First slide

Surface phenomenon

Question

You mix AgNO₃ and KI in the following molar ratio
AgNO₃ KI
I. 1 1
II. 1 2
III. 2 1
Fixed layer of the electrical double layer migrates to cathode in

Moderate

A

Only I
B

Only II
C

Only III
D

Both II and III

Solution

I. $\underset{1}{{AgNO}_{3}} + \underset{1}{K} I ⟶ AgI ↓ + {KNO}_{3}$
Being in equal stoichiometric ratio, Agl is precipitated. It carries no charge and, hence no migration
II. $\underset{1}{{AgNO}_{3}} + \underset{2}{KI} ⟶ AgI + {KNO}_{3}$
KI is in excess hence Agl adsorbs common ion of KI (ie. I^-)
$AgI + KI ⟶ [AgI] I^{-} ⋮ K^{+}$
Fixed layer Mobile layer
Fixed layer carries negative charge and moves to anode plate.
III. ${\underset{2}{AgNO}}_{} + \underset{1}{KI} ⟶ AgI + {KNO}_{3}$
Agl is precipitated and AgNO₃ being in excess, Ag⁺ is adsorbed on AgI.
$AgI + {AgNO}_{3} ⟶ [AgI] {Ag}^{+} : {NO}_{3}^{-} Fixed layer Mobile layer$
Fixed layer carries positive charge and migrates to cathode (-ve) plate.

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