The electrolytic cells A, B, and C are connected with identical bulbs in separate circuits . Electrolytic cell A contains Sodium chloride solution and electrolytic cell B contains acetic acid solution. The electrolytic cell C contains distilled water. (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 justify your answer
Answers
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(a) Sodium Chloride (NaCl) is a more conducting electrote as the ions in this are more freely, and as we Know, the more free the ions are, the more electric current then make flow.
(b) Acetic Acid i.e is little low in this because it, has free ions, but it moves not as freely as in NaCl , so in this case the bulb would glow but, would be less brighter as in comparison to the first one
(c) In Distilled water, bulb does not glow at all, because in this , the electrotes does not let the free ions move, so in this case, the bulb does not glow at all
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1. Electricity is conducted by charged particles such as electron,proton,ions etc.
2. Sodium chloride solution and acetic acid solution contains ions such as Na+,Cl-,CH3COO- and H+.
3. Sugar solution does not contain any ions so it does not conduct electricity.