Electrolytes and Electrolysis Questions in English

(D) $1 \ Faraday$ is defined as the total charge carried by $1 \ mole$ of electrons.
Since $1 \ mole$ of any substance contains $6.022 \times 10^{23}$ particles,$1 \ Faraday$ of electricity corresponds to $6.022 \times 10^{23}$ electrons.

(C) In the electrolysis of aqueous sodium chloride $(NaCl)$ with inert electrodes,the solution contains $Na^{+}$,$Cl^{-}$,$H^{+}$,and $OH^{-}$ ions.
At the cathode,the reduction of water is preferred over the reduction of $Na^{+}$ ions because the standard reduction potential of water is higher:
$2H_2O(l) + 2e^{-} \rightarrow H_2(g) + 2OH^{-}(aq) ; E^{\circ} = -0.83 \ V$
$Na^{+}(aq) + e^{-} \rightarrow Na(s) ; E^{\circ} = -2.71 \ V$
Thus,$H_2$ gas is produced at the cathode.
At the anode,although the oxidation potential of water is higher,the oxidation of $Cl^{-}$ ions is preferred due to overpotential effects:
$2Cl^{-}(aq) \rightarrow Cl_2(g) + 2e^{-} ; E^{\circ} = -1.36 \ V$
$2H_2O(l) \rightarrow O_2(g) + 4H^{+}(aq) + 4e^{-} ; E^{\circ} = -1.23 \ V$
Therefore,$Cl_2$ gas is produced at the anode.

(C) In the electrolysis of brine solution ($NaCl$ aqueous),the solution contains $Na^{+}$,$Cl^{-}$,$H^{+}$,and $OH^{-}$ ions.
At the cathode,$H^{+}$ ions (from water) have a higher reduction potential than $Na^{+}$ ions,so $H_2$ gas is liberated.
Reaction at cathode: $2H_2O(l) + 2e^{-} \rightarrow H_2(g) + 2OH^{-}(aq)$.
At the anode,$Cl^{-}$ ions are oxidized to $Cl_2$ gas.
Reaction at anode: $2Cl^{-}(aq) \rightarrow Cl_2(g) + 2e^{-}$.
Therefore,$H_2$ gas is liberated at the cathode.

(C) According to Faraday's first law of electrolysis,the mass $(w)$ deposited is given by the formula: $w = Z \cdot Q$,where $Z$ is the electrochemical equivalent and $Q$ is the quantity of electricity in Coulombs.
When $Q = 1 \ C$,the mass deposited becomes $w = Z \cdot 1 = Z$.
Therefore,$1 \ C$ of electricity deposits an amount of substance equal to its electrochemical equivalent.

(B) The reduction reaction is: $Al^{3+} + 3e^- \rightarrow Al$.
Moles of $Al = \frac{\text{Mass}}{\text{Atomic mass}} = \frac{4.5 \times 10^{-5} \text{ g}}{27 \text{ g/mol}} = 1.666 \times 10^{-6} \text{ mol}$.
Moles of electrons required $= 3 \times \text{moles of } Al = 3 \times 1.666 \times 10^{-6} = 5 \times 10^{-6} \text{ mol}$.
Number of electrons $= \text{moles of electrons} \times N_A = 5 \times 10^{-6} \times 6.022 \times 10^{23} \approx 3.011 \times 10^{18}$.

(B) Anode: The electrode of an electrochemical cell at which oxidation occurs.
Cathode: The electrode of an electrochemical cell at which reduction occurs.
Electrolysis of molten $NaCl$ or fused $NaCl$:
Sodium ions $(Na^+)$ migrate to the cathode,where they gain electrons and are reduced to sodium metal:
$Na^+ + e^- \longrightarrow Na_{(s)}$
Chlorine ions $(Cl^-)$ migrate to the anode. They give up their electrons to the anode and are oxidized to chlorine gas:
$Cl^- \longrightarrow \frac{1}{2} Cl_{2(g)} + e^-$
Overall reaction:
$2 NaCl_{(l)} \longrightarrow 2 Na_{(s)} + Cl_{2(g)}$
Therefore,the product obtained at the anode is $Cl_{2(g)}$.

(A) In solid sodium chloride $(NaCl)$,the ions are held together by strong electrostatic forces in a rigid crystal lattice.
Since the ions are not free to move,solid $NaCl$ cannot conduct electricity.
In contrast,aqueous $KCl$ contains free ions,while graphite and copper metal contain free electrons,allowing them to conduct electricity.

