Electrolytes and Electrolysis Questions in English

(B) During the electrolysis of water,the reaction at the cathode is: $2H^+ (aq) + 2e^- \rightarrow H_2 (g)$.
Since hydrogen ions $(H^+)$ are positively charged cations,they migrate towards the negatively charged electrode,which is the cathode,to get reduced and liberated as hydrogen gas.

(C) Electrolyte solutions or molten salts conduct electricity through the process of electrolysis.
In electrolytes,the electric current is carried by the movement of positive and negative ions,not by electrons.
For example,in a lead-acid battery,the electrolyte is dilute $H_2SO_4$,which dissociates into positive $H^+$ ions and negative $SO_4^{2-}$ ions.

(B) In electroplating,the object to be plated (the spoon) must act as the cathode.
During electrolysis,the metal ions from the electrolyte are reduced at the cathode and deposit on its surface.
Therefore,the spoon is placed at the position of the cathode.

(A) The formula for electric charge is $q = I \times t$,where $I$ is current in amperes and $t$ is time in hours.
Therefore,$1 \text{ Ampere hour} = 1 \text{ A} \times 1 \text{ h} = 3600 \text{ Coulombs}$.
Thus,Ampere hour is a unit of quantity of charge.

(B) The total charge $Q$ passed is given by $Q = I \times t = 1 \ A \times 10 \ s = 10 \ C$.
The deposition of $Cu^{2+}$ ions at the cathode follows the reaction: $Cu^{2+} + 2e^- \rightarrow Cu$.
This shows that $2$ moles of electrons are required to deposit $1$ mole of $Cu$.
The charge required to deposit $1$ ion of $Cu^{2+}$ is $2 \times e^- = 2 \times 1.6 \times 10^{-19} \ C = 3.2 \times 10^{-19} \ C$.
The number of $Cu^{2+}$ ions deposited by $10 \ C$ charge is $\frac{10 \ C}{3.2 \times 10^{-19} \ C/ion} \approx 3.125 \times 10^{19} \ ions$.
Thus,the number of copper ions is about $3.1 \times 10^{19}$.

(B) Molten sodium chloride $(NaCl)$ exists in a liquid state where the ionic lattice breaks down,resulting in the presence of free $Na^+$ and $Cl^-$ ions. These free ions are responsible for the conduction of electricity.

(A) Ionic compounds consist of ions that are free to move in a molten state or in an aqueous solution. Due to the presence of these mobile ions,they act as good electrolytes and conduct electricity effectively.

(B) An electrolyte is a substance that dissociates into ions when dissolved in a solvent like water,thereby allowing the solution to conduct electricity.
Therefore,the correct option is $(b)$.

(D) An electrolyte is a substance that dissociates into ions in an aqueous solution or in a molten state,allowing it to conduct electricity.
$ (D) $ $ MgCl_2 $ is an ionic compound.
In an aqueous solution,it dissociates as follows:
$ MgCl_2 (aq) \rightleftharpoons Mg^{2+} (aq) + 2Cl^{-} (aq) $
Chloroform,benzene,and toluene are covalent compounds and do not dissociate into ions.

(C) The electrical conductivity of a solution depends on the concentration of ions present in the solution.
$H_2SO_4$ is a strong acid that undergoes complete ionization in aqueous solution:
$H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
Since it provides the highest concentration of ions compared to the other weak acids listed (Boric acid,Acetic acid,and Phosphoric acid),it is the best conductor of electricity.

(C) The reaction that absorbs energy is known as an endothermic reaction.
Electrolysis of water is an endothermic process because it requires the input of electrical energy to break the chemical bonds in water molecules.
The chemical equation is: $H_2O_{(l)} \xrightarrow{\text{electrolysis}} H_{2_{(g)}} + \frac{1}{2}O_{2_{(g)}}$.

(D) The electrolysis of molten sodium chloride $(NaCl)$ involves the following reactions:
At the cathode: $2Na^+ + 2e^- \rightarrow 2Na$
At the anode: $2Cl^- \rightarrow Cl_2 + 2e^-$
The overall reaction is: $2NaCl \text{ (molten)} \xrightarrow{\text{Electrolysis}} 2Na \text{ (cathode)} + Cl_2 \text{ (anode)}$
Therefore,the correct option is $D$.

