Electrolytes and Electrolysis Questions in English

(C) During the electrolysis of aqueous sodium chloride $(NaCl(aq))$,the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ (Reduction of water occurs because $E^\circ_{H_2O/H_2} > E^\circ_{Na^+/Na}$).
At the anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ (Oxidation of chloride ions occurs).
Thus,$H_2$ gas is liberated at the cathode and $Cl_2$ gas is liberated at the anode.

(C) The electrolysis of carnallite $(KCl \cdot MgCl_2 \cdot 6H_2O)$ in the molten state results in the deposition of $Mg$ at the cathode and $Cl_2$ gas at the anode.
$K^{+} + e^- \to K$; $E^o = -2.93 \; V$
$Mg^{2+} + 2e^- \to Mg$; $E^o = -2.37 \; V$
Since the reduction potential of $Mg^{2+}$ is higher (less negative) than that of $K^{+}$,$Mg^{2+}$ is preferentially reduced to $Mg$ at the cathode.

(B) The formula for charge $(Q)$ is given by $Q = I \times t$.
Given current $(I)$ = $1.8 \, A$.
Time $(t)$ = $1.36 \, \text{minutes} = 1.36 \times 60 \, \text{seconds} = 81.6 \, \text{seconds}$.
Therefore,$Q = 1.8 \, A \times 81.6 \, \text{s} = 146.88 \, C$.
Rounding to the nearest whole number,we get $147 \, C$.

(C) In the $Castner-Kellner$ process,a brine solution ($NaCl$ aqueous) is electrolyzed using a mercury cathode and a carbon anode.
At the mercury cathode,sodium ions $(Na^+)$ are reduced to form sodium metal,which then dissolves in the mercury to form sodium amalgam ($Na-Hg$ alloy).
This sodium amalgam is subsequently reacted with water to produce sodium hydroxide $(NaOH)$,hydrogen gas,and mercury.

(C) The total charge $Q$ passed is given by the formula $Q = I \times t$.
Given $I = 1 \ A$ and $t = 60 \ s$,we have $Q = 1 \times 60 = 60 \ C$.
The charge on a single electron is $1.60 \times 10^{-19} \ C$.
The number of electrons $n$ is calculated as $n = \frac{Q}{e} = \frac{60}{1.60 \times 10^{-19}}$.
$n = 37.5 \times 10^{19} = 3.75 \times 10^{20}$ electrons.

(B) Metals that are highly reactive,such as those in the $s$-block (e.g.,$Ca$,$Na$,$K$),have a very negative standard reduction potential.
When an aqueous solution of their salts is electrolyzed,water is reduced at the cathode instead of the metal ion.
The reaction at the cathode is $H_2O(l) + e^- \rightarrow \frac{1}{2} H_2(g) + OH^-(aq)$.
Therefore,these metals cannot be obtained by the electrolysis of their aqueous salt solutions; they are typically extracted by the electrolysis of their molten salts.

(C) The reduction reaction of water during electrolysis is: $2 H_2O + 2 e^- \longrightarrow H_2 + 2 OH^-$.
From the stoichiometry,$2 \ mole$ of $H_2O$ require $2 \ mole$ of electrons ($2 \ F$ of charge).
Therefore,$1 \ mole$ of $H_2O$ requires $1 \ F$ of charge.
For $3 \ mole$ of $H_2O$,the theoretical charge required is $3 \ F$.
Given the current efficiency is $75 \ \%$,the actual charge required $(Q)$ is calculated as:
$Q \times 0.75 = 3 \ F$
$Q = \frac{3}{0.75} = 4 \ F$.

(D) In the electrolytic extraction of sodium (Downs process),the mixture of $NaCl$,$CaCl_2$,and $NaF$ (or $Na_3AlF_6$) is used to lower the melting point of $NaCl$.
During electrolysis,the reduction potential of $Na^{+}$ is higher than that of $K^{+}$ and $Ca^{2+}$ (meaning $Na^{+}$ is easier to reduce than $K^{+}$ and $Ca^{2+}$).
Since $K$ and $Ca$ are more reactive than $Na$,their ions have a higher discharge potential,meaning they remain in the molten state while $Na^{+}$ ions are reduced to metallic $Na$ at the cathode.

