Electrolytes and Electrolysis Questions in English

(A) During the electrolysis of $CuSO_{4(aq)}$ with copper electrodes,the copper anode undergoes oxidation: $Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^-$.
At the cathode,the copper ions from the solution undergo reduction: $Cu^{2+}_{(aq)} + 2e^- \to Cu_{(s)}$.
Thus,copper dissolves at the anode and deposits at the cathode.
Since oxidation occurs at the anode and reduction at the cathode,the Reason correctly explains the Assertion.

$i$. At the cathode,$Ag^+$ ions are reduced to $Ag_{(s)}$. At the anode,$Ag$ electrode is oxidized to $Ag^+_{(aq)}$.
$ii$. At the cathode,$Ag^+$ ions are reduced to $Ag_{(s)}$. At the anode,$H_2O$ is oxidized to $O_{2_{(g)}}$.
$iii$. At the cathode,$H^+$ ions are reduced to $H_{2_{(g)}}$. At the anode,$H_2O$ is oxidized to $O_{2_{(g)}}$.
$iv$. At the cathode,$Cu^{2+}$ ions are reduced to $Cu_{(s)}$. At the anode,$Cl^-$ ions are oxidized to $Cl_{2_{(g)}}$.

(A) Given current $I = 0.5 \ A$ and time $t = 2 \ hours = 2 \times 3600 \ s = 7200 \ s$.
The total charge $Q$ flowing through the wire is given by $Q = I \times t = 0.5 \ A \times 7200 \ s = 3600 \ C$.
The charge of one electron is $e = 1.602 \times 10^{-19} \ C$.
The number of electrons $n$ is calculated as $n = \frac{Q}{e} = \frac{3600 \ C}{1.602 \times 10^{-19} \ C} \approx 2.247 \times 10^{22} \approx 2.25 \times 10^{22}$.
Thus,$2.25 \times 10^{22}$ electrons flow through the wire.

(N/A) $(i)$ At cathode: $Ag_{(aq)}^{+} + e^{-} \longrightarrow Ag_{(s)}$ (Silver deposits).
At anode: $Ag_{(s)} \longrightarrow Ag_{(aq)}^{+} + e^{-}$ (Silver dissolves).
$(ii)$ At cathode: $Ag_{(aq)}^{+} + e^{-} \longrightarrow Ag_{(s)}$ (Silver deposits).
At anode: $2H_2O_{(l)} \longrightarrow O_{2_{(g)}} + 4H_{(aq)}^{+} + 4e^{-}$ (Oxygen gas is evolved).
$(iii)$ At cathode: $H_{(aq)}^{+} + e^{-} \longrightarrow \frac{1}{2} H_{2_{(g)}}$ (Hydrogen gas is evolved).
At anode: $2H_2O_{(l)} \longrightarrow O_{2_{(g)}} + 4H_{(aq)}^{+} + 4e^{-}$ (Oxygen gas is evolved).
$(iv)$ At cathode: $Cu_{(aq)}^{2+} + 2e^{-} \longrightarrow Cu_{(s)}$ (Copper deposits).
At anode: $2Cl_{(aq)}^{-} \longrightarrow Cl_{2_{(g)}} + 2e^{-}$ (Chlorine gas is evolved).

(N/A) Chlorine is obtained as a by-product in the electrolysis of molten $NaCl$ and in the electrolysis of brine $(NaCl_{(aq)})$ in the Chlor-alkali process.
In the electrolysis of molten $NaCl$:
At anode: $2Cl^-_{(melt)} \longrightarrow Cl_{2(g)} + 2e^-$
At cathode: $2Na^+_{(melt)} + 2e^- \longrightarrow 2Na_{(s)}$
If an aqueous solution of $NaCl$ (brine) is subjected to electrolysis,$Cl_2$ gas is obtained at the anode,but at the cathode,$H_2$ gas is evolved instead of $Na$ metal. This is because the standard reduction potential of $Na^+$ $(E^{\circ} = -2.71 \ V)$ is much more negative than that of $H_2O$ $(E^{\circ} = -0.83 \ V)$. Therefore,$H_2O$ is preferentially reduced at the cathode.
At cathode: $2H_2O_{(l)} + 2e^- \longrightarrow H_{2(g)} + 2OH^-_{(aq)}$
At anode: $2Cl^-_{(aq)} \longrightarrow Cl_{2(g)} + 2e^-$

