$500ml$ of $0.3M{AgNO}_3$ are added to $600ml$ of $0.5M$ KI solution. The ions which will move towards the cathode and anode respectively are

The complete reaction is as follows:

${\mathrm{AgNO}}_3+\mathrm{KI}\to \mathrm{AgI}+{\mathrm{KNO}}_3$

Milliequivalance of silver nitrate  is equal to 0.3x500=150

600 ml is the volume of 0.5 m potassium iodide. 

Iodide of potassium milliequivalence = 0.5x600=300

Silver nitrate is a limiting reagent since there is fewer milliequivalent silver nitrate present. 

At the conclusion of the process, 150 milliequivalents of potassium iodide will still be present. 

Iodide ion will be adsorb to create a negatively charged sol, which will then draw potassium ion and create a mobile ion. 

Because of this, a mobile layer of potassium ions will be drawn to the cathode and a fixed layer to the anode. 

Hence the correct option is (C).