(D) Carnallite is a double salt with the chemical formula $KCl \cdot MgCl_2 \cdot 6H_2O$.
When fused carnallite is subjected to electrolysis,it undergoes decomposition.
The magnesium ions $(Mg^{2+})$ are reduced at the cathode to form magnesium metal $(Mg)$,and chloride ions $(Cl^-)$ are oxidized at the anode to form chlorine gas $(Cl_2)$.
The reaction is: $KCl \cdot MgCl_2$ $\xrightarrow{\text{electrolysis}} K^+ + Mg^{2+} + 3Cl^-$ $\rightarrow Mg + Cl_2 + KCl$.

(B) During the electrolysis of an aqueous solution of cupric bromide $(CuBr_2)$,the ions present are $Cu^{2+}$,$Br^-$,$H^+$,and $OH^-$.
At the cathode,the reduction reaction occurs. The discharge potential of $Cu^{2+}$ $(E^0 = +0.34 \ V)$ is lower than that of $H^+$ ions $(E^0 = 0.00 \ V)$.
Therefore,$Cu^{2+}$ ions are preferentially reduced at the cathode to form metallic copper $(Cu)$.
The reaction at the cathode is: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.

(B) The electrolysis of an aqueous solution of potassium succinate is known as $Kolbe's$ electrolysis.
During this process,the dicarboxylate ion undergoes decarboxylation at the anode to form an alkene.
The reaction is: $KOOC-CH_2-CH_2-COOK + 2H_2O \xrightarrow{\text{electrolysis}} CH_2=CH_2 + 2CO_2 + H_2 + 2KOH$.
Thus,the major product $(A)$ is ethene $(CH_2=CH_2)$.

(A) The electrolysis of an aqueous solution of sodium acetate is known as the Kolbe electrolysis reaction.
The overall reaction is: $2CH_3COONa + 2H_2O \rightarrow CH_3-CH_3 + 2CO_2 + H_2 + 2NaOH$.
At the anode,the acetate ion $(CH_3COO^-)$ undergoes oxidation to form ethane and carbon dioxide $(CO_2)$.
The reaction at the anode is: $2CH_3COO^- \rightarrow CH_3-CH_3 + 2CO_2 + 2e^-$.
Therefore,$CO_2$ gas is evolved at the anode.

(A) solution of sodium sulphate $(Na_2SO_4)$ in water is electrolysed using inert platinum electrodes.
The ions present in the solution are $Na^+$,$SO_4^{2-}$,$H^+$,and $OH^-$.
At the cathode,the reduction reaction with a higher reduction potential occurs. Since $H^+$ ions have a higher reduction potential than $Na^+$ ions,$H_2$ gas is liberated:
$2H^+ + 2e^- \rightarrow H_2 \uparrow$
At the anode,the oxidation reaction with a lower discharge potential occurs. Since $OH^-$ ions have a lower discharge potential than $SO_4^{2-}$ ions,$O_2$ gas is liberated:
$4OH^- \rightarrow 2H_2O + O_2 + 4e^-$
Thus,the products at the cathode and anode are $H_2$ and $O_2$ respectively.

(D) The electrolysis of brine ($NaCl$ aqueous solution) is known as the Chlor-alkali process.
The chemical reaction is: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.
As shown in the reaction,sodium hydroxide $(NaOH)$,chlorine gas $(Cl_2)$,and hydrogen gas $(H_2)$ are all produced as products.
Therefore,the correct answer is all of these.

(D) The formula for charge is $Q = i \times t$.
Given:
Current $i = 0.010 \ mA = 0.010 \times 10^{-3} \ A = 1.0 \times 10^{-5} \ A$.
Time $t = 1000 \ hours = 1000 \times 3600 \ s = 3.6 \times 10^6 \ s$.
Substituting the values:
$Q = (1.0 \times 10^{-5} \ A) \times (3.6 \times 10^6 \ s) = 36 \ C$.

(B) In the electrolysis of fused $NaCl$,the ions present are $Na^{+}$ and $Cl^{-}$.
At the anode (positive electrode),anions are attracted and undergo oxidation.
Therefore,the chloride ion $(Cl^{-})$ loses an electron to form chlorine gas:
$Cl^{-} \to \frac{1}{2} Cl_2 + e^{-}$

(C) In an aqueous solution of sodium chloride $(NaCl)$,the ions present are $Na^{\oplus}$,$Cl^{\Theta}$,$H^{\oplus}$,and $OH^{\Theta}$.
During electrolysis,the ions with lower discharge potential are preferentially discharged at the electrodes.
At the cathode,$H^{\oplus}$ ions have a lower discharge potential than $Na^{\oplus}$ ions,so $H^{\oplus}$ ions are reduced to $H_2$ gas.
At the anode,$Cl^{\Theta}$ ions are preferentially discharged over $OH^{\Theta}$ ions to form $Cl_2$ gas.
Therefore,the ions discharged at the electrodes are $H^{\oplus}$ and $Cl^{\Theta}$.