(N/A) $(i)$ At cathode: $Ag^+_{(aq)} + e^- \rightarrow Ag_{(s)}$. At anode: $Ag_{(s)} \rightarrow Ag^+_{(aq)} + e^-$.
$(ii)$ At cathode: $Ag^+_{(aq)} + e^- \rightarrow Ag_{(s)}$. At anode: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-$.
$(iii)$ At cathode: $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$. At anode: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-$.
$(iv)$ At cathode: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$. At anode: $2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-$.

(A) rectifier is an electrical device that converts alternating current $(AC)$,which periodically reverses direction,into direct current $(DC)$,which flows in only one direction. This process is known as rectification.

(A) The process of obtaining chlorine from brine (sea water) is as follows:
$2Cl^-_{(aq)} + 2H_2O_{(l)} \rightarrow 2OH^-_{(aq)} + H_{2(g)} + Cl_{2(g)}$
For this reaction,$\Delta G^{\ominus} = +422 \ kJ$. Using the formula $\Delta G^{\ominus} = -nFE^{\ominus}$,we calculate $E^{\ominus} = -2.2 \ V$.
Thus,by applying an external $EMF$ greater than $2.2 \ V$,the reaction proceeds in the forward direction. However,in electrolysis,additional potential is required to overcome certain hindering reactions.
Therefore,chlorine gas is obtained by electrolysis,while $H_2$ and aqueous $NaOH$ are obtained as by-products.
Electrolysis of molten $NaCl$ can also be performed,but in that case,$Na$ metal is obtained instead of $NaOH$.

(A) Electrolysis of molten $NaCl$ is the process where an electric current is passed through molten $NaCl$ to decompose it into its constituent elements.
At the cathode: $Na^+ + e^- \rightarrow Na(l)$
At the anode: $2Cl^- \rightarrow Cl_2(g) + 2e^-$
Overall reaction: $2NaCl(l) \rightarrow 2Na(l) + Cl_2(g)$.

(B) $(i)$ In electrolytes,current flows due to the movement of ions,not electron displacement. Thus,this statement is $False$ $(F)$.
$(ii)$ When current flows through a metal (metallic conductor),it is an electronic conduction process where electrons move,but the composition of the metal does not change. Thus,this statement is $False$ $(F)$.
$(iii)$ Electrolysis involves the passage of $DC$ current through an electrolytic solution,which causes chemical reactions at the electrodes (oxidation and reduction). Thus,this statement is $True$ $(T)$.
Therefore,the correct sequence is $F, F, T$.

(N/A) One of the simplest electrolytic cells consists of two copper strips dipping in an aqueous solution of copper sulphate $(CuSO_4)$.
Construction: It consists of two copper electrodes immersed in an aqueous solution of copper sulphate. One electrode acts as the anode and the other as the cathode.
Working: When a $DC$ voltage is applied to the two electrodes,$Cu^{2+}$ ions migrate to the cathode (negatively charged) and undergo reduction:
$Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$
Copper metal is deposited on the cathode.
At the anode,copper metal is oxidized into $Cu^{2+}$ ions:
$Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^{-}$
Thus,copper is dissolved (oxidized) at the anode.
Overall reaction: $Cu_{(s)} (\text{anode}) \rightarrow Cu_{(s)} (\text{cathode})$ (Impure $\rightarrow$ Pure).
Uses: This process is the basis for the electrolytic refining of copper,where impure copper is converted into high-purity copper. The impure copper is used as the anode,which dissolves upon passing current,and pure copper is deposited at the cathode.

(B) coulometer is a standard electrolytic cell used to determine the quantity of electricity passed through a circuit.
It works by measuring the amount of metal (generally silver or copper) deposited or consumed at the electrodes during electrolysis.
This was historically significant because constant current sources were not available during Faraday's time.
However,coulometers are now considered obsolete.

