > I^{-} > A g > C l^{-}$

$L i > I^{-} > N a > C l^{-} > A g$

$I^{-} > A g > C l^{-} > L i > N a$

answer is B.

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Detailed Solution

The reducing strength of an ionic species is related to its tendency to lose electrons and get oxidized. A higher reducing strength means the ionic species is more likely to get oxidized, and hence it can reduce other species. The reducing strength increases as the oxidation state of the element becomes more negative or as its electronegativity decreases.

Based on the given values of E^Θ for the various ionic species, we can arrange them in decreasing order of reducing strength as follows:

Ag⁺ > Cl^- > I^- > Na⁺> Li⁺

Explanation:

The most positive E^Θ value is for Ag⁺/Ag, indicating that Ag⁺ is the strongest oxidizing agent among the given species, and hence has the highest reducing strength. Similarly, Cl^-/Cl₂has a positive E^Θ value, but lower than that of Ag⁺/Ag, indicating that Cl- is a weaker oxidizing agent compared to Ag+ and has a lower reducing strength.

I^-/I₂ also has a positive E^Θ value, but lower than that of Cl^-/Cl₂, indicating that I^- is an even weaker oxidizing agent compared to Cl^-and has a lower reducing strength.

Na+/Na has a negative E^Θ value, indicating that Na⁺ is a reducing agent rather than an oxidizing agent, and hence has a lower reducing strength compared to the positively charged species.

Li+/Li has the most negative EΘ value among the given species, indicating that Li⁺ is the weakest reducing agent and has the lowest reducing strength.

Therefore, the decreasing order of reducing strength of the given ionic species is Ag⁺ > Cl^- > I^- > Na⁺ > Li⁺.

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