Electrolytes and Electrolysis Questions in English

(A) The reduction reaction for the deposition of aluminium from its molten salt is given by:
$Al^{3+} + 3e^- \to Al$.
According to the stoichiometry of the reaction,$1 \ mole$ of $Al^{3+}$ ions requires $3 \ moles$ of electrons to be reduced to $1 \ mole$ of $Al$ metal.
Therefore,$3 \ moles$ of electrons are required.

(C) Electric current is defined as the rate of flow of charge.
Since charge is measured in $Coulombs$ $(C)$ and time in $seconds$ $(s)$,the unit of electric current is $Coulomb/second$ $(C/s)$.
$1 \ C/s$ is defined as $1 \ Ampere$ $(A)$.
Therefore,the $SI$ unit of electric current is $Ampere$.

(A) In the molten or fused state,$NaCl$ dissociates into its constituent ions: $NaCl(l) \rightarrow Na^+(l) + Cl^-(l)$.
Since the ions are free to move,fused $NaCl$ acts as an electrolyte and conducts electricity.
$CO_2$ is a covalent molecule,$Br_2$ is a non-polar covalent liquid,and $Si$ is a semiconductor,but fused $NaCl$ is a classic ionic conductor.

(D) In electrolytic conductors,current is carried by the movement of ions,not by a stream of electrons. Electrons flow through metallic conductors,whereas ions move towards electrodes in electrolytic solutions to form new products. Therefore,the statement that a single stream of electrons flows from cathode to anode is not applicable to electrolytic conductors.

(B) In solid state,$NaCl$ exists as a crystal lattice where ions are held by strong electrostatic forces and are not free to move.
Since there are no free ions or free electrons to carry charge,$NaCl$ (solid) does not conduct electricity.

(D) . When the cathode is removed,the external electric circuit is broken,and the electric field across the electrolyte vanishes. In the absence of an electric field,the ions in the solution no longer experience a directional force and move randomly due to thermal energy.

(B) $Pt$ electrodes are inert and do not participate in the reaction at either electrode.
Dilute $H_{2}SO_{4}$ contains $H^{+}$ ions from both $H_{2}SO_{4}$ and $H_{2}O$.
At the cathode,$H^{+}$ ions gain electrons to form hydrogen gas:
$2H^{+} + 2e^{-} \rightarrow H_{2}(g)$
Since this process involves the gain of electrons,it is defined as a reduction reaction.
Therefore,the cathodic reaction in the electrolysis of dilute $H_{2}SO_{4}$ with $Pt$ electrodes is reduction.

(C) In electrolytic cells,the electrodes must be chemically inert so that they do not react with the electrolyte or the products of electrolysis.
$Na$ is a highly reactive alkali metal that reacts vigorously with water.
$S$ (sulfur) is a non-metal and a poor conductor of electricity.
$Hg$ (mercury) is a liquid metal and can be used as an electrode in specific cases (like the Castner-Kellner cell),but $Pt$ (platinum) and graphite are the standard inert electrodes used in aqueous electrolytic cells because they are chemically stable and conduct electricity well.
Therefore,$Pt$ and graphite are the most suitable choices for general electrolytic cells.

(A) During the electrolysis of dilute $H_2SO_4$ using platinum electrodes,the following reactions occur:
At the cathode: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$
At the anode: $2OH^- (aq) \rightarrow \frac{1}{2} O_2 (g) + H_2O (l) + 2e^-$
Thus,$H_2$ gas is evolved at the cathode and $O_2$ gas is evolved at the anode. Therefore,option $A$ is correct.

(D) Anode reaction involves oxidation,which is the loss of electrons.
$A, B,$ and $C$ represent oxidation reactions (loss of electrons).
$D$ represents the reduction reaction (gain of electrons),where $Zn^{2+}$ is reduced to $Zn$.

(C) In an electrolytic cell,electrical energy is converted into chemical energy through a non-spontaneous redox reaction.

(A) In the process of electroplating,the object to be coated is always made the $Cathode$ (negative electrode).
Metal ions from the electrolyte solution are reduced at the $Cathode$ to form a thin layer of metal on the surface of the article.

(C) The electrolyte used for gold plating is potassium dicyanoaurate$(I)$,which is represented as $K[Au(CN)_2]$.
This complex salt provides a stable source of gold ions for the electroplating process.

(B) The metals $Mg$ (Magnesium) belongs to the $s$-block elements (alkaline earth metals).
Due to their highly negative standard reduction potentials,the reduction of water occurs more readily than the reduction of $Mg^{2+}$ ions at the cathode during the electrolysis of an aqueous solution.
Consequently,$H_2$ gas is evolved at the cathode instead of the deposition of $Mg$ metal.
In contrast,$Cu$,$Cr$,and $Ni$ have higher reduction potentials and can be deposited at the cathode from their aqueous solutions.