(B) Nowadays,we have constant current $I$ sources available,and the quantity of electricity $Q$ passed is given by:
$Q = I \times t$
Where:
$I$ = current in ampere $(A)$
$t$ = time in second $(s)$ for the passage of current
$Q$ = quantity of electricity in Coulomb $(C)$

(N/A) There are two types of electrodes:
$(a)$ Inert electrode: The electrode which does not participate in the chemical reaction and acts only as a source or sink for electrons is called an inert electrode. Inert electrodes do not dissolve or participate in the reaction; their mass remains constant. Examples include $Pt$ and $Au$.
Example: In the electrolysis of $CuCl_{2}$ solution using $Pt$ electrodes,the $Pt$ electrode does not undergo oxidation. Instead,$Cl^{-}$ ions are oxidized at the anode to produce $Cl_{2}$ gas.
$2Cl_{(aq)}^{-} \rightarrow Cl_{2(g)} + 2e^{-}$
$(b)$ Reactive electrode: Electrodes that take part in the chemical reaction are known as reactive electrodes. When these electrodes participate,their atoms may dissolve into the solution or deposit on the electrode,causing a change in their mass. Examples include $Cu$,$Zn$,and $Al$.
Example: When the electrolysis of $CuCl_{2}$ solution is carried out using $Cu$ electrodes,the $Cu$ anode undergoes oxidation,forming $Cu^{2+}$ ions that dissolve into the solution.
$Cu_{(s)} \rightarrow Cu_{(aq)}^{2+} + 2e^{-}$
Conclusion: The products of electrolysis differ depending on whether the electrodes are inert or reactive.

(N/A) Reduction reactions occur on the surface of the cathode. If more than one species is present near the cathode,the species with the higher standard reduction potential $(E^{\ominus})$ value undergoes reduction.
For an aqueous $NaCl$ solution,both $Na^{+}$ ions from $NaCl$ and $H^{+}$ ions from $H_2O$ are present near the cathode.
$NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
$H_2O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$
The possible reduction reactions are:
$(i) \ Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\ominus} = -2.71 \ V$
$(ii) \ H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} \quad E^{\ominus} = 0.00 \ V$
Since the $E^{\ominus}$ value for the reduction of $H^{+}$ (or $H_2O$) is higher than that of $Na^{+}$,the reduction of water (or $H^{+}$) is thermodynamically favored.
Therefore,the reaction occurring at the cathode is:
$H_2O_{(l)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} + OH^{-}_{(aq)}$
As a result,$H_2$ gas is evolved at the cathode,and $OH^{-}$ ions accumulate,making the solution basic,which can be confirmed by the pink color produced with phenolphthalein.

(A) For molten $NaCl$,the dissociation is: $NaCl \longrightarrow Na^+ + Cl^-$
At the cathode (reduction): $Na^+ + e^- \longrightarrow Na$
At the anode (oxidation): $Cl^- \longrightarrow \frac{1}{2} Cl_{2(g)} + e^-$
Overall reaction: $NaCl \longrightarrow Na + \frac{1}{2} Cl_{2(g)}$

(N/A) $NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
At the anode,both $Cl^{-}$ ions and $H_{2}O$ molecules are present.
Two possible oxidation reactions are:
$(i) \ Cl^{-}_{(aq)} \rightarrow \frac{1}{2} Cl_{2(g)} + e^{-} \quad E^{\circ} = 1.36 \ V$
$(ii) \ 2 H_{2}O_{(l)} \rightarrow O_{2(g)} + 4 H^{+}_{(aq)} + 4 e^{-} \quad E^{\circ} = 1.23 \ V$
Generally,the reaction with the lower $E^{\circ}$ value is preferred. However,due to the overpotential of oxygen,the oxidation of water is kinetically hindered. Consequently,the oxidation of $Cl^{-}$ ions occurs at the anode,resulting in the evolution of $Cl_{2}$ gas.

