Question

A 0.200 M KOH solution is electrolysed for 1.5 hours using a current of $8.00A{}^\circ $. How many moles were prepared at the anode?
(A)- 0.48
(B)- 0.224
(C)- 0.112
(D)- $2.24\times {{10}^{-2}}$

Answer 1

Hint: Electrolysis of potassium hydroxide melts potassium hydroxide to potassium producing oxygen and water. The anodic and cathodic process for the electrolysis of KOH is given as-
Cathodic process- $4{{K}^{+}}+4{{e}^{-}}\to 4K$ (Reduction)
Anodic process- $4O{{H}^{-}}-4{{e}^{-}}\to {{O}_{2}}+2{{H}_{2}}O$ (Oxidation)

Complete answer:
-A simple electrolysis unit consists of an anode and cathode connected through an external power supply and immersed in a conducting electrolyte. A direct current (DC) is applied to the unit due to which electrons flow the negative terminal of the DC power source to the cathode where they are consumed by hydrogen ions (protons) to form hydrogen atoms.
-The positive ions in the water electrolysis move towards the cathode whereas the negatively charged ions move toward the anode.
-1 mol of oxygen is produced from 4 moles of hydroxide ion.
According to Ohm’s law,
$Q=It$
Where q = charge, I = current, t = time
Charge $=8A\times 1.5\times 60\times 60=43200C$
$4\times 96500C$ produces ${{O}_{2}}$= 1mol
Therefore, 43200C will produce ${{O}_{2}}=\dfrac{43200}{4\times 96500}=0.112mol$

Hence the correct answer is option C.

Note:
Potassium hydroxide (KOH) is commonly known as caustic soda which is the largest-volume potassium chemical for non-fertiliser use and is produced by the electrolysis of potassium chloride with membrane cell technology producing co-products as chlorine and hydrogen. Potassium salt in the form of brine is electrolysed using membrane cell technology to produce potassium hydroxide, chlorine and hydrogen. KOH is available in both liquid and dry forms for various applications such as fertilizers, batteries, soaps and detergents. It is a highly corrosive and reactive chemical which can be irritating to the skin, eyes and gastrointestinal tract.