(B) $Mg$ and $Al$ cannot be obtained by the electrolysis of aqueous solutions of their salts.
This is because the reduction potential of $H_2O$ is higher than that of $Mg^{2+}$ and $Al^{3+}$ ions.
Consequently,$H_2$ gas is liberated at the cathode instead of the metal.

(A) The primary concern regarding the use of mercury in the Castner-Kellner process for $NaOH$ production is environmental and health safety.
Mercury $(Hg)$ is highly toxic and poses significant risks of bioaccumulation in the food chain.
While $Hg$ is a good conductor of electricity,this property is actually utilized in the cell; however,the environmental hazards associated with mercury leakage and disposal are the main reasons for its phase-out in modern industrial applications.

(C) The electroplating of $Cr$ is performed primarily because $Cr$ provides a hard,corrosion-resistant,and aesthetically pleasing (decorative) finish to the base metal. This process is widely used to prevent corrosion and improve the appearance of objects.

(A) Electroplating is a process of coating a metal surface with a thin layer of another metal to prevent corrosion or improve appearance.
$Ni$ (Nickel),$Zn$ (Zinc),and $Au$ (Gold) are commonly used for electroplating due to their resistance to corrosion and aesthetic appeal.
$Fe$ (Iron) is highly reactive and prone to rusting,making it unsuitable for use as a protective plating material on other metals.

(B) To prevent the rusting of iron,a process called galvanization is used.
In this process,a layer of $Zn$ (zinc) is deposited on the iron surface to protect it from corrosion.

(C) The electrolysis of a solution of sodium chloride $(NaCl)$ produces chlorine $(Cl_2)$ and sodium hydroxide $(NaOH)$.
The chlorine reacts with ethyl alcohol $(C_2H_5OH)$ to form chloral $(CCl_3CHO)$.
Subsequently,the chloral reacts with the sodium hydroxide produced during electrolysis to yield chloroform $(CHCl_3)$ and sodium formate $(HCOONa)$.

(D) Anodizing is an electrochemical process that converts the metal surface into a decorative,durable,corrosion-resistant,anodic oxide finish.
In the case of aluminium,the metal is made the anode in an electrolytic cell containing an electrolyte (like dilute sulphuric acid).
When current is passed,oxygen is evolved at the anode,which reacts with the aluminium surface to form a protective layer of aluminium oxide $(Al_2O_3)$.
Therefore,anodized aluminium refers to aluminium with an electrolytically deposited layer of aluminium oxide on its surface.

(A) cyclotron is a device used to accelerate charged particles. It uses electric and magnetic fields to increase the kinetic energy of charged particles. Since a neutron is a neutral particle (having no charge),it cannot be accelerated by the electric field in a cyclotron.

(B) An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent,such as water.
$NaCl$ (salt) dissociates into ions ($Na^+$ and $Cl^-$) in water,allowing it to conduct electricity.
Glucose,sugar,and urea are non-electrolytes because they do not dissociate into ions in aqueous solution.

(B) The dissociation of $AlCl_3$ in solution is given by: $AlCl_3 \rightarrow Al^{+3} + 3Cl^-$.
From the stoichiometry of the reaction,$1 \text{ mole}$ of $AlCl_3$ produces $1 \text{ mole}$ of $Al^{+3}$ ions and $3 \text{ moles}$ of $Cl^-$ ions.
Therefore,the number of $Al^{+3}$ ions is $1/3$ of the number of $Cl^-$ ions.

(D) An electrolyte is a substance that dissociates into ions in solution.
Strong electrolytes dissociate completely into ions,whereas weak electrolytes dissociate only partially.
$1$. Urea $(NH_2CONH_2)$ and Sugar $(C_{12}H_{22}O_{11})$ are non-electrolytes as they do not dissociate into ions.
$2$. Ammonium hydroxide $(NH_4OH)$ is a weak base and thus a weak electrolyte.
$3$. Sodium acetate $(CH_3COONa)$ is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$. Salts are generally strong electrolytes because they dissociate completely in aqueous solution.
Therefore,sodium acetate is the strongest electrolyte among the given options.

(A) The electrolysis of molten $NaCl$ produces sodium metal $(Na)$.
However,the electrolysis of an aqueous solution of $NaCl$ (brine) produces $NaOH$,$Cl_2$,and $H_2$ gas.
Given the context of industrial production,$NaOH$ is the primary product of the chlor-alkali process involving aqueous $NaCl$.

(B) During the preparation of $NaOH$ by the chlor-alkali process,$H_2$ gas is liberated at the cathode.