(A) The electrolysis of brine ($NaCl$ solution) is known as the Chlor-alkali process.
During this process,$NaCl$ dissociates into $Na^+$ and $Cl^-$ ions.
At the cathode,water is reduced to produce $H_2$ gas and $OH^-$ ions: $2H_2O + 2e^- \to H_2 + 2OH^-$.
At the anode,$Cl^-$ ions are oxidized to produce $Cl_2$ gas: $2Cl^- \to Cl_2 + 2e^-$.
The $Na^+$ ions combine with $OH^-$ ions to form $NaOH$.
The overall reaction is: $2NaCl(aq) + 2H_2O(l) \to 2NaOH(aq) + Cl_2(g) + H_2(g)$.

(B) During the electrolysis of an aqueous solution of sodium chloride $(NaCl)$,the following reactions occur:
At the cathode,water molecules are reduced in preference to sodium ions because the reduction potential of water is higher than that of $Na^+$ ions.
The reaction at the cathode is: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.
Thus,$H_2$ gas is liberated at the cathode.
The overall reaction is: $2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{Electrolysis}} 2NaOH(aq) + Cl_2(g) + H_2(g)$.

(D) $2NaCl \xrightarrow{\text{Electric current}} 2Na^{+} + 2Cl^{-}$
At anode (oxidation): $2Cl^{-} \rightarrow Cl_2 + 2e^{-}$
At cathode (reduction): $2Na^{+} + 2e^{-} \rightarrow 2Na$
Thus,the anodic reaction is the oxidation of chloride ions.

(A) The correct answer is $A$.
Down’s cell is specifically designed and used for the industrial electrolysis of fused $NaCl$ to produce metallic sodium and chlorine gas.

(B) Sodium is extracted by the electrolytic reduction of molten or fused sodium chloride (Downs process).
On passing electricity through molten sodium chloride,decomposition occurs,and sodium metal is formed at the cathode,while chlorine gas is formed at the anode.
The chemical reaction is:
$2 NaCl \text{ (fused)} \xrightarrow{\text{Electrolysis}} 2 Na + Cl_2$

(C) During the electrolysis of an aqueous solution of $NaCl$ (brine),the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \to H_2(g) + 2OH^-(aq)$
At the anode: $2Cl^-(aq) \to Cl_2(g) + 2e^-$
Thus,hydrogen gas is evolved at the cathode and chlorine gas is evolved at the anode.

(B) Chlorine is commercially produced by the electrolysis of brine ($NaCl$ solution) in the Castner-Kellner cell or diaphragm cell.
The chemical reaction is:
$2NaCl_{(aq)} + 2H_2O_{(l)} \xrightarrow{\text{Electrolysis}} 2NaOH_{(aq)} + Cl_{2(g)} + H_{2(g)}$
Thus,the correct option is $(B)$.

(A) The electrolysis of an aqueous solution of $NaCl$ (brine) is used for the commercial production of $Cl_2$ gas.
$2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.
Aluminium is extracted by the electrolysis of fused $Al_2O_3$ in cryolite.
Calcium is extracted by the electrolysis of fused $CaCl_2$.
Bromine is extracted by the oxidation of bromide ions from sea water.

(C) The electrolysis of an aqueous solution of potassium acetate is known as the $Kolbe's$ electrolysis method.
In this process,$2CH_3COOK + 2H_2O$ $\xrightarrow{\text{Electrolysis}} CH_3-CH_3 + 2CO_2 \uparrow \text{ (at Anode)} + 2KOH + H_2 \uparrow \text{ (at Cathode)}$.
Thus,ethane is obtained at the anode.

(C) An electrolyte is a substance that dissociates into ions when dissolved in a solvent or when melted.
In the solid state,an electrolyte consists of a crystal lattice where ions are held together by strong electrostatic forces.
Therefore,it does possess ions even in the solid state,although they are not mobile,which prevents the conduction of electricity until the substance is dissolved or melted.

(D) When an electrolyte is dissolved in water,the high dielectric constant of water weakens the electrostatic forces of attraction between the ions of the solute.
This allows the ions to separate and become hydrated,resulting in dissociation into $Na^{+}$ and $Cl^{-}$ ions (in the case of $NaCl$).
Therefore,the correct reason is that the forces of electrostatic attraction are broken down by water.

(D) An electrolyte conducts electricity due to the presence of free ions. $ \text{When an electrolyte is fused or dissolved in a solvent, it dissociates into ions.} $
$ \text{These ions act as charge carriers, allowing the flow of electric current through the solution or melt.} $

(B) The standard reduction potential of $Mg^{2+}/Mg$ is $-2.37 \ V$,which is much lower than the reduction potential of water $(E^o = -0.83 \ V)$.
In an aqueous solution,water molecules are reduced more easily than $Mg^{2+}$ ions.
Therefore,during electrolysis,$H_2$ gas is evolved at the cathode instead of $Mg$ metal.
Thus,$Mg$ cannot be obtained by the electrolysis of its aqueous salt solution.