(C) $(1)$ Diamond: Diamond does not conduct electricity. It is an insulator because it has no free electrons or ions to carry charge. The carbon atoms in diamond form a strong covalent network,preventing the free flow of electrons.
$(2)$ Sulphur (solid): Sulphur in its solid form also does not conduct electricity. It is a non-metal and does not have free electrons or charged particles to facilitate conduction.
$(3)$ Molten $NaCl$: Molten $NaCl$ (sodium chloride in its liquid form) conducts electricity. When $NaCl$ is heated to its melting point,it dissociates into free sodium $(Na^+)$ and chloride ions $(Cl^-)$,which are able to move freely and carry an electric current.
$(4)$ Crystalline $NaCl$: Crystalline $NaCl$ (solid sodium chloride) does not conduct electricity. While it contains ions ($Na^+$ and $Cl^-$),they are not free to move in the solid state. For conduction,the ions need to be mobile,which happens only in the molten state.

(A) Urea is a non-electrolyte and does not dissociate into ions in water.
Therefore,its electrical conductivity is nearly the same as that of distilled water.
Sodium chloride $(NaCl)$ and sodium hydroxide $(NaOH)$ are strong electrolytes,while acetic acid $(CH_3COOH)$ is a weak electrolyte,all of which increase the conductivity of water significantly.

(A) The reduction half-reaction for $Cr^{3+}$ to $Cr_{(s)}$ is given by:
$Cr^{3+} + 3e^{-} \rightarrow Cr_{(s)}$
From the stoichiometry of the balanced equation,$1 \ mole$ of $Cr^{3+}$ ions requires $3 \ moles$ of electrons to be reduced to $1 \ mole$ of $Cr_{(s)}$ metal.

(B) The reduction half-reaction is: $Zn^{2+} + 2e^{-} \rightarrow Zn_{(s)}$
From the stoichiometry of the reaction,$1 \ mol$ of $Zn^{2+}$ ions requires $2 \ mol$ of electrons for reduction.
Therefore,for $3 \ mol$ of $Zn^{2+}$ ions,the required moles of electrons are $3 \times 2 = 6 \ mol$.

(B) During the electrolysis of highly concentrated $H_2SO_4$ (approx. $50 \%$),the oxidation of $HSO_4^-$ ions occurs at the anode.
The reaction is: $2HSO_4^- \rightarrow H_2S_2O_8 + 2e^-$.
The product formed is peroxodisulphuric acid $(H_2S_2O_8)$,also known as Marshall's acid.
Therefore,the correct option is $B$.

(A) During the electrolysis of a highly concentrated $H_2SO_4$ solution,the concentration of $SO_4^{2-}$ ions is very high.
Due to this high concentration,the oxidation of $SO_4^{2-}$ ions becomes more favorable than the oxidation of water at the anode.
The reaction occurring at the anode is: $2SO_4^{2-}{_{\text{(aq)}}} \rightarrow S_2O_8^{2-}{_{\text{(aq)}}} + 2e^{-}$.
Therefore,the correct option is $A$.

(C) During the electrolysis of an aqueous solution of sodium chloride ($NaCl$ (aq)),the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.
At the anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$.
The remaining ions in the solution are $Na^+$ and $OH^-$,which combine to form $NaOH(aq)$.
Therefore,the products obtained are $NaOH, Cl_2$ and $H_2$.

(B) The electrolysis of aqueous $CuSO_4$ with inert electrodes involves the following reactions: \\ At cathode: $Cu^{2+} (aq) + 2e^- \rightarrow Cu (s)$ \\ At anode: $2H_2O (l) \rightarrow O_2 (g) + 4H^+ (aq) + 4e^-$ \\ As the reaction proceeds,$H^+$ ions are produced in the solution,which leads to the formation of $H_2SO_4$. \\ The increase in the concentration of $H^+$ ions causes the $pH$ of the solution to decrease.

(B) During the electrolysis of concentrated aqueous $NaCl$ (brine),the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$
At the anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$
The net reaction is: $2NaCl(aq) + 2H_2O(l) \rightarrow 2NaOH(aq) + Cl_2(g) + H_2(g)$
Since $NaOH$ is a strong base,the resulting solution is basic in nature.
Therefore,it turns red litmus into blue.

