Tag: electrolytic cells and electrolysis

Questions Related to electrolytic cells and electrolysis

During the electrolysis  of aqueous zinc nitrate:

  1. Zinc plates out at the cathode

  2. Zinc plates out at the anode

  3. Hydrogen gas $H _{2}$ is evolved at the anode

  4. Oxygen gas $O _{2}$ is evolved at the anode


Correct Option: A,D

Which of the following aqueous solutions remain alkaline after electrolysis ?

  1. $CH _{3}COONa$

  2. $KNO _{3}$

  3. $NaCl$

  4. $LiF$


Correct Option: A,C
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.

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


Correct Option: A,D
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.

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


Correct Option: D
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:

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

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

  1. True

  2. False


Correct Option: A
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 ${ SO } _{ 4 }^{ 2- }$ 

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


Correct Option: B
Explanation:

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

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