Class 10 Chemistry Chapter 6 Electrolysis

Key Concept: Electrode Reactions

b) Lead metal ($Pb$)
[Solution Description]
During the electrolysis of molten lead bromide, $Pb^{2+}$ ions migrate to the cathode and gain electrons to form lead metal. The reaction is: $Pb^{2+} + 2e^- \rightarrow Pb$. Greyish white metal lead is observed at the cathode.

Key Concept: Ions in Molten Salts

b) To allow the ions to become mobile and conduct electricity
[Solution Description]
In solid lead bromide, the ions ($Pb^{2+}$ and $Br^-$) are held together by strong electrostatic forces and are not free to move. Heating it to the molten state provides enough energy to overcome these forces, making the ions mobile and allowing them to conduct electricity.

Key Concept: Anode Reaction in Electrometallurgy

c) Chlorine gas ($Cl_2$)
[Solution Description]
In the electrolysis of magnesium chloride ($MgCl_2$), chloride ions ($Cl^-$) lose electrons at the anode to form chlorine gas ($Cl_2$). The reaction is: $2Cl^- - 2e^- \rightarrow Cl_2$.

Key Concept: Direction of Current Density

a) Opposite to electron flow[Solution Description]Current density is a vector quantity whose direction is the same as the flow of positive charge (conventional current direction). It always points along the direction of the electric field ($\vec{E}$) in an ohmic conductor.

Key Concept: Electrolysis of Molten Lead Bromide

d) Bromine
[Solution Description]
In the electrolysis of molten lead bromide (\$ PbBr_2 \$), bromide ions (\$ Br^- \$) are oxidized at the anode to form bromine gas (\$ Br_2 \$). The reaction is:
$2Br^- \rightarrow Br_2 + 2e^-$
Hence, bromine gas is liberated at the anode.

Key Concept: Electrochemical Series

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] Zinc is higher than copper in the electrochemical series, which means it has a greater tendency to lose electrons and form positive ions. Therefore, zinc can displace copper from its salt solution, as shown by the reaction $Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$. The reason correctly explains why the assertion is true.

Key Concept: Nature of Electrodes

c) Inert electrodes
[Solution Description]
Inert electrodes, such as graphite or platinum, do not take part in the electrolytic reaction. They remain chemically unchanged during electrolysis.

Key Concept: First Law of Electrolysis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The First Law of Electrolysis states that the mass of a substance liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. Mathematically, this is expressed as $m \propto Q$ where $Q = I \times t$.
The assertion claims that doubling the current ($I$) while keeping time ($t$) constant will increase the mass of lead deposited. Since $Q = I \times t$, doubling $I$ while $t$ remains the same indeed doubles $Q$, and thus doubles the mass ($m$) according to $m \propto Q$.
The reason correctly restates the First Law of Electrolysis ($m \propto I \times t$) and logically explains why doubling the current leads to doubling the mass of lead deposited under the given conditions.
Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Nature of Electrodes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that inert electrodes like platinum do not participate in the electrolytic reaction, which is true as per the syllabus. The reason claims that platinum is inert because it does not dissolve or deposit ions during electrolysis, which is also true and correctly explains why platinum is classified as an inert electrode. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Electroplating

c) The electrolyte must contain ions of the metal to be plated.
[Solution Description]
The conditions for successful electroplating include:
1) The article to be plated is placed at the cathode.
2) The metal to be deposited (silver here) acts as the anode.
3) The electrolyte must contain silver ions.
4) A direct current (D.C.) must be used, not alternating current (A.C.).
5) Low current for a longer duration is preferred for uniform coating.
Therefore, all these conditions are necessary for proper electroplating.

Key Concept: Electroplating of an article with nickel

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In nickel electroplating, the article to be plated (iron) is made the cathode, and a pure nickel block is used as the anode. The electrolyte is a solution of nickel sulphate ($NiSO_4$). At the anode, nickel loses electrons to form $Ni^{2+}$ ions, which dissolve into the solution: $Ni - 2e^- \rightarrow Ni^{2+}$. This ensures a continuous supply of $Ni^{2+}$ ions for deposition on the cathode. The assertion correctly states the dissolution of the nickel anode, and the reason accurately explains the electrochemical process behind it (loss of electrons by nickel atoms). Thus, both are true, and the reason correctly explains the assertion.