(C) In the electrolysis of a dilute aqueous solution of $NaCl$,water undergoes oxidation and reduction at the electrodes.
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: $2H_2O(l) \rightarrow O_{2(g)} + 4H^+(aq) + 4e^-$.
Thus,$H_{2(g)}$ is produced at the cathode and $O_{2(g)}$ is produced at the anode.

(C) During the electrolysis of $CuSO_{4(aq)}$ using graphite 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 $H^+$ ions in the solution.
Since $pH = -\log[H^+]$,an increase in $[H^+]$ leads to a decrease in $pH$.
Therefore,the $pH$ of the solution becomes less than $7.0$.

(C) During the electrolysis of a highly concentrated aqueous solution of $H_2SO_4$,the oxidation of the hydrogen sulfate ion $(HSO_4^-)$ occurs at the anode.
The reaction is: $2HSO_4^- (aq) \rightarrow S_2O_8^{2-} (aq) + 2H^+ (aq) + 2e^-$.
This results in the formation of the peroxodisulfate ion $(S_2O_8^{2-})$,also known as Marshall's acid anion.

(D) The formula for charge $(Q)$ is given by $Q = I \times t$,where $I$ is the current in amperes and $t$ is the time in seconds.
Given: $I = 0.5 \ A$,$t = 2 \ hours = 2 \times 3600 \ s = 7200 \ s$.
Calculation: $Q = 0.5 \ A \times 7200 \ s = 3600 \ C$.
Therefore,the total charge that flows through the wire is $3600 \ C$.

(A) In the electrolysis of a concentrated aqueous solution of $NaCl$ (brine),the ions present are $Na^+$,$Cl^-$,$H^+$,and $OH^-$.
At the anode,$Cl^-$ ions are preferentially oxidized over $OH^-$ ions due to their higher concentration,producing $Cl_2$ gas: $2Cl^- \rightarrow Cl_2 + 2e^-$.
At the cathode,$H^+$ ions are preferentially reduced over $Na^+$ ions because the reduction potential of $H^+$ is higher than that of $Na^+$,producing $H_2$ gas: $2H^+ + 2e^- \rightarrow H_2$.
Therefore,$Cl_2$ is produced at the anode and $H_2$ is produced at the cathode.

(A) The extraction of chlorine from brine solution ($NaCl$ solution) is based on the oxidation of chloride ions $(Cl^{-})$ to chlorine gas $(Cl_2)$ during electrolysis.
The reaction at the anode is: $2Cl^{-}_{(aq)} \rightarrow Cl_{2(g)} + 2e^{-}$.

(D) During the electrolysis of brine ($NaCl$ aqueous solution) using inert electrodes:
At anode: $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$
At cathode: $2 H_2 O + 2 e^{-} \longrightarrow H_2 + 2 OH^{-}$
Overall reaction: $2 NaCl + 2 H_2 O \longrightarrow Cl_2 + H_2 + 2 NaOH$
Thus,$Cl_2$ gas is liberated at the anode and $H_2$ gas is liberated at the cathode.

(A) The electrolysis of an aqueous solution of $CuSO_{4}$ with inert electrodes involves the following reactions:
At the cathode: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$
At the anode: $2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-}$
Since $H^{+}$ ions are produced at the anode,the concentration of $H^{+}$ ions in the solution increases.
As $pH = -\log[H^{+}]$,an increase in $[H^{+}]$ leads to a decrease in the $pH$ of the solution.

(D) In an aqueous solution of sodium chloride,the following ions are present: $Na^{+}$,$Cl^{-}$,$H^{+}$,and $OH^{-}$.
At the anode,oxidation occurs. The two possible oxidation reactions are:
$1$) $2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-} ; E^{\circ} = 1.23 \ V$
$2$) $Cl^{-}_{(aq)} \rightarrow \frac{1}{2}Cl_{2(g)} + e^{-} ; E^{\circ} = 1.36 \ V$
Although the standard electrode potential for water oxidation is lower,the oxidation of $Cl^{-}$ is kinetically preferred due to the overpotential of oxygen evolution on the electrode surface. Thus,$Cl^{-}$ ions are preferentially oxidized at the anode.

(D) During the electrolysis of an aqueous solution of $NaF$ using inert electrodes,water molecules are preferentially oxidized at the anode rather than fluoride ions because the oxidation potential of water is lower than that of fluoride ions.
At the cathode: $2H_{2}O + 2e^{-} \longrightarrow H_{2} + 2OH^{-}$.
At the anode: $2H_{2}O \longrightarrow O_{2} + 4H^{+} + 4e^{-}$.
Even if $F_{2}$ were produced,it would react with water to form $O_{2}$ as follows: $2F_{2} + 2H_{2}O \longrightarrow 4HF + O_{2}$.
Therefore,the final product obtained at the anode is $O_{2}$.

