Tag: electrolysis and its applications

Questions Related to electrolysis and its applications

The resulting solution obtained at the end of electrolysis of concentrated aqueous solution of $NaCl:$

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

  1. Turns blue litmus into red

  2. Turns red litmus into blue

  3. Remains colourless with phenolphthalein

  4. The colour of red or blue litmus does not change

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Correct Option: B
Explanation:

During the electrolysis of aqueous sodium chloride solution, the products are $NaOH , Cl _2$ and $H _2$.
$NaCl(aq) + H _2O(l) \rightarrow Na^+ (aq) +OH^-(aq) +\frac{1}{2} H _2(g) +\frac{1}{2}Cl _2(g)$
Due to formation of base, it turns res litmus into blue.

Which of the following batteries are responsible for direct environmental hazard?

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

  1. Alkaline dry cell

  2. Mercury cell

  3. $Ni-Cd$ battery

  4. Lithium battery

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Correct Option: A,B,C
Explanation:

Alkaline batteries are prone to leaking potassium hydroxide, a caustic agent that can cause respiratory, eye and skin irritation. Potassium, if it leaks, can cause severe chemical burns thereby affecting the eyes and skin.

Mercury can even be absorbed through the skin. These harmful substances permeate into the soil, groundwater and surface water through landfills and also release toxins into the air when they are burnt in municipal waste combustors.

For humans, both lead and cadmium can be taken only by ingestion or inhalation.  Moreover, cadmium is easily taken up by plant roots and accumulates in fruits, vegetables and grass. The impure water and plants in turn are consumed by animals and human beings, who then fall prey to a host of ill-effects.

Since Li-ion batteries contain less of toxic metals than other types of batteries which may contain lead or cadmium, they are generally categorized as non-hazardous waste. 

A, B and C are correct options.

An acidic solution of $C{u^{2 + }}$ salt containing $0.4\,g$ of $C{u^{2 + }}$ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml and the current at $1.2$ amp. Find the volume of gases evolved at anode and Cathode at NTP during the entire electrolysis. 

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

  1. $78.20\,ml,\,\,99.78\,ml$

  2. $58.48\,ml,\,\,78.20\,ml$

  3. $98.56\,ml,\,\,58.24\,ml$

  4. $78.20\,ml,\,\,58.48\,ml$

Show answer
Correct Option: C
Explanation:
Solution:- (C) $98.56 mL, \; 58.24 \; mL$
Assuming ${Cu}^{+2}$ salt to be $CuS{O} _{4}$, the reactions occuring at the electrodes would be-
At anode:-
${H} _{2}O \longrightarrow 2 {H}^{+} + \cfrac{1}{2} {O} _{2} + 2 {e}^{-}$
At cathode:- 
${Cu}^{+2} + 2 {e}^{-} \longrightarrow Cu$
Equivalent weight of ${Cu}^{+2} = 31.5 \; g$
$0.4 \; g$ of ${Cu}^{+2} = \cfrac{0.4}{31.5} = 0.0127 \; g$ equivalent
At the same time the oxygen deposited at anode.
Equivalent weight of oxygen $= 8 \; gm$
Mass of oxygen deposited $= 0.0127 \times \cfrac{8}{32} = 0.0031 \text{ mol}$
After the complete deposition of copper, the reactions would be-
At anode:-
${H} _{2}O \longrightarrow 2 {H}^{+} + \cfrac{1}{2} {O} _{2} + 2 {e}^{-}$
At cathode:-
$2 {H} _{2}O + 2{e}^{-} \longrightarrow {H} _{2} + 2 {OH}^{-}$

Given:-
$I = 1.2 \; A$
$t = 7 \text{ min} = 7 \times 60 = 420 \; s$
Amount of charge passed $= I \times t \ = 1.2 \times 7 \times 60 = 504 \; C$
Therefore,
Amount of oxygen liberated $= \cfrac{1}{96500} \times 504 = 0.00523 \; g$ equivalent $= \cfrac{8}{32} \times 0.00523 = 0.0013 \text{ mol}$
Amount of hydrogen liberated $= 0.00523 \; g$ equivalent $= \cfrac{1}{2} \times 0.00523 = 0.0026 \text{ mol}$
Now,
Gas evolved at anode $= {O} _{2}$
Total no. of moles of ${O} _{2}$ evolved $= 0.0031 + 0.0013 = 0.0044 \text{ mol}$
$\therefore$ Volume of gas evolved at anode $= 0.00447 \times 22400 = 98.56 \; mL$
Gas evolved at cathode $= {H} _{2}$
Total no. of moles of ${H} _{2}$ evolved $= 0.0026 \text{ mol}$
$\therefore$ Volume of gas evolved at cathode $= 0.0026 \times 22400 = 58.24 \; mL$
Hence the volume of gases evolved at anode and cathode at NTP during the entire electrolysis $98.56 \; mL$ and $58.24 \; mL$ respectively.

During electrolysis of brine solution, product obtained at cathode and anode are:

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

  1. hydrogen and chlorine gas respectively

  2. chlorine and hydrogen gas respectively

  3. sulphate and chlorine gas respectively

  4. sodium and hydrogen gas respectively

Show answer
Correct Option: A
Explanation:

$At\ cathode:{ 2H } _{ 2 }O+{ 2e }^{ - }\rightarrow { H } _{ 2 }+{ 2OH }^{ - }\ At\ anode:{ 2Cl }^{ - }\rightarrow { Cl } _{ 2 }+{ 2e }^{ - }$

Conductivity of $NaCl$ is more:

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

  1. in a aqueous solution than in alcohol

  2. in alcohol than in a aqueous solution

  3. both in aqueous solution and in alcohol

  4. none of the above

Show answer
Correct Option: A
Explanation:

 The conductivity of NaCl is more in an aqueous solution than in alcohol.
Water is a highly polar solvent. In water, NaCl completely dissociates into sodium ions and chloride ions.
$\displaystyle NaCl \rightarrow Na^+  +  Cl^-$
The electrical conductivity of aqueous NaCl solution is due to a presence of these ions.
Alcohol is less polar than water. Hence, the degree of ionization of NaCl in alcohol is much lower than that in water.   Hence, a lesser number of current carrying ions are present in alcohol. Hence, $NaCl$ in alcohol has lower conductivity than $NaCl$ in an aqueous solution.

3%solution of glucose is isotonic with 1% solution of a non-volatile non-electrolyte substance the molecular mass of the substance would be 

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

  1. 30

  2. 60

  3. 90

  4. 120

Show answer
Correct Option: B
Explanation:

since the two solution are isotonic,

therefore, $C _1 RT = C _2 RT$
$C _1 = C _2$
let  mass of solution = 100 g
so mass of glucose = 3g
and mass of non-volatile substance = 1g
so, we get
$\dfrac{3}{180} = \dfrac{1}{M}$
$M = 60$
hence, molecular mass of non-volatile substance = 60

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} $

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Correct Option: A