Question

electrolytic cells A, B and C are connected with identical bulbs in separate
as shown in the diagram. Electrolytic cell A contains sodium chloride
ton and Electrolytic cell B contains acetic acid solution. The electrolytic
cell C contains distilled water.
Distilled
Electrolyte Ceil A
Electrolyte
Electrolyte Cell
Set Up
Set-Up C
(a) In which Set-up will the bulb glow the brightest?
(b) In which Set-up will the glow of the bulb be quite dim?
(c) in which Set-up will the bulb not glow at all?​

Answer 1

Answer:

a). NaCl

b). Acetic acid

c). Distilled water

Explanation:

1). See in case of NaCl since it is a very good electrolyte so it dissociates into ions very easily hence more no. of electrons flow through the cell so the bulb glows brightest.

NaCl --->Na(+) + Cl(-)

2). In case of acetic acid the since it is a weak acid so it behaves like a weak electrolyte hence it does not dissociates into ions easily. so the bulb glows dim

3). Whereas in case of distilled water i.e, it is pure form of water hence it does not dissociate into ions. (for water to dissociate it should be acidified with sulphuric acid). so the bulb does not glow at all in this case.

Answer 2

Answer:

In my opinion

a) A

b) B

c) C

Explanation:

A) It is because in A case the bulb will glow more lighter because it has sodium chloride and electrolytes and also it has a high number of free mobile ions present.

B) In case B the bulb will glow dim because it has acetic acid and it is a solution of weak electrolytes and it has a very little amount of free mobile ions

C) In case C the bulb will not glow at all because it has distilled water and we know that it comes in non electrolytes that is why.