(A) Solid $NaCl$ is a bad conductor of electricity because the ions are held in a rigid crystal lattice and are not free to move to carry charge.
In contrast,$Cu$ is a metallic conductor,while fused $NaCl$ and brine solution contain free ions that allow them to conduct electricity.

(C) In Kolbe's electrolysis of sodium acetate $(CH_3COONa)$:
At Anode (Oxidation): $2CH_3COO^{-} \rightarrow C_2H_6 + 2CO_2 + 2e^-$
At Cathode (Reduction): $2H_2O + 2e^- \rightarrow H_2 + 2OH^{-}$
Therefore,the product at cathode $(X)$ is $H_2$ and the products at anode $(Y)$ are $C_2H_6$ and $CO_2$.

(B) The aqueous solution of $K_2SO_4$ dissociates as: $K_2SO_4(aq) \rightarrow 2K^+(aq) + SO_4^{2-}(aq)$.
Water also undergoes self-ionization: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
At the cathode,the reduction potential of $H^+$ is higher than that of $K^+$,so $H^+$ ions are reduced: $2H^+(aq) + 2e^- \rightarrow H_2(g)$.
At the anode,the oxidation potential of $OH^-$ (or $H_2O$) is higher than that of $SO_4^{2-}$,so $H_2O$ is oxidized: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$.
Therefore,$H_2$ gas is produced at the cathode and $O_2$ gas is produced at the anode.

(C) The aqueous solution of $CuCl_2$ dissociates as: $CuCl_2 \longrightarrow Cu^{2+}_{(aq)} + 2 Cl^{-}_{(aq)}$.
At the cathode,$Cu^{2+}$ ions have a higher reduction potential than $H_2O$ molecules,so $Cu^{2+}$ ions are reduced to $Cu$ metal:
$Cu^{2+}_{(aq)} + 2 e^{-} \longrightarrow Cu_{(s)}$.
At the anode,$Cl^{-}$ ions are oxidized to $Cl_2$ gas:
$2 Cl^{-}_{(aq)} \longrightarrow Cl_{2(g)} + 2 e^{-}$.

(B) During the electrolysis of aqueous $CuSO_4$ using inert $Pt$ electrodes,the ions present in the solution are $Cu^{2+}$,$SO_4^{2-}$,$H^+$,and $OH^-$.
At the cathode,$Cu^{2+}$ ions have a higher reduction potential than $H^+$ ions,so $Cu^{2+}$ ions are reduced to $Cu$ metal: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
At the anode,$OH^-$ ions are oxidized in preference to $SO_4^{2-}$ ions,leading to the evolution of oxygen gas: $4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-$.
Therefore,'$X$' (anode product) is $O_2$ and '$Y$' (cathode product) is $Cu$.

(C) During the electrolysis of molten $NaCl$,$Na$ metal is deposited at the cathode,and $Cl_2$ gas is liberated at the anode.
At cathode: $Na^+ + e^- \rightarrow Na$
At anode: $2Cl^- \rightarrow Cl_2 + 2e^-$
Thus,Statement $I$ is correct.
When aqueous $CuSO_4$ is electrolyzed using $Pt$ electrodes,$Cu$ is deposited at the cathode and $O_2$ is liberated at the anode.
At cathode: $Cu^{2+} + 2e^- \rightarrow Cu$
At anode: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$
Since $O_2$ is liberated at the anode,not the cathode,Statement $II$ is incorrect.
Therefore,Statement $I$ is correct but Statement $II$ is incorrect.

(C) The chemical reaction at the anode is:
$2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$
Total charge passed $Q = I \times t = 3 \ A \times (10 \times 60 \ s) = 1800 \ C$.
According to the stoichiometry,$4 \ mol$ of electrons produce $1 \ mol$ of $O_2$ gas.
$4 \times 96500 \ C$ of charge produces $22.4 \ L$ of $O_2$ at $STP$.
Therefore,$1800 \ C$ of charge produces volume $V = \frac{22.4 \times 1800}{4 \times 96500} \approx 0.1045 \ L$.
The closest value is $0.1 \ L$.