(B) During the electrolysis of an aqueous solution of sodium sulphate $(Na_2SO_4)$,water is reduced at the cathode and oxidized at the anode instead of $Na^{+}$ and $SO_4^{2-}$ ions.
Cathode reaction: $2H_2O + 2e^- \to H_2(g) + 2OH^{-}(aq)$
Anode reaction: $H_2O(l) \to 2H^{+}(aq) + \frac{1}{2}O_2(g) + 2e^-$
Thus,$H_2$ gas is evolved at the cathode.

(A) In an electrolysis process,oxidation occurs at the anode and reduction occurs at the cathode. Therefore,the correct sequence is $A$.

(A) The correct answer is $A$.
Sugar $(C_{12}H_{22}O_{11})$ is a covalent compound.
It does not dissociate into ions in an aqueous solution.
Since electricity conduction in aqueous solutions requires the presence of free ions,sugar solution does not conduct electricity and thus does not undergo electrolysis.
In contrast,$NaCl$,$NaBr$,and $CH_3COONa$ are ionic compounds that dissociate into ions in water,allowing them to undergo decomposition via electrolysis.

(C) An electric current causes decomposition in an aqueous solution only if the solute is an electrolyte,meaning it dissociates into ions in water.
$AgNO_3$ (Silver nitrate) is an ionic compound that dissociates into $Ag^+$ and $NO_3^-$ ions in an aqueous solution,making it an electrolyte.
Urea,glucose,and ethyl alcohol are non-electrolytes and do not dissociate into ions; therefore,they do not decompose under the passage of an electric current.
Thus,the correct option is $(C)$.

(A) During the electrolysis of an aqueous solution of sodium sulphate $(Na_2SO_4)$,the ions present are $Na^+$,$SO_4^{2-}$,$H^+$,and $OH^-$.
At the cathode,$H^+$ ions are reduced in preference to $Na^+$ ions because the reduction potential of $H^+$ is higher than that of $Na^+$.
The reaction at the cathode is: $2H_2O(l) + 2e^- \to H_2(g) + 2OH^-(aq)$.
At the anode,$OH^-$ ions (or $H_2O$) are oxidized in preference to $SO_4^{2-}$ ions.
The reaction at the anode is: $2H_2O(l) \to O_2(g) + 4H^+(aq) + 4e^-$.
Therefore,the products at the cathode and anode are $H_2$ and $O_2$ respectively.

(C) 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: $2H_2O (l) \rightarrow O_2 (g) + 4H^+ (aq) + 4e^-$
Thus,$O_2$ gas is evolved at the anode.

(A) In fused $NaCl$,the electrolyte dissociates as $NaCl \rightarrow Na^+ + Cl^-$.
At the anode,negatively charged chloride ions $(Cl^-)$ migrate and undergo oxidation to form chlorine gas: $2Cl^- \rightarrow Cl_2 + 2e^-$.
Therefore,chloride ions are oxidized at the anode.

(B) During the electrolysis of aqueous $HCl$,the following reactions occur at the electrodes:
At the cathode: $2H^+_{(aq)} + 2e^- \to H_{2(g)}$
At the anode: $2Cl^-_{(aq)} \to Cl_{2(g)} + 2e^-$
Therefore,$H_2$ gas is produced at the cathode and $Cl_2$ gas is produced at the anode.

(B) In the reaction $Na^{+} + e^- \to Na$,the oxidation state of sodium decreases from $+1$ to $0$.
Since there is a gain of electrons and a decrease in oxidation number,this process is defined as reduction.

(B) The correct answer is $B$. In an aqueous solution of $NaCl$,the ions present are $Na^+$,$Cl^-$,$H^+$,and $OH^-$.
At the cathode,$H_2O$ is reduced in preference to $Na^+$ because the reduction potential of water is higher than that of $Na^+$.
Cathode reaction: $2H_2O(l) + 2e^- \to H_2(g) + 2OH^-(aq)$.
At the anode,$Cl^-$ is oxidized to $Cl_2$ gas.
Anode reaction: $2Cl^-(aq) \to Cl_2(g) + 2e^-$.

(B) In the molten state,$NaCl$ dissociates into its constituent ions:
$NaCl(l) \rightleftharpoons Na^{+}(l) + Cl^{-}(l)$.
These free ions are responsible for the electrical conductivity of molten $NaCl$.