(B) The reduction of $H^+$ ions at a $Hg$ cathode requires a higher overpotential (voltage) compared to a $Pt$ cathode. Due to this high overpotential,the reduction of $Na^+$ ions to $Na$ metal becomes more favorable at the $Hg$ cathode,which then reacts with $Hg$ to form sodium amalgam $(Na-Hg)$.

(B) During the electrolysis of an aqueous solution of $NaCl$ using $Pt$ electrodes,the reduction of water occurs at the cathode because the reduction potential of $H_2O$ is higher than that of $Na^+$. The reaction at the cathode is: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$. Thus,$H_2$ gas is produced at the cathode.

(D) During the electrolysis of an aqueous solution of $Na_2SO_4$ using inert electrodes,water undergoes electrolysis.
At the cathode,the reduction of water occurs: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.
At the anode,the oxidation of water occurs: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$.
Therefore,the products obtained at the cathode and anode are $H_2$ and $O_2$,respectively.

(D) The electrolysis of brine ($NaCl$ solution) involves the following reactions:
$NaCl \rightleftharpoons Na^+ + Cl^-$
$H_2O \rightleftharpoons H^+ + OH^-$
At the cathode: $2H^+ + 2e^- \to H_2$ (Since the reduction potential of $H^+$ is higher than that of $Na^+$).
At the anode: $2Cl^- - 2e^- \to Cl_2$ (Since the oxidation potential of $Cl^-$ is higher than that of $OH^-$).
In the solution: $Na^+ + OH^- \to NaOH$.
Thus,$NaOH$,$Cl_2$,and $H_2$ are produced,but $O_2$ is not produced.

(B) The dissociation of aqueous sodium sulfate is: $Na _2 SO _4(a q) \rightarrow 2 Na ^{+}(a q)+ SO _4^{2-}(a q)$
Water also dissociates slightly: $H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}$.
At the cathode,the reduction potential of $H^+$ ions is higher than that of $Na^+$ ions.
Therefore,$H^+$ ions are reduced to form hydrogen gas: $2H^+_{(aq)} + 2e^- \to H_{2(g)}$.

(B) During the electrolysis of aqueous $CuSO_4$ using platinum electrodes,the following reactions occur:
At the cathode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$
At the anode: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$
As the reaction proceeds,$H^+$ ions are produced at the anode,which increases the concentration of $H^+$ in the solution.
Since $pH = -\log[H^+]$,an increase in $[H^+]$ leads to a decrease in the $pH$ of the solution.

(C) The electrolysis of water follows the reaction: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$.
According to the stoichiometry,the molar ratio of $H_2$ gas to $O_2$ gas produced is $2:1$.
Since the volume of gas is directly proportional to the number of moles under the same conditions of temperature and pressure (Avogadro's Law),the ratio of volumes will also be $2:1$.
Given volume of $O_2 = 2.24 \ dm^3$.
Therefore,volume of $H_2 = 2 \times 2.24 \ dm^3 = 4.48 \ dm^3$.

(A) During the electrolysis of aqueous $CuSO_4$ solution using inert platinum electrodes,the following reactions occur:
At the cathode: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. Thus,copper is deposited at the cathode.
At the anode: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$. Oxygen gas is evolved at the anode.
Therefore,copper is deposited only at the cathode.

(D) In an electrolytic cell,the movement of ions is driven by the potential difference applied across the electrodes.
When the cathode is removed,the circuit is broken.
Since the circuit is no longer complete,the electric field that drives the migration of ions towards the electrodes ceases to exist.
Consequently,the directional movement of both positive and negative ions stops.

(D) During the process of electrolysis,an electric current is passed through an electrolyte solution or melt.
This causes the cations (positively charged ions) to migrate towards the cathode and the anions (negatively charged ions) to migrate towards the anode.
At the electrodes,these ions gain or lose electrons to become neutral atoms or molecules,a process known as discharge.
Therefore,the ions become discharged.

(B) During the electrolysis of molten $NaCl$,the ions present are $Na^{+}$ and $Cl^{-}$.
At the anode (positive electrode),oxidation occurs.
The $Cl^{-}$ ions migrate to the anode and lose electrons to form chlorine gas:
$2Cl^{-} \rightarrow Cl_2 + 2e^{-}$.
Therefore,$Cl^{-}$ is oxidized at the anode.

(D) Molten $KCl$ $(potassium \ chloride)$ is a good conductor of electricity because it contains free mobile ions ($K^+$ and $Cl^-$) that allow the flow of electric current. In contrast,diamond is a covalent network solid with no free electrons,and crystalline $NaCl$ and $BaSO_4$ have ions fixed in a lattice structure,preventing conduction.

(D) When an electrolyte $AB$ dissociates in an aqueous solution,it splits into its constituent ions:
$AB_{(aq)} \rightarrow A^{+}_{(aq)} + B^{-}_{(aq)}$
Therefore,both cations $(A^{+})$ and anions $(B^{-})$ are produced.