(B) During the electrolysis of aqueous $NaCl$,the following oxidation reactions are possible at the anode:
$(i)$ $Cl^{-}_{(aq)} \rightarrow \frac{1}{2} Cl_{2_{(g)}} + e^{-} \quad E^{o} = 1.36 \ V$
$(ii)$ $2H_{2}O_{(l)} \rightarrow O_{2_{(g)}} + 4H^{+}_{(aq)} + 4e^{-} \quad E^{o} = 1.23 \ V$
Although the oxidation of water has a lower standard electrode potential $(E^{o} = 1.23 \ V)$ compared to chloride ions $(E^{o} = 1.36 \ V)$,the oxidation of water requires an additional 'overpotential' to occur at a significant rate.
Due to this overpotential of oxygen,the oxidation of water is kinetically hindered,and the oxidation of $Cl^{-}$ ions becomes the preferred reaction at the anode,resulting in the evolution of $Cl_{2}$ gas.

(N/A) Inert electrodes are electrodes that do not participate in the chemical reaction.
In an aqueous $NaCl$ solution,the ionization occurs as: $NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$.
At the cathode,the competition is between the reduction of $Na^{+}$ and the reduction of $H_{2}O$ (or $H^{+}$):
$Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\ominus} = -2.71 \ V$
$2H_{2}O_{(l)} + 2e^{-} \rightarrow H_{2(g)} + 2OH^{-}_{(aq)} \quad E^{\ominus} = -0.83 \ V$
Since the reduction potential of water is higher,$H_{2}$ gas is produced at the cathode.
At the anode,the competition is between the oxidation of $Cl^{-}$ and the oxidation of $H_{2}O$:
$2Cl^{-}_{(aq)} \rightarrow Cl_{2(g)} + 2e^{-} \quad E^{\ominus} = 1.36 \ V$
$2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-} \quad E^{\ominus} = 1.23 \ V$
Although the oxidation potential of water is lower,due to the overpotential of oxygen,$Cl^{-}$ is preferentially oxidized to $Cl_{2}$ gas.
Overall reaction:
$2NaCl_{(aq)} + 2H_{2}O_{(l)} \rightarrow Cl_{2(g)} + H_{2(g)} + 2NaOH_{(aq)}$

(N/A) The ionization of $H_2SO_4 \rightarrow 2H^{+}{_{\text{(aq)}}} + SO_4^{2-}{_{\text{(aq)}}}$.
Near the anode,both $SO_4^{2-}$ ions and $H_2O$ molecules are present. Therefore,oxidation of either $SO_4^{2-}$ or $H_2O$ is possible.
$1$. Oxidation of $H_2O$: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^- \quad E^{\circ} = 1.23 \ V$
$2$. Oxidation of $2SO_4^{2-}{_{\text{(aq)}}} \rightarrow S_2O_8^{2-}{_{\text{(aq)}}} + 2e^{-} \quad E^{\circ} = 1.96 \text{ V}$
Since the standard electrode potential $(E^{\circ})$ for the oxidation of water is lower than that of the sulfate ion,water is preferentially oxidized at the anode in dilute solutions,producing oxygen gas. However,in highly concentrated $H_2SO_4$ solutions,the oxidation of $SO_4^{2-}$ to form peroxodisulfate $(S_2O_8^{2-})$ becomes possible.

(N/A) $(i)$ The nature of the material being electrolysed.
$(ii)$ The type of electrodes used: If the electrode is inert (e.g.,$Pt$ or $Au$),it does not participate in the reaction. If the electrode is reactive,it participates in the electrode reaction.
$(iii)$ The standard electrode potentials of the species present: Species with a higher $E^{\circ}$ value are reduced at the cathode,and species with a lower $E^{\circ}$ value are oxidized at the anode.
Example: In the electrolysis of aqueous $NaCl$ with inert electrodes,$H_{2}O$ is reduced at the cathode instead of $Na^{+}$ because $H_{2}O$ has a higher reduction potential.
$(iv)$ The concentration of the electrolyte: According to the Nernst equation,changes in concentration affect the electrode potential,which can alter the products.
Example: In the electrolysis of dilute $H_{2}SO_{4}$,water is oxidized to $O_{2}$. However,in highly concentrated $H_{2}SO_{4}$,$SO_{4}^{2-}$ ions are oxidized to $S_{2}O_{8}^{2-}$ according to the reaction: $2SO_{4}^{2-} \rightarrow S_{2}O_{8}^{2-} + 2e^{-}$.

