a) At 291 K, the molar conductivities at infinite dilution of NH$_4$Cl, NaOH and NaCl are 129.8, 217.4 and 108.9 ohm$^{-1}$ cm$^2$ mol$^{-1}$ respectively. If the molar conductivity of a centinormal solution of NH$_4$OH is 9.33 ohm$^{-1}$ cm$^2$ mol$^{-1}$, what is the degree of dissociation of the NH$_4$OH solution? b) The standard reduction potential values of three metallic cations X, Y, Z are 0.52, −3.03 and −1.18 V respectively. What will be the order of reducing power of the corresponding metals?

CBSE Class 12 · 2006 · All India Compartment · Redox Reactions and ElectrochemistryShort Answer

a) At 291 K, the molar conductivities at infinite dilution of NH

_{4}

Cl, NaOH and NaCl are 129.8, 217.4 and 108.9 ohm

^{- 1}

^{2}

mol

^{- 1}

respectively. If the molar conductivity of a centinormal solution of NH

_{4}

OH is 9.33 ohm

^{- 1}

^{2}

mol

^{- 1}

, what is the degree of dissociation of the NH

_{4}

OH solution?
b) The standard reduction potential values of three metallic cations X, Y, Z are 0.52, −3.03 and −1.18 V respectively. What will be the order of reducing power of the corresponding metals?

Answer: (a) α ≈ 0.039 (3.9%) (b) X < Z < Y

Solution

(a) By Kohlrausch's law:

Λ_{m}^{\circ} (N H_{4} OH) = Λ_{m}^{\circ} (N H_{4} Cl) + Λ_{m}^{\circ} (NaOH) - Λ_{m}^{\circ} (NaCl)

= 129.8 + 217.4 - 108.9 = 238.3 oh m^{- 1} c m^{2} mo l^{- 1}

Degree of dissociation:

α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{9.33}{238.3}

α = 0.039 (3.9%)

(b) The more negative the reduction potential, the stronger the metal as a reducing agent:

X < Z < Y

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CBSE Class 12 · 2006 · All India Compartment · Redox Reactions and ElectrochemistryShort Answer

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