(B) The reduction reaction at the cathode is: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.
From the stoichiometry,$1 \, mol$ of $H_2$ requires $2 \, mol$ of electrons.
Therefore,$0.01 \, mol$ of $H_2$ requires $0.01 \times 2 = 0.02 \, mol$ of electrons.
Total charge $Q = n \times F = 0.02 \, mol \times 96500 \, C \, mol^{-1} = 1930 \, C$.
Given current $I = 10 \, mA = 10 \times 10^{-3} \, A = 0.01 \, A$.
Using the formula $Q = I \times t$,we get $t = \frac{Q}{I} = \frac{1930 \, C}{0.01 \, A} = 193000 \, s$.
This can be written as $19.3 \times 10^4 \, s$.

(C) In an aqueous solution of $NaBr$,the ions present are $Na^+$,$Br^-$,$H^+$,and $OH^-$.
At the cathode,the reduction potential of $H_2O$ $(-0.83 \ V)$ is higher than that of $Na^+$ $(-2.71 \ V)$,so $H_2$ gas is evolved: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.
At the anode,the oxidation potential of $Br^-$ is higher than that of $H_2O$,so $Br_2$ is produced: $2Br^-(aq) \rightarrow Br_2(l) + 2e^-$.
Consequently,$H_2$ gas is produced at the cathode,$Br_2$ is produced at the anode,and $NaOH$ remains in the solution.

(C) The formula for total charge is $Q = I \times t$.
Given current $I = 100 \ mA = 100 \times 10^{-3} \ A = 0.1 \ A$.
Given time $t = 30 \ minutes = 30 \times 60 \ seconds = 1800 \ s$.
Substituting the values: $Q = 0.1 \ A \times 1800 \ s = 180 \ C$.
Therefore,the total charge used is $180 \ C$.

(A) During electrolysis,an electric current is passed through an electrolyte solution.
Due to the applied potential difference,the positively charged ions (cations) are attracted towards the negatively charged electrode (cathode).
Similarly,the negatively charged ions (anions) are attracted towards the positively charged electrode (anode).
Therefore,anions move towards the anode and cations move towards the cathode.

(C) In molten $NaCl$,the dissociation is: $NaCl(l) \rightarrow Na^{+} + Cl^{-}$.
At the cathode (reduction occurs): $Na^{+} + e^{-} \rightarrow Na_{(s)}$.
At the anode (oxidation occurs): $Cl^{-} \rightarrow 1/2 Cl_2 + e^{-}$ or $2Cl^{-} \rightarrow Cl_2 + 2e^{-}$.

(B) During the process of electrolysis,the cathode is the negatively charged electrode.
Cations are positively charged ions that migrate towards the cathode.
At the cathode,these cations gain electrons to become neutral species (reduction).
Therefore,the species that becomes neutral at the cathode is the cation.

(C) The process in which a chemical change occurs by the passage of an electric current through an electrolyte is known as $Electrolysis$. During this process,electrical energy is converted into chemical energy.

(A) The deposition of metals at the cathode depends on the standard reduction potential $(E^{\circ})$. The metal with the higher reduction potential is reduced first at the cathode.
Comparing the given values: $E^{\circ}_{Ag^+/Ag} (0.80 \ V) > E^{\circ}_{Hg_2^{2+}/2Hg} (0.79 \ V) > E^{\circ}_{Cu^{2+}/Cu} (0.34 \ V) > E^{\circ}_{Mg^{2+}/Mg} (-2.37 \ V)$.
Therefore,the order of deposition is $Ag > Hg > Cu$.
Note: $Mg$ will not be deposited from an aqueous solution because the reduction of $H^+$ ions to $H_2$ gas occurs more readily than the reduction of $Mg^{2+}$ ions.

(B) $Na$ (sodium) is a highly reactive alkali metal and $S$ (sulfur) is a non-metal that does not conduct electricity well and reacts with water or the electrolyte.
Therefore,they cannot be used as inert electrodes.
$Hg$ (mercury),$Pt$ (platinum),and graphite are commonly used as electrodes in aqueous electrolytic cells because they are relatively inert or stable under electrolysis conditions.

(A) The dissociation of sodium sulfate in water is: $Na _2 SO _4(a q) \rightarrow 2 Na ^{+}(a q)+ SO _4^{2-}(a q)$
Water also dissociates slightly: $H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}$.
At the cathode,$H^+$ ions are reduced in preference to $Na^+$ ions because the reduction potential of $H^+$ is higher: $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$.
At the anode,$OH^-$ ions are oxidized in preference to $SO_4^{2-}$ ions: $4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^-$.
Therefore,the products at the cathode and anode are $H_2$ and $O_2$,respectively.