(B) During the electrolysis of an aqueous solution of copper sulphate using copper electrodes,both $Cu^{2+}$ and $H^{+}$ ions move towards the cathode.
Since the reduction potential of $Cu^{2+}$ is higher than that of $H^{+}$,$Cu^{2+}$ ions are reduced in preference to $H^{+}$ ions.
Thus,copper metal is deposited at the cathode according to the following reaction:
$Cu^{2+}_{(aq)} + 2e^{-} \longrightarrow Cu_{(s)}$

(A) In the electrolysis of molten $CuCl_2$,the ions present are $Cu^{2+}$ and $Cl^{-}$.
During electrolysis,anions move towards the anode (positive electrode) and cations move towards the cathode (negative electrode).
At the anode,oxidation occurs,where chloride ions lose electrons to form chlorine gas:
$2 Cl^{-} \longrightarrow Cl_{2(g)} + 2 e^{-}$

(B) The electrolysis of cryolite $(Na_3AlF_6)$ involves the following reactions:
$Na_3AlF_6 \rightleftharpoons 3NaF + AlF_3$
$4AlF_3 \rightleftharpoons 4Al^{3+} + 12F^-$
At the cathode: $4Al^{3+} + 12e^- \rightarrow 4Al$
At the anode: $12F^- \rightarrow 6F_2 + 12e^-$
Thus,the molar ratio of $Al$ to $F_2$ produced is $4:6$,which simplifies to $2:3$.

(B) During the electrolysis of $50 \% H_2SO_4$ using platinum electrodes,the concentration of the acid is high enough that the oxidation of the hydrogen sulfate ion $(HSO_4^-)$ occurs at the anode instead of the oxidation of water.
The reaction at the anode is: $2HSO_4^- \longrightarrow H_2S_2O_8 + 2e^-$.
Thus,the product obtained at the anode is peroxodisulfuric acid $(H_2S_2O_8)$.

(C) In the Castner-Kellner cell process,electrolysis of brine ($NaCl$ solution) is performed.
In this process,a mercury cathode and a carbon or graphite anode are used.
At the anode,$Cl^-$ ions are oxidized to produce $Cl_2$ gas: $2Cl^- \rightarrow Cl_2 + 2e^-$.
At the cathode,$Na^+$ ions are reduced and dissolve in mercury to form sodium amalgam $(Na-Hg)$: $Na^+ + e^- + Hg \rightarrow Na-Hg$.
Therefore,the statement that mercury acts as the anode and carbon as the cathode is incorrect; it is the reverse.
Thus,the correct option is $C$.

(B) The Castner process involves the electrolysis of fused sodium chloride $(NaCl)$.
At the cathode,sodium ions are reduced: $Na^{+} + e^{-} \longrightarrow Na$.
At the anode,chloride ions are oxidized to chlorine gas: $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$.
Therefore,the reaction occurring at the anode is $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$.

(A) Brine is an aqueous solution of sodium chloride $(NaCl)$.
During the electrolysis of brine,the ions present are $Na^{+}$,$Cl^{-}$,$H^{+}$,and $OH^{-}$.
At the anode,oxidation occurs. Between $Cl^{-}$ and $OH^{-}$ ions,$Cl^{-}$ ions are preferentially oxidized to form chlorine gas due to their lower discharge potential in concentrated solutions.
The reaction at the anode is: $Cl^{-}_{(aq)} \rightarrow \frac{1}{2} Cl_{2(g)} + e^{-}$.

(D) During the electrolysis of aqueous $CuSO_4$ solution using $Cu$ electrodes,the following ions are present in the solution: $Cu^{2+}$,$H^{+}$,$SO_{4}^{2-}$,and $OH^{-}$.
At the cathode,$Cu^{2+}$ ions are reduced: $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$.
At the anode,the $Cu$ electrode itself undergoes oxidation because it is an active electrode: $Cu_{(s)} - 2e^{-} \rightarrow Cu^{2+}_{(aq)}$.
Therefore,the correct option is $D$.

(B, C, D) When water is added to molten $NaCl$ during electrolysis,the following occurs:
$1$. The reaction $2Na + 2H_2O \longrightarrow 2NaOH + H_2$ takes place.
$2$. Hydrogen gas $(H_2)$ is evolved at the cathode because the discharge potential of $H^+$ ions is lower than that of $Na^+$ ions.
$3$. Caustic soda $(NaOH)$ is formed as a byproduct.
$4$. The reaction between sodium and water is highly exothermic,which can lead to the ignition of the evolved hydrogen gas,making a fire likely.
Therefore,options $B$,$C$,and $D$ are correct.