(C) During the electrolysis of an aqueous solution of $KCl$,the ions present are $K^+$,$Cl^-$,$H^+$,and $OH^-$.
At the cathode,the reduction potential of $H^+$ $(E^0 = 0.00 \ V)$ is higher than that of $K^+$ $(E^0 = -2.93 \ V)$,so $H_2$ gas is liberated at the cathode.
At the anode,the oxidation of $Cl^-$ occurs to form $Cl_2$ gas.
Therefore,the electrolysis of $KCl_{(aq)}$ results in the liberation of $H_2$ at the cathode and $Cl_2$ at the anode.

(A) During the electrolysis of aqueous $CuSO_4$ solution using copper electrodes:
At the anode,oxidation occurs.
Since copper is an active electrode,it undergoes oxidation instead of the water or sulfate ions.
The copper anode loses electrons to form $Cu^{2+}$ ions,which then move into the solution.
The reaction is: $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-$.

(D) The relationship between charge,current,and time is given by the formula: $Q = I \times t$
Given:
Charge $(Q) = 3.21 \times 10^6 \ C$
Current $(I) = 500 \ A$
Substituting the values into the formula:
$3.21 \times 10^6 = 500 \times t$
Solving for time $(t)$ in seconds:
$t = \frac{3.21 \times 10^6}{500} = 6420 \ s$
To convert the time into minutes:
$t = \frac{6420}{60} \ min = 107 \ min$

(C) During the electrolysis of aqueous $H_2SO_4$ using $Pt$ electrodes,the following ions are present: $H^+$,$SO_4^{2-}$,and $OH^-$ (from water).
At the anode,oxidation occurs. The competing reactions are:
$1$) $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$,$E^o = +1.23 \ V$
$2$) $2SO_4^{2-}(aq) \rightarrow S_2O_8^{2-}(aq) + 2e^-$,$E^o = +1.96 \ V$
Since the oxidation potential of water is higher (or the reduction potential of $O_2$ is lower),water is preferentially oxidized at the anode.
Therefore,$O_2$ gas is evolved at the anode.

(C) solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are $H_2$ and $O_2$ respectively.
At the cathode,hydrogen ions $(H^+)$ have a higher reduction potential than sodium ions $(Na^+)$,so they are reduced to hydrogen gas:
$2H^+ + 2e^- \rightarrow H_2 \uparrow$
At the anode,hydroxide ions $(OH^-)$ have a lower discharge potential than sulphate ions $(SO_4^{2-})$,so they are oxidized to oxygen gas:
$4OH^- \rightarrow 2H_2O + O_2 + 4e^-$

(B) Highly reactive metals with reduction potentials significantly lower than that of hydrogen (e.g.,$Mg, Al, Na, Ca$) cannot be obtained by the electrolysis of their aqueous salt solutions.
This is because water molecules are reduced at the cathode in preference to these metal cations $(2H_2O + 2e^- \rightarrow H_2 + 2OH^-)$.
Among the given options,$Magnesium$ $(Mg)$ has a standard reduction potential $(E^o = -2.37 \ V)$ which is much lower than that of hydrogen $(E^o = 0.00 \ V)$.
Therefore,$Magnesium$ is obtained by the electrolysis of its molten salt,not its aqueous solution.

(D) The electrolysis of a concentrated aqueous solution of $NaCl$ (brine) involves the following reactions:
At the anode: $2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-$
At the cathode: $2H_2O_{(l)} + 2e^- \rightarrow H_{2(g)} + 2OH^-_{(aq)}$
The overall reaction is: $2NaCl_{(aq)} + 2H_2O_{(l)} \rightarrow 2NaOH_{(aq)} + H_{2(g)} + Cl_{2(g)}$
Thus,the products formed are $Cl_{2(g)}$,$NaOH_{(aq)}$,and $H_{2(g)}$.
Hence,the correct option is $D$.