For $A$. Electrolysis of water:
At the anode (oxidation): $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$
At the cathode (reduction): $4H_2O(l) + 4e^- \rightarrow 2H_2(g) + 4OH^-(aq)$
Overall reaction: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$
For $B$. Electrolysis of molten $NaCl$:
At the anode (oxidation): $2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$
At the cathode (reduction): $2Na^+(l) + 2e^- \rightarrow 2Na(l)$
Overall reaction: $2NaCl(l) \rightarrow 2Na(l) + Cl_2(g)$

(N/A) For $A$ (Concentrated $NaCl$ solution):
At cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$
At anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$
Overall reaction: $2NaCl(aq) + 2H_2O(l) \rightarrow 2Na^+(aq) + 2OH^-(aq) + H_2(g) + Cl_2(g)$
For $B$ (Dilute $NaCl$ solution):
At cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$
At anode: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$
Overall reaction: $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$

(N/A) . Electrolysis of $H_2SO_4$:
$1$. Dilute $H_2SO_4$: The reaction is essentially the electrolysis of water.
At cathode: $4H^+ + 4e^- \rightarrow 2H_2(g)$
At anode: $2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$
$2$. Concentrated $H_2SO_4$: The $HSO_4^-$ ions are discharged at the anode.
At cathode: $2H^+ + 2e^- \rightarrow H_2(g)$
At anode: $2HSO_4^- \rightarrow H_2S_2O_8 + 2e^-$
$B$. Oxidation of $SO_4^{2-}$:
$SO_4^{2-}$ ions can be oxidized to peroxydisulphate $(S_2O_8^{2-})$ at high potentials.
$2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$

(C) The reduction potential of an ion determines its ease of reduction at the cathode. Higher (more positive) reduction potential means easier reduction.
For pair $(i)$: The reduction potential of $H_2O$ $(-0.83 \ V)$ is higher than that of $Na^+$ $(-2.71 \ V)$. Thus,$H_2O$ is reduced more easily than $Na^+$.
For pair (ii): The reduction potential of $H^+$ $(0.00 \ V)$ is significantly higher than that of $Na^+$ $(-2.71 \ V)$. Thus,$H^+$ is reduced more easily than $Na^+$.
Since both pairs contain a species that is reduced more easily than $Na^+$,both $(i)$ and (ii) represent pairs where one species is reduced more easily at the cathode.

(C) The ease of oxidation at the anode depends on the standard oxidation potential and the concentration of the ions.
For $(iii)$,$Cl^{-}$ has a lower oxidation potential than $H_2O$,but due to the overpotential of oxygen,$Cl^{-}$ is preferentially oxidized over $H_2O$ in concentrated solutions.
For $(iv)$,$SO_4^{2-}$ has a very high oxidation potential,making it much harder to oxidize than $H_2O$. Therefore,$H_2O$ is oxidized more easily than $SO_4^{2-}$.
Comparing the pairs,the pair $(iii)$ $Cl^{-}$ and $H_2O$ involves species where oxidation is competitive and significant,whereas $(iv)$ involves a species $(SO_4^{2-})$ that is practically impossible to oxidize in aqueous solution compared to $H_2O$. Thus,the pair $(iii)$ is the correct answer regarding the ease of oxidation.

(N/A) Electrolysis is a process in which a chemical decomposition is produced by passing an electric current through a liquid or solution containing ions.
During this process,electrical energy is converted into chemical energy.
With the help of electrolysis,various substances can be produced,such as:
$1$. Production of metals like $Na$,$Mg$,$Al$ from their fused salts.
$2$. Production of gases like $H_2$ and $O_2$ from water.
$3$. Electroplating of metals (e.g.,coating of $Cu$ or $Ag$).
$4$. Production of chemicals like $NaOH$ and $Cl_2$ via the chlor-alkali process.

(A) In the electrolysis of $CuSO_4$ solution using $Cu$ electrodes,the electrodes are active.
At the anode,the $Cu$ metal undergoes oxidation: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$.
At the cathode,the $Cu^{2+}$ ions from the solution undergo reduction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
Thus,the net reaction involves the transfer of copper from the anode to the cathode.

