CBSE Class 12 · 2016 · All India · Redox Reactions and ElectrochemistryLong Answer

(a) The conductivity of a 0.001 mol L

^{- 1}

solution of CH

_{3}

COOH is

3.905 \times 1 0^{- 5}

S cm

^{- 1}

. Calculate its molar conductivity and degree of dissociation (α).
Given: λ°(H

^{+}

) = 349.6 S cm

^{2}

mol

^{- 1}

and λ°(CH

_{3}

COO

^{-}

) = 40.9 S cm

^{2}

mol

^{- 1}

.
(b) Define electrochemical cell. What happens if the external potential applied becomes greater than the E°_cell of the electrochemical cell?

Solution

(a) Molar conductivity:

Λ_{m} = \frac{κ \times 1000}{C} = \frac{3.905 \times 1 0 ^{- 5} \times 1000}{0.001}

Λ_{m} = 39.05 S c m^{2} mo l^{- 1}

By Kohlrausch's law:

Λ_{m}^{\circ} = 349.6 + 40.9 = 390.5 S c m^{2} mo l^{- 1}

Degree of dissociation:

α = \frac{39.05}{390.5}

α = 0.1 (10%)

(b) An electrochemical (galvanic) cell is a device that converts the chemical energy of a spontaneous redox reaction into electrical energy.
If the applied external potential exceeds

E_{ce l l}^{\circ}

, electrons are forced in the opposite direction: the cell reaction reverses and the cell functions as an electrolytic cell.

CBSE Class 12 Board · 2016 · All India · Subjective

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CBSE Class 12 · 2016 · All India · Redox Reactions and ElectrochemistryLong Answer

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