(B) The electrolysis of brine ($NaCl$ solution) produces $NaOH$,$Cl_2$,and $H_2$ as products.
When $Cl_2$ gas is passed through cold and dilute $NaOH$ solution,it forms sodium hypochlorite $(NaOCl)$.
$2NaOH + Cl_2 \rightarrow NaCl + NaOCl + H_2O$.
$NaCl$,$NaOCl$,and $H_2O$ are formed in this reaction.
$H_2$ gas is a primary product of electrolysis but does not react with the other products to form a new compound in this context.
Therefore,$H_2$ is not a compound obtained by mixing the electrolysis products.

(A) Chlorine $(Cl_2)$ is extracted industrially by the electrolysis of an aqueous solution of sodium chloride $(NaCl)$,which is known as the Chlor-alkali process.
$2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.

(B) The chemical reaction for the electrolysis of fused $MgCl_2$ is: $MgCl_2 \rightarrow Mg + Cl_2$.
According to the stoichiometry,$1 \ mol$ of $Mg$ $(24 \ g)$ is produced along with $1 \ mol$ of $Cl_2$ gas.
At $STP$,$1 \ mol$ of any gas occupies $22.4 \ L$.
Therefore,$24 \ g$ of $Mg$ corresponds to $22.4 \ L$ of $Cl_2$ gas.
For $6.5 \ g$ of $Mg$,the volume of $Cl_2$ gas produced is: $V = \frac{22.4 \times 6.5}{24} \ L$.
$V = 6.0666... \ L \approx 6.067 \ L$.

(A) During electrolysis,the ion with the higher standard reduction potential $(E^0)$ is reduced at the cathode.
For $MgCl_2$,the ions present are $Mg^{2+}$ and $H^+$ (from water). The reduction potential of $H^+/H_2$ $(0.00 \ V)$ is higher than that of $Mg^{2+}/Mg$ $(-2.37 \ V)$.
Therefore,$H^+$ ions are reduced to $H_2$ gas at the cathode.
In the other options ($AuCl_3$,$CuCl_2$,$AgNO_3$),the metal ions ($Au^{3+}$,$Cu^{2+}$,$Ag^+$) have higher reduction potentials than $H^+$,so the respective metals will be deposited at the cathode instead of $H_2$ gas.

(C) During the electrolysis of aqueous $KI$ solution,the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$. Thus,$H_2$ gas is released at the cathode.
At the anode: $2I^-(aq) \rightarrow I_2(s) + 2e^-$. Thus,$I_2$ is formed at the anode,not $O_2$.
Since $OH^-$ ions are produced in the solution,the concentration of $OH^-$ increases,which means $pOH$ decreases and $pH$ increases.
Potassium $(K)$ is not deposited because the reduction potential of $H_2O$ is higher than that of $K^+$.
Therefore,statements $(a)$ and $(c)$ are correct.

(A) For $CuCl_2$:
$1$. Electrolysis of fused $CuCl_2$: $Cu^{2+} + 2e^- \rightarrow Cu$ (at cathode) and $2Cl^- \rightarrow Cl_2 + 2e^-$ (at anode).
$2$. Electrolysis of concentrated aqueous $CuCl_2$: $Cu^{2+} + 2e^- \rightarrow Cu$ (at cathode) and $2Cl^- \rightarrow Cl_2 + 2e^-$ (at anode).
Since the products at both electrodes are the same ($Cu$ and $Cl_2$),$CuCl_2$ is the correct answer.

(B) The electrolysis of aqueous $NaOH$ involves the following reactions:
Anode: $4 OH^{-} \rightarrow O_{2(g)} + 2 H_{2}O + 4 e^{-}$
Cathode: $4 H_{2}O + 4 e^{-} \rightarrow 2 H_{2(g)} + 4 OH^{-}$
The overall reaction is $2 H_{2}O \rightarrow 2 H_{2(g)} + O_{2(g)}$.
From the stoichiometry,the molar ratio of $H_{2}$ to $O_{2}$ is $2 : 1$.
Since the volume of gas is directly proportional to the number of moles at $NTP$,the volume of $H_{2}$ liberated at the cathode will be twice the volume of $O_{2}$ liberated at the anode.
Given volume of $O_{2} = 2.8 \ L$.
Volume of $H_{2} = 2 \times 2.8 \ L = 5.6 \ L$.
Therefore,the correct option is $B$.

(B) Dry hydrogen chloride $(HCl)$ is a covalent compound. When dissolved in a non-polar solvent like methylbenzene (toluene),it does not ionize to form $H^+$ and $Cl^-$ ions.
Since there are no free ions present in the solution,it cannot conduct electricity.
Therefore,it is a non-conductor of electricity.