(A) The $pH$ of the solution will rise as $NaOH$ is formed in the electrolytic cell.
During the electrolysis of brine $(NaCl_{(aq)})$,the reduction of $Na^+$ ions does not occur at the cathode because water is more easily reduced.
The reduction reaction at the cathode is:
$2 H_2O_{(l)} + 2 e^- \rightarrow H_{2(g)} + 2 OH^-_{(aq)}$
As the reaction proceeds,the concentration of $OH^-$ ions increases in the solution.
Since $pOH = -\log_{10} [OH^-]$,an increase in $[OH^-]$ leads to a decrease in $pOH$.
Given the relationship $pH + pOH = 14$,a decrease in $pOH$ results in an increase in $pH$.
Therefore,the $pH$ of the brine solution increases during electrolysis.

(N/A) There is no change in the $pH$ value. The $pH$ value depends on the $H^{+}$ ion concentration of the solution.
In the electrolysis of dilute $H_{2}SO_{4}$ solution,the following reactions occur:
$H_{2}SO_{4} \rightarrow 2H^{+} + SO_{4}^{2-}$
At the cathode,$H^{+}$ ions are reduced:
$2H^{+} + 2e^{-} \rightarrow H_{2(g)}$ $(i)$
At the anode,$H_{2}O$ is oxidized rather than $SO_{4}^{2-}$ ions:
$H_{2}O_{(l)} \rightarrow \frac{1}{2}O_{2(g)} + 2H^{+} + 2e^{-}$ $(ii)$
Overall reaction = $(i)$ + $(ii)$:
$H_{2}O_{(l)} \rightarrow H_{2(g)} + \frac{1}{2}O_{2(g)}$
Since the net reaction involves the consumption of $H_{2}O$ and the production of $H_{2}$ and $O_{2}$ gases,the concentration of $H^{+}$ ions in the solution remains constant. Therefore,the $pH$ of the solution does not change.

(A) $CH_3COOH$ is a weak electrolyte,which undergoes partial dissociation in solution.
On dilution,the degree of dissociation $(\alpha)$ increases significantly according to Ostwald's dilution law,leading to a sharp increase in the number of ions and thus a drastic increase in molar conductivity $(\Lambda_m)$.
$CH_3COONa$ is a strong electrolyte,which is already completely dissociated in solution.
On dilution,the number of ions remains constant,and the increase in $\Lambda_m$ is only due to the decrease in inter-ionic attractions,which results in a gradual increase.

(C) During the electrolysis of dilute $H_2SO_4$,the following reactions occur:
At the cathode: $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$
At the anode: $2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-$
Since the oxidation potential of water is higher than that of the sulphate ion $(SO_4^{2-})$,water is oxidized at the anode to release oxygen gas.

(B) During the electrolysis of a concentrated acidified sulphate solution,the oxidation of sulphate ions occurs at the anode.
The anode reaction is: $2S{O_{4}}^{2-}_{(aq)} \rightarrow S_{2}{O_{8}}^{2-}_{(aq)} + 2e^{-}$.
The cathode reaction is: $2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)}$.
The resulting peroxodisulphuric acid $(H_{2}S_{2}O_{8})$ is formed by the combination of $S_{2}O_{8}^{2-}$ ions with $H^{+}$ ions in the solution.

(A) Sodium hydroxide $(NaOH)$ is prepared commercially by the electrolysis of brine ($NaCl$ solution) in a $Castner-Kellner$ cell.
At the cathode,sodium ions are reduced to sodium amalgam $(Na-Hg)$:
$Na^{+} + e^{-} \xrightarrow{Hg} Na-Hg$
At the anode,chloride ions are oxidized to chlorine gas:
$Cl^{-} \longrightarrow \frac{1}{2}Cl_{2} + e^{-}$
The sodium amalgam is then treated with water to produce sodium hydroxide and molecular hydrogen gas as a byproduct:
$2Na(amalgam) + 2H_{2}O \longrightarrow 2NaOH + H_{2} + 2Hg$

(A) The electrolysis of concentrated aqueous $NaCl$ (brine) involves the following half-reactions:
At the cathode: $2 H_2O(l) + 2 e^{-} \longrightarrow H_2(g) + 2 OH^{-}(aq)$
At the anode: $2 Cl^{-}(aq) \longrightarrow Cl_2(g) + 2 e^{-}$
The overall reaction is: $2 NaCl(aq) + 2 H_2O(l) \longrightarrow 2 NaOH(aq) + H_2(g) + Cl_2(g)$
Since $NaOH$ is produced in the solution,the concentration of $OH^{-}$ ions increases,which leads to an increase in the $pH$ of the solution.
Thus,the correct option is $A$.

