Tag: chemical changes

Quizzes (7) Questions (139)

Questions Related to chemical changes

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 ${ H } _{ 2 }$ & ${ O } _{ 2 }$

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

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes.

  1. $\text{1M CuSO} _4$ solution, $\text{1M CuCl} _2$ solution
  2. $\text{1M KCl}$ solution, $\text{1M Kl}$ solution
  3. $\text{1M AgNO} _3$ solution, $\text{1M Cu(NO} _3) _2$ solution
  4. $\text{1M KCl}$ solution, $\text{1M NaCl}$ solution
  5. $\text{1M CuBr} _2$ solution, $\text{1M CuSO} _4$ solution
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Electrolysis of 1M KCl and 1M NaCl using inert electrodes both yield potassium or sodium ions remaining in solution while hydrogen gas is evolved at the cathode and chlorine gas at the anode, making their electrolysis products identical and indistinguishable.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

On passing one friday of electric charge through a dilute solution of an acid, the volume of hydrogen obtained at S.T.P. is :

  1. 22400 ML.

  2. 1120 mL

  3. 2240 mL

  4. 11200 ML.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

One Faraday (96500 C) of charge deposits or releases 1 gram equivalent of a substance. For H2 gas (n=2), 1 Faraday releases 0.5 moles of H2. Volume at STP = 0.5 * 22400 mL = 11200 mL.