(C) In the electrolysis of fused alkali metal salts,the alkali metals are very strong reducing agents.
They react with the mercury cathode to form amalgams,but more importantly,they are highly reactive and would react with the electrolyte or the environment.
However,the primary reason they are not used in the fused state is that alkali metals are extremely strong reducing agents and would react with the components of the cell or the electrolyte itself,making the process inefficient or dangerous.
Actually,the most direct reason is that alkali metals are strong reducing agents and would react with the electrolyte or the cathode material in a way that prevents the isolation of the pure metal.

(B) The electrode reaction for the liberation of chlorine gas is: $2Cl^- \rightarrow Cl_2 + 2e^-$.
The molar mass of $Cl_2$ is $71 \ g/mol$.
The number of moles of $Cl_2$ liberated is $n = \frac{710 \ g}{71 \ g/mol} = 10 \ mol$.
According to the reaction,$1 \ mol$ of $Cl_2$ requires $2 \ mol$ of electrons.
Therefore,$10 \ mol$ of $Cl_2$ requires $20 \ mol$ of electrons.
The total charge $Q$ required is $Q = n \times F = 20 \ mol \times 96500 \ C/mol = 1930000 \ C = 1.93 \times 10^6 \ C$.

(D) In the electrolysis of dilute $H_2SO_4$ using $Pt$ electrodes,$O_2$ is evolved at the anode due to the oxidation of water.
However,when using $Cu$ electrodes,the anode itself is active and undergoes oxidation instead of water.
At the anode: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$
At the cathode: $2H^+(aq) + 2e^- \rightarrow H_2(g)$
Since $Cu$ is oxidized at the anode,oxygen gas is not liberated.

(C) During the electrolysis of an aqueous solution of $KI$,the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$. Thus,$H_2$ gas is released at the cathode.
At the anode: $2I^-(aq) \rightarrow I_2(s) + 2e^-$. Thus,$I_2$ is formed at the anode,not $O_2$.
Since $OH^-$ ions are produced at the cathode,the concentration of $OH^-$ increases,which means the $pH$ of the solution increases and the $pOH$ decreases.
Therefore,statements $(a)$ and $(c)$ are correct.

(A) The standard reduction potential of $Mg$ $(E^{\circ} = -2.37 \ V)$ is much lower than that of water $(E^{\circ} = -0.83 \ V)$.
Hence,$Mg^{2+}$ ions in the aqueous solution cannot be reduced at the cathode; instead,water will be reduced to liberate $H_2$ gas.
$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$

(D) Initial moles of $Sn^{2+} = \text{Molarity} \times \text{Volume} = 0.60 \ M \times 0.50 \ L = 0.30 \ mol$.
The reaction at the cathode is $Sn^{2+} + 2e^- \rightarrow Sn(s)$.
Total charge passed $Q = I \times t = 4.60 \ A \times (30.0 \times 60 \ s) = 8280 \ C$.
Moles of electrons passed $= \frac{8280}{96500} \approx 0.0858 \ mol$.
Moles of $Sn^{2+}$ deposited $= \frac{0.0858}{2} = 0.0429 \ mol$.
Remaining moles of $Sn^{2+} = 0.30 - 0.0429 = 0.2571 \ mol$.
Final concentration of $Sn^{2+} = \frac{0.2571 \ mol}{0.50 \ L} = 0.514 \ M$.
Note: The provided options suggest a calculation error in the original question's source. Based on the standard calculation,the result is $0.514 \ M$.

(B) The electrolysis of an aqueous solution of sodium benzoate (Kolbe's electrolysis) yields benzene. The reaction is: $C_6H_5COONa + H_2O \rightarrow C_6H_6 + CO_2 + NaOH + H_2$.

(D) Metals that are highly reactive,such as those at the top of the electrochemical series (e.g.,$Al$,$Mg$,$Na$,$Ca$),cannot be obtained by the electrolysis of their aqueous salt solutions.
During the electrolysis of an aqueous solution of an $Al$ salt,$H_2O$ is reduced at the cathode instead of $Al^{3+}$ ions because the reduction potential of $H_2O$ is higher than that of $Al^{3+}$.
Therefore,$Al$ is extracted by the electrolytic reduction of its molten oxide $(Al_2O_3)$ dissolved in molten cryolite $(Na_3AlF_6)$.