(D) The electrolysis of brine ($NaCl$ solution) involves the following reactions:
$NaCl_{(aq)} \longrightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
At the anode: $2Cl^{-}_{(aq)} \longrightarrow Cl_{2_{(g)}} + 2e^-$
At the cathode: $2H_2O_{(\ell)} + 2e^- \longrightarrow H_{2_{(g)}} + 2OH^{-}_{(aq)}$
As a result,$H_2$ gas is evolved at the cathode,and $OH^{-}$ ions are produced in the solution near the cathode,leading to the formation of $NaOH$.

(D) Brine is an aqueous solution of $NaCl$.
$NaCl_{(aq)} \rightarrow Na^+_{(aq)} + Cl^-_{(aq)}$
At the cathode,water is reduced in preference to $Na^+$ ions:
$2H_2O_{(l)} + 2e^- \rightarrow H_{2(g)} + 2OH^-_{(aq)}$
At the anode,chloride ions are oxidized:
$2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-$
The remaining ions in the solution are $Na^+$ and $OH^-$,which combine to form $NaOH_{(aq)}$.
Therefore,$H_2$,$Cl_2$,and $NaOH$ are produced,while $HCl$ is not formed during the electrolysis of brine.

(C) The electrolysis of concentrated sulphuric acid $(H_2SO_4)$ is used to produce peroxydisulphuric acid $(H_2S_2O_8)$.
At the anode,the oxidation of hydrogen sulphate ions $(HSO_4^-)$ occurs:
$2HSO_4^{-} \rightarrow H_2S_2O_8 + 2e^{-}$

(A) In the electrolysis of an aqueous solution of $AgNO_3$ using silver electrodes $(A)$,the anode dissolves $(Ag \rightarrow Ag^+ + e^-)$,so no $O_2$ is evolved.
In the electrolysis of an aqueous solution of $AgNO_3$ using platinum electrodes $(B)$,the oxidation of water occurs at the anode $(2H_2O \rightarrow O_2 + 4H^+ + 4e^-)$,evolving $O_2$ gas.
In the electrolysis of a dilute solution of $H_2SO_4$ using platinum electrodes $(C)$,the oxidation of water occurs at the anode $(2H_2O \rightarrow O_2 + 4H^+ + 4e^-)$,evolving $O_2$ gas.
In the electrolysis of a high concentration solution of $H_2SO_4$ using platinum electrodes $(D)$,the oxidation of $SO_4^{2-}$ ions occurs at the anode $(2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-)$,forming peroxodisulphuric acid instead of $O_2$ gas.
Therefore,$O_2$ is evolved in cases $(B)$ and $(C)$.

(B) Statement $(I)$ is correct: Metals with higher standard reduction potential $(E^{\theta})$ are deposited first at the cathode. The order of $E^{\theta}$ values is $Ag^{+} (0.80 \ V) > Hg_2^{2+} (0.79 \ V) > Cu^{2+} (0.24 \ V) > Mg^{2+} (-2.37 \ V)$. Thus,the sequence of deposition is $Ag, Hg, Cu$.
Statement $(II)$ is incorrect: At the cathode,the reduction of water $(2H_2O + 2e^- \rightarrow H_2 + 2OH^-)$ occurs instead of $Mg^{2+}$ reduction because $E^{\theta}_{H_2O/H_2} > E^{\theta}_{Mg^{2+}/Mg}$. Therefore,$H_2$ gas is evolved at the cathode,not oxygen gas (oxygen gas is evolved at the anode).

(A) In the electrolysis of aqueous $AgNO_3$ solution using a silver anode and a platinum cathode:
At the cathode,$Ag^+$ ions are reduced to metallic silver: $Ag^+ (aq) + e^- \rightarrow Ag (s)$.
At the anode,the silver electrode itself undergoes oxidation because it is an active electrode: $Ag (s) \rightarrow Ag^+ (aq) + e^-$.
Therefore,the products are $Ag$ at the cathode and $Ag^+$ ions at the anode.

(D) $1$. In the electrolysis of molten $NaCl$,the reactions are: At cathode: $Na^+ + e^- \rightarrow Na_{(s)}$; At anode: $2Cl^- \rightarrow Cl_{2(g)} + 2e^-$.
$2$. In the electrolysis of aqueous $NaCl$ (brine),the reactions are: At cathode: $2H_2O + 2e^- \rightarrow H_{2(g)} + 2OH^-_{(aq)}$; At anode: $2Cl^- \rightarrow Cl_{2(g)} + 2e^-$.
$3$. Comparing both processes,$Cl_{2(g)}$ gas is liberated at the anode in both cases.

(C) The electrolysis of molten $NaCl$ is a non-spontaneous process because the standard cell potential $(E^\circ_{cell})$ is negative.
Therefore,the decomposition of $NaCl$ into $Na_{(s)}$ and $Cl_{2(g)}$ is not spontaneous.
Electrical energy is required to drive this non-spontaneous reaction.
At the anode,$Cl^-$ ions are oxidized to $Cl_2$ gas.
At the cathode,$Na^+$ ions are reduced to $Na$ metal.

(D) During the electrolysis of aqueous $NaCl$,the ions present in the solution are $Na^+$,$Cl^-$,$H^+$,and $OH^-$.
At the cathode,the reduction potential of water is higher than that of $Na^+$ ions.
Therefore,water is reduced in preference to $Na^+$ ions.
The reaction at the cathode is: $2H_2O_{(l)} + 2e^- \rightarrow H_{2(g)} + 2OH^-_{(aq)}$.

(B) During the electrolysis of fused (molten) $NaCl$,the electrolyte dissociates into $Na^+$ and $Cl^-$ ions.
At the cathode,$Na^+$ ions undergo reduction: $Na^+ + e^- \rightarrow Na_{(s)}$.
At the anode,$Cl^-$ ions undergo oxidation: $2Cl^- \rightarrow Cl_{2(g)} + 2e^-$.
Therefore,the product formed at the cathode is metallic sodium,$Na_{(s)}$.

(D) During the electrolysis of molten $NaCl$,the ions present are $Na^{+}$ and $Cl^{-}$.
At the anode,oxidation occurs,which involves the loss of electrons.
The $Cl^{-}$ ions migrate to the anode and undergo oxidation to form chlorine gas:
$2 Cl^{-} \rightarrow Cl_{2(g)} + 2 e^{-}$

(D) During electrolysis,the reduction reaction occurs at the cathode,where metal ions gain electrons to form metal atoms. Therefore,the metal is deposited at the cathode,not the anode. Thus,option $D$ is incorrect.

(C) The total charge $q$ passed is given by the formula $q = I \times t$.
Given $I = 1 \ A$ and $t = 16.1 \ \text{minutes} = 16.1 \times 60 \ \text{seconds} = 966 \ \text{seconds}$.
Thus,$q = 1 \times 966 = 966 \ \text{Coulombs}$.
The number of electrons $n$ is calculated using $q = n \times e$,where $e = 1.602 \times 10^{-19} \ \text{C}$.
$n = \frac{q}{e} = \frac{966}{1.602 \times 10^{-19}} \approx 6.022 \times 10^{21}$ electrons.

(B) The total charge $Q$ flowing through the wire is given by $Q = I \times t$.
Given $I = 1.5 \ A$ and $t = 3 \ hours = 3 \times 3600 \ s = 10800 \ s$.
So,$Q = 1.5 \times 10800 = 16200 \ C$.
The number of electrons $n$ is given by $n = \frac{Q}{e}$,where $e = 1.6 \times 10^{-19} \ C$.
$n = \frac{16200}{1.6 \times 10^{-19}} = 1.0125 \times 10^{23} \approx 1.01 \times 10^{23}$.

(B) During the electrolysis of aqueous $NaCl$ solution,the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$
At the anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$
Thus,$Cl_2$ gas is released at the anode.