Tag: electrolytic cells and electrolysis

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Questions Related to electrolytic cells and electrolysis

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

Which of the following aqueous solutions remain alkaline after electrolysis ?

  1. $CH _{3}COONa$
  2. $KNO _{3}$
  3. $NaCl$
  4. $LiF$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

a. Reduction potential of $H _{2}O\, >$ Reduction potential of $Na^{\oplus}$
Hence, 
Cathode : $2H _{2}O\, +\, 2e^{-}\, \rightarrow\, 2 \overset{\ominus}{O}H\, +\,\begin{matrix} H _{2}\ (solution\, is\, basic)\end{matrix}$

Anode : $CH _{3}\, COO^{\ominus}\, \overset{Kolbe's\, electrolysis}{\rightarrow}\, C _{2}H _{6}\, (Ethane)\, +\, 2CO _{2}$

b. $K^{\oplus}$ does not undergo reduction but reduction of $H _{2}O$ occurs to give $\overset{\ominus}{O}\, H$ ion and $H _{2}(g)$.
Similarly, $NO^{\ominus} _{3}$ ion does not undergo oxidation but oxidation of $H _{2}O$ occurs to give $H^{\oplus}$ ions and $O _{2}(g)$. 
$H^{\oplus}$ and $\overset {\ominus}{O}H$ ions get neutralised and pH = 7 (neutral solution).
c. $Na^{\oplus}$ ions do not undergo reduction but reduction of $H _{2}O$ occurs to give $\overset{\ominus}{O}H$ ions and $H _{2}(g)$. (Hence, solution is basic). $Cl^{\ominus}$ undergoes oxidation to give $Cl _{2}(g)$. 
d. Same explanation as in (b).

Hence, options A and C are correct.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

A solution containing $Na^{\oplus},\, NO _{3}^{\ominus},\, Cl^{\ominus}$, and $SO _{4}^{2-}$ ions, all at unit concentrations, is electrolyzed between nickel anode and plantinum cathode. As the current is passed through the cell :

  1. pH of the cathode increases

  2. Oxygen is the major product at anode

  3. Nickel is deposited at cathode

  4. Chlorine is the major product at anode
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

At cathode : Reduction of $Na^{\oplus}$ does not occur but reduction of $H _{2}O$ occurs to give $\overset{\ominus}{O}H$ and $H _{2} (g)$, so pOH decreases and pH increases. Hence, option A is correct


At anode : Oxidation of $Cl^{\ominus}$ ions occur to give $Cl _{2} (g)$. Likewise, oxidation of $NO^{\ominus} _{3}$ and $SO _{4}^{2-}$ does not occur, but oxidation of $H _{2}O$ occurs to give $H^{\oplus}$ ions and $H _{2}$ (g). So pH decreases at anode. Also, chlorine forms hence, option D is correct.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

Two platinum electrodes were immersed in a solution of $CuSO _4$ and electric current was passed through the solution. After some time, it was found that colour of $CuSO _4$ disappeared with evolution of gas at the electrode. The colorless solution contains:

  1. Platinum sulphate

  2. Copper hydroxide

  3. Copper sulphate

  4. Sulphuric acid
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$CuSO _4(aq)\, \xrightarrow{Electrolsis} \, Cu^{2+}(aq)\, +\, SO _4^{2-}(aq)$

At cathode: $Cu^{2+}(aq)\, +\, 2e^-\, \rightarrow\, Cu\, (reduction)$

The blue color of $CuSO _4$ disappears due to the deposition of Cu on Pt electrode.

At anode: $H _2O\, \rightarrow\, 2H^{\oplus}\, +\, 2e^-\, \frac{1}{2} O _2(g)$

Since oxidation potential of $H _2O$ > oxidation potential of $SO _4^{2-}$, so oxidation of $H _2O$ occurs and $O _2(g)$ is evolved at anode.

The colourless solution is due to the formation of $H _2SO _4$ as follows:

$2H^{\oplus}\, (from\, anode)\, +\, SO _4^{2-}\, \rightarrow\, H _2SO _4$

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

A dilute solution of sulphuric acid during electrolysis liberate O$ _2$ gas at the anode. 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Because  ${ SO } _{ 4 }^{ 2- }$ mobility is very less as compared to ${ OH }^{ - }$ . So instead of ${ SO } _{ 4 }^{ 2- },{ OH }^{ - }$ undergo oxidation to form ${ O } _{ 2 }$ at anode. 

Also oxidation potential of ${ OH }^{ - }$ is more than ${ SO } _{ 4 }^{ 2- }$ 

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

Which of the following methods can be used to separate hydrogen and oxygen in water?

  1. Boiling

  2. Electrolysis

  3. Distillation

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Electrolysis of ${ H } _{ 2 }O$ helps in separation of ${ H } _{ 2 }O$ to ${ H } _{ 2 }$ & ${ O } _{ 2 }$

${ 2H } _{ 2 }O\longrightarrow { 2H } _{ 2 }+{ O } _{ 2 }$