(A) The electrolysis of an aqueous solution of $NaCl$ (brine) yields $Cl_2$ gas at the anode,$H_2$ gas at the cathode,and $NaOH$ in the solution.
$2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) + Cl_2(g)$.
Sodium $(Na)$ and Aluminum $(Al)$ are obtained by the electrolysis of their molten salts,not aqueous solutions,because water would be reduced preferentially at the cathode.

(D) Anodizing is an electrochemical process that converts the metal surface into a decorative,durable,corrosion-resistant,anodic oxide finish.
In the case of aluminum,the metal is made the anode in an electrolytic cell containing an electrolyte (usually sulfuric acid).
When current is passed,oxygen is evolved at the anode,which reacts with the aluminum surface to form a thick,protective layer of aluminum oxide $(Al_2O_3)$.

(A) Chlorine $(Cl_2)$ is produced industrially by the electrolysis of brine,which is an aqueous solution of sodium chloride $(NaCl)$.
In this process,$2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g) + Cl_2(g)$.
Bromine is obtained from sea water by oxidation of bromide ions.
Aluminum is obtained by the electrolytic reduction of molten alumina $(Al_2O_3)$ in cryolite.
Sodium is obtained by the electrolysis of fused $NaCl$ (Downs process),not aqueous solution,because $H_2$ would be produced instead of $Na$ in aqueous solution.

(B) The electrolysis of molten $NaCl$ (Downs process) involves the following reactions:
At the cathode: $Na^+ + e^- \rightarrow Na(l)$
At the anode: $2Cl^- \rightarrow Cl_2(g) + 2e^-$
Thus,metallic sodium $(Na)$ is produced at the cathode and chlorine gas is produced at the anode.

(B) In the chlor-alkali process,an aqueous solution of sodium chloride (brine) is electrolyzed.
The chemical reaction is: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$.
At the anode,chloride ions are oxidized to form chlorine gas $(Cl_2)$.
At the cathode,water molecules are reduced to form hydrogen gas $(H_2)$ and hydroxide ions $(OH^-)$.
Therefore,$H_2$ gas is produced at the cathode.

(C) The balanced chemical equation for the electrolysis of water is:
$2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$
According to the stoichiometry of the reaction,$2 \text{ moles}$ of $H_2$ are produced for every $1 \text{ mole}$ of $O_2$.
Under the same conditions of temperature and pressure,the volume of gas is directly proportional to the number of moles.
Therefore,the volume of $H_2$ liberated is twice the volume of $O_2$ liberated.
$\text{Volume of } H_2 = 2 \times \text{Volume of } O_2 = 2 \times 2.24 \ dm^3 = 4.48 \ dm^3$.

(B) The reduction half-reaction for $MnO_4^-$ to $MnO_2$ in an acidic medium is given by:
$MnO_4^- + 4H^{+} + 3e^- \to MnO_2 + 2H_2O$
From the balanced equation,it is clear that $3 \ mol$ of electrons are required to reduce $1 \ mol$ of $MnO_4^-$.
Since the charge of $1 \ mol$ of electrons is $1 \ F$,the total charge required is $3 \ F$.

(C) In an aqueous solution,$NaBr$ dissociates as: $NaBr(aq) \to Na^{+}(aq) + Br^{-}(aq)$.
Water also dissociates slightly: $H_2O(l) \rightleftharpoons H^{+}(aq) + OH^{-}(aq)$.
At the cathode,the reduction of water is preferred over $Na^{+}$ ions: $2H_2O(l) + 2e^{-} \to H_2(g) + 2OH^{-}(aq)$.
At the anode,the oxidation of $Br^{-}$ ions is preferred over water: $2Br^{-}(aq) \to Br_2(l) + 2e^{-}$.
The $Na^{+}$ ions remain in the solution and combine with $OH^{-}$ ions to form $NaOH$.
Thus,the final products are $H_2(g)$,$Br_2(l)$,and $NaOH(aq)$.

(A) Water is a covalent compound,hence pure water is a weak electrolyte and feebly ionised,making it a poor conductor of electricity.
Addition of a small amount of acid or alkali provides ions that facilitate the flow of current,thereby increasing the conductivity of water for electrolysis.
Thus,both Assertion and Reason are correct,and the Reason is the correct explanation of the Assertion.

(D) Specific conductance (conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of the solution.
On increasing dilution,the number of ions per unit volume decreases,which leads to a decrease in specific conductance.
Therefore,the Assertion is incorrect.
Regarding the Reason,while the degree of ionisation of a weak electrolyte increases with dilution,the number of ions per unit volume decreases,not increases.
Thus,both the Assertion and the Reason are incorrect.