Key Concept: First Law of Electrolysis, Electrochemical Equivalent

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The First Law of Electrolysis states that the mass ($m$) of a substance deposited is directly proportional to the quantity of electricity passed ($Q = I \times t$). Here, the Assertion claims that doubling the current ($I$) while keeping time ($t$) constant will double the mass of silver deposited. This is correct because $m \propto I$ when $t$ is constant. The Reason correctly states the formula $m = Z \times I \times t$, which justifies the Assertion since $Z$ is a constant for silver. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Electrochemical Series

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the electrochemical series, a metal higher in the series can displace a metal below it from its compound. Copper ($Cu$) is placed above silver ($Ag$) in the electrochemical series, which means copper has a higher tendency to lose electrons compared to silver. Therefore, copper can displace silver from silver nitrate solution as per the reaction: $Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag$. Both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Concentration of Ions in the Electrolyte

b) Because $Cl^-$ has a higher concentration than $OH^-$
[Solution Description]
In concentrated $NaCl$ solution, the concentration of $Cl^-$ is much higher compared to $OH^-$. Even though $OH^-$ is lower in the electrochemical series, the high concentration of $Cl^-$ causes it to be preferentially discharged at the anode.

Key Concept: Volume Ratio of Gases

c) 250 mL
[Solution Description]
According to the reaction:
$2H_2O \rightarrow 2H_2 + O_2$
The molar ratio of $H_2$ to $O_2$ is 2:1. Since volume is proportional to moles (Avogadro's Law), the volume ratio is also 2:1.
Given: Volume of $H_2$ = 500 mL.
Thus, volume of $O_2$ produced will be:
$\text{Volume of } O_2 = \frac{500}{2} = 250 \text{ mL}$

Key Concept: Electrolysis of copper sulphate solution with copper electrodes, electrode reactions

d) The copper anode dissolves into the solution.
[Solution Description]
When copper electrodes are used, the anode itself participates in the reaction. Instead of $OH^-$ or $SO_4^{2-}$ discharging, the copper atoms from the anode lose electrons and dissolve into the solution as $Cu^{2+}$ ions: $Cu - 2e^- \rightarrow Cu^{2+}$. This ensures that the concentration of $Cu^{2+}$ ions in the solution remains constant, so the blue colour does not fade.

Key Concept: Current Density, Electrolysis of Aqueous Copper Sulphate

b) $J = 2 \, \text{A/cm}^2$, $m = 11.84 \, \text{g}$
[Solution Description]
Step 1: Calculate current density ($J$) using $J = \frac{I}{A}$.
$J = \frac{10 \, \text{A}}{5 \, \text{cm}^2} = 2 \, \text{A/cm}^2$
Step 2: Calculate total charge ($Q$) passed in 1 hour using $Q = I \times t$.
$Q = 10 \, \text{A} \times 3600 \, \text{s} = 36000 \, \text{C}$
Step 3: Calculate moles of electrons ($n_e$) using $n_e = \frac{Q}{F}$.
$n_e = \frac{36000 \, \text{C}}{96500 \, \text{C/mol}} = 0.373 \, \text{mol}$
Step 4: Since each $\text{Cu}^{2+}$ ion requires 2 electrons, moles of Cu deposited ($n_{Cu}$) is:
$n_{Cu} = \frac{0.373 \, \text{mol}}{2} = 0.1865 \, \text{mol}$
Step 5: Calculate mass of Cu deposited using $m = n_{Cu} \times M_{Cu}$.
$m = 0.1865 \, \text{mol} \times 63.5 \, \text{g/mol} = 11.84 \, \text{g}$

Key Concept: Ventilation principles

c) Air exchange rate
[Solution Description]
Proper ventilation requires maintaining adequate airflow and removing contaminants. The most critical factor among the below options will be the correct answer.

Key Concept: Electrochemical Series and Selective Discharge

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In the electrolysis of dilute $CuSO_4$, the ions present are $Cu^{2+}$ and $H^+$ (from water). According to the electrochemical series, since copper ($Cu$) is below hydrogen ($H_2$), $Cu^{2+}$ is preferentially discharged at the cathode over $H^+$. Thus, the Assertion (A) is true because copper gets deposited. The Reason (R) is also true and correctly explains why $Cu^{2+}$ is discharged instead of $H^+$.

Key Concept: Electrolytic Dissociation

c) Solid state
[Solution Description]
Solid sodium chloride is an electrovalent compound held together by electrostatic forces of attraction between $Na^+$ and $Cl^-$ ions. In the solid state, these ions are not free to move, hence it cannot conduct electricity.

Key Concept: Applications of Electrolysis: Electroplating

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the article to be electroplated is placed at the cathode, which is correct as per the conditions for electroplating. The reason explains that metal ions from the electrