CBSE Class 12 · 2007 · Delhi · Redox Reactions and ElectrochemistryLong Answer

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i) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is

0.146 \times 1 0^{- 3}

S cm

^{- 1}

?
ii) Predict the products of electrolysis in the following: a solution of H

_{2}

_{4}

with platinum electrodes.

Answer: (i) G* = 0.219 cm⁻¹ (ii) H₂ at cathode, O₂ at anode.

Solution

(i) Cell constant:

G^{*} = κ \times R = 0.146 \times 1 0^{- 3} \times 1500

G^{*} = 0.219 c m^{- 1}

(ii) Dilute H

_{2}

_{4}

with Pt electrodes:
Cathode:

2 H^{+} + 2 e^{-} \to H_{2}

— hydrogen gas.
Anode:

2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}

— oxygen gas (SO

_{4}^{2 -}

is not discharged).

CBSE Class 12 Board · 2007 · Delhi · Subjective

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