Question

2. Three electrolytic cells A, B and C are connected with identical bulbs in separate circuits as shown in the diagram. Electrolytic cell A contains sodium chloride solution and Electrolytic cell B contains acetic acid solution. The electrolytic cell C contains distilled water, (.) (.) 4.) Sodium Chloride solution Acetic acid Distilled water Electrolyte Cell A Set-Up A Electrolyte Cell B Electrolyte Cell C Set-Up B 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? Justify your answers.​

Answer 1

Answer:

$(a)$  In electrolyte cell A, the bulb glows the brightest·

$(b)$ In electrolyte cell B, the glow of the bulb be quite dim·

$(c)$ In electrolyte cell C, the bulb does not glow at all·

Explanation:

$(a)$  In electrolyte cell A, the bulb glows the brightest·

Because here the electrolyte is sodium chloride solution·Sodium chloride, NaCl is a strong electrolyte that can completely dissociate into its respective cation and anion· If there is five NaCl solid present in the solution, it will dissociate and give five sodium cation and five chloride anion·

That is,

     $5\ NaCl_(_s_)$       ⇄      $5\ Na^{+}_(_a_q_) \ \ \ +\ \ \ 5\ Cl^{-}_(_a_q_)$

There will be no NaCl solid left in the solution after its dissociation and it is said to be $100$% dissociated or ionized· So the amount of the ion in the solution will be comparatively higher· Since these ions are charged one, the positively charged $Na^{+}$ ion will move towards the negative electrode and the negatively charged $Cl^-$ ion will move towards the positive electrode· So this movement of the ions produces the current· Since the electrolyte is a strong electrolyte, the current produced will be very higher· So the bulb glows in brightest in this setup·

$(b)$ In electrolyte cell B, the glow of the bulb be quite dim·

Because here the electrolyte used is acetic acid· Acetic acid, $CH_3COOH$ is a weak electrolyte that does not undergo a complete dissociation in the solution·  

       $CH_3COOH$    ⇄    $CH_3COO^-\ \ \ +\ \ \ H^+$

After the dissociation, the solution contains both the dissociated ions and the  $CH_3COOH$ molecules· Only the movement of charged ions can create the electric current· Since the number of charged ions produced after the dissociation is less, the current produced will also be less· So, the glow of the bulb will be quite dim in this setup·

$(c)$ In electrolyte cell C, the bulb does not glow at all·

Because in this setup, the distilled water is used as electrolyte· Distilled water is the purest form of water, that is free from any other impurities and ions·  It contains only neutral molecules which are chargeless species·  As we know that, only the movement of the charged molecules can conduct electricity, distilled water is unable to conduct electricity· So the bulb will not glow in this setup·