Tag: chemistry

Questions Related to chemistry

During the purification of copper by electrolysis:

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

  1. the anode used is made of copper ore

  2. pure copper is deposited on the cathode

  3. the impurities such as $Ag,\space Au$ present in solution as ions

  4. concentration of $CuSO _4$ solution remains constant during dissolution of $Cu$

Show answer
Correct Option: A,B,D
Explanation:
Anode : - Impure copper or crude copper.
Cathode : Pure copper
Electrolyte : $15\%$ $CuSO _{4}$ solution $+5\%$ $H _{2} SO _{4}$
When electric current is passed through the electrolyte, the angle anode gradually dissolve and pure copper is deposited on the cathode which gradually grow in size. The impurities like, Fe,Zn,Ni, etc, dissolved in the solution as  while gold, silver, platinum settle down the anode as anode mud.
Reaction Occurring are as follows.
$CuSO _{4}\rightleftharpoons Cu^{+2}+SO _{4}^{-2};$ At anode : $Cu-2e\rightarrow Cu^{+2}$
                                                Cathode : $Cu^{+2}+2e\rightarrow Cu$

When an aqueous concentrate solution of lithium chloride is electrolysed using inert electrodes:

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

  1. $Cl _2$ is liberated at the anode

  2. $Li$ is deposited at the cathode

  3. as the current flows, $pH$ of the solution around the cathode remains constant

  4. as the current flows, $pH$ of the solution around the cathode increases

Show answer
Correct Option: A,D
Explanation:

As chloride ion have more oxidation potential than hydroxide ion, so $Cl _2$ is liberated at the anode.
And reduction potential of hydrogen ion is more so hydrogen gas produces at cathode so as the current flows, pH of the solution around the cathode increases.

Oxygen and hydrogen gas are produced at the anode and cathode respectively, during the electrolysis of fairly concentrated aqueous solution of:

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

  1. $K _2SO _4$

  2. $AgNO _3$

  3. $H _2SO _4$

  4. $NaOH$

Show answer
Correct Option: A,C,D
Explanation:

As the reduction potential of $K^+$ and $ Na^+$ are lower than hydrogen ion, hydrogen ion gets preferably reduced and hydrogen gas is produced at the cathode.
Also, as oxidation potential of $OH^-$ is higher than sulphate ion, it will oxidise first and oxygen is evolved at the anode.

Hence, a fairly concentrated electrolytic aqueous solution of $K _2SO _4,\ H _2SO _4$ and $NaOH$ produces oxygen and hydrogen gas.

$ A \,Tl^{+} |Tl$ couple was prepared by saturating $ 0.10 M-KBr $ with TlBr and allowing $ Tl^{+}$ ions form the insoluble bromide to equilibrate. This couple was observed to have a potential $ -0.444 V $ with respect to $ PB^{2+} | Pb $ couple in which $ Pb^{2+}$ was 0.10 M. What is the $K _{sp} $ of $ TlBr.$ [Given :$ E _{Pb^{2+}|Pb}^{0} = -0.126 V, E _{Tl^{+}|Tl}^{0} = -0.336 V,$
$ log 2.5 = 0.4, 2.303 RT/F = 0.06]$

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

  1. $ 4.0 \times 10^{-6}$

  2. $ 2.5 \times 10^{-4}$

  3. $ 4.0 \times 10^{-5} $

  4. $ 6.3 \times 10^{-3} $

Show answer
Correct Option: A

During the electrolysis  of aqueous zinc nitrate:

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

  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

Show answer
Correct Option: A,D

Which of the following aqueous solutions remain alkaline after electrolysis ?

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

  1. $CH _{3}COONa$

  2. $KNO _{3}$

  3. $NaCl$

  4. $LiF$

Show answer
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 :

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

  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

Show answer
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:

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

  1. Platinum sulphate

  2. Copper hydroxide

  3. Copper sulphate

  4. Sulphuric acid

Show answer
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. 

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

  1. True

  2. False

Show answer
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?

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

  1. Boiling

  2. Electrolysis

  3. Distillation

  4. None of the above

Show answer
Correct Option: B
Explanation:

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

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