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Equivalent, Molar Conductivity and Cell Constant

The E∘ for Cl...

Question

The $E^{\circ}$ for $C l_{2} / C l^{-}$ is $+ 1.36$ , for $I_{2} / I^{-}$ 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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Solution

The alkali metals are strong reducing agents

The standard electrode potential $(E_{o})$ measures the reducing

Power, Higher the negative $E_{o}$ value , higher will be reducing power.

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

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Similar questions

Arrange the following as per instructions given in the brackets.

(a) Mg. Cl, Na, S, Si (decreasing order of atomic size)

(b) Cs, Na, Li, K, Rb (increasing metallic character)

(c) Na, K, Cl, S, Si (increasing ionisation potential)

(d) Cl, F, Br, I (increasing electron affinity)

(e) Cs, Na, Li, K, Rb (decreasing electronegativity)

The order of decreasing ionisation enthalpy in alkali metals is
(a) Na> Li > K> Rb (b) Rb< Na< K< Li
(c) Li > Na> K> Rb (d) K< Li < Na < Rb

Q. Choose the correct answer form the alternatives given.

E^{\circ}

C l_{2} / C l^{-} = + 1.36, I_{2} / I^{-} = + 0.53, A g^{+} / A g = + 0.79, N a^{+} / N a = - 2.71

L i^{+} / L i = - 3.04 V

What is the order of decreasing reducing strength for these species?

Q. The correct order of atomic radii for the following species is

N a, L i, R b,

K

Q. arrange the following in increasing order of atomic radii
O ,Na ,Cl ,Li

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Equivalent, Molar Conductivity and Cell Constant

Standard VII Chemistry

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