(A) During the electrolysis of aqueous $CuSO_4$ with inert electrodes (like platinum),the following reactions occur:
At the cathode,$Cu^{2+}$ ions are reduced to $Cu$ metal,but if the concentration is low or considering the standard reduction potentials,$H_2O$ can also be reduced. However,typically $Cu$ deposits. If the question implies the evolution of gases,we consider the water decomposition reactions:
Cathode: $2H_2O + 2e^- \to H_2(g) + 2OH^-(aq)$
Anode: $H_2O \to 2H^+(aq) + \frac{1}{2}O_2(g) + 2e^-$
Thus,the gas evolved at the cathode is $H_2$ and at the anode is $O_2$.

(C) Electrolysis is a process that uses electrical energy to drive a non-spontaneous chemical reaction.
It is widely used in industries for $Electroplating$ (coating one metal with another) and $Electrorefining$ (purifying metals).
Therefore,both $(a)$ and $(b)$ are correct applications of electrolysis.

(D) Electrolysis is defined as a process of decomposing ionic compounds into their elements by passing a direct electric current through the compound in a fluid form.
During this process,cations migrate to the cathode and gain electrons to become neutral atoms (reduction).
Anions migrate to the anode and lose electrons to become neutral atoms (oxidation).
Thus,both ions are discharged at their respective electrodes.

(B) During electrolysis,the cathode is negatively charged,so it attracts positively charged ions called $Cations$. Therefore,$Cations$ are discharged at the cathode,while $Anions$ are discharged at the anode.

(A) The electrolysis of molten anhydrous $CaCl_2$ involves the following reactions:
At the cathode: $Ca^{2+} + 2e^- \rightarrow Ca$
At the anode: $2Cl^- \rightarrow Cl_2 + 2e^-$
Therefore,$Calcium$ is produced at the cathode.

(D) Fused potassium chloride $(KCl)$ consists of free mobile ions in the molten state,which allows it to conduct electricity.
In contrast,diamond is a covalent network solid with no free electrons,while crystalline sodium chloride and barium sulphate are ionic solids where ions are held in a rigid lattice and are not free to move.

(C) The correct answer is $(C)$.
$A$ $Coulometer$ (also known as a voltameter) is an electrochemical cell used for the measurement of the quantity of electricity passed through an electrolyte by measuring the amount of substance deposited or liberated at the electrodes.

(A) When an electric current is passed through an electrolyte solution,the ions experience an electrostatic force due to the applied potential difference.
Cations (positively charged ions) are attracted towards the cathode (negatively charged electrode).
Anions (negatively charged ions) are attracted towards the anode (positively charged electrode).
Therefore,the correct behavior is that anions move towards the anode and cations move towards the cathode.

(D) During the electrolysis of $CuSO_4$ 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 $Cu^{2+}$ ions are removed from the solution to form copper metal at the cathode,the blue colour of the solution fades.
The remaining ions in the solution are $H^+$ and $SO_4^{2-}$,which combine to form $H_2SO_4$ (sulphuric acid).
Thus,the final colourless solution contains sulphuric acid.

(A) The relationship between charge $(Q)$,current $(I)$,and time $(t)$ is given by the formula: $Q = I \times t$.
Since the unit of charge is Coulomb $(C)$,the unit of current is Ampere $(A)$,and the unit of time is second $(s)$,we have: $1 \ C = 1 \ A \times 1 \ s$.
Therefore,Coulomb is equal to ampere $\times$ second.

(D) The current $I$ is given as $10^{-6} \ A$,which means $10^{-6} \ C$ of charge passes through the cross-section per second.
The charge of a single electron is $e = 1.602 \times 10^{-19} \ C$.
The number of electrons $n$ passing per second is calculated using the formula $n = \frac{I}{e}$.
$n = \frac{10^{-6} \ C/s}{1.602 \times 10^{-19} \ C} \approx 6.24 \times 10^{12} \ \text{electrons/s}$.
Rounding to the nearest provided option,the correct answer is $6 \times 10^{12}$.

(D) An aqueous solution decomposes on passing an electric current if it contains ions that can undergo oxidation and reduction at the electrodes. This process is known as electrolysis.
$Canesugar$,$Urea$,and $Methanol$ are non-electrolytes and do not dissociate into ions in aqueous solution.
$Potassium \ iodide$ $(KI)$ is an ionic compound that dissociates into $K^+$ and $I^-$ ions in water,making it an electrolyte that undergoes electrolysis.

(B) The quantity of electricity $(Q)$ is defined as the product of current $(I)$ and time $(t)$.
$Q = I \times t$
Given $I = 1 \ A$ and $t = 1 \ s$,we get $Q = 1 \ A \times 1 \ s = 1 \ C$.
Therefore,the quantity of electricity is known as $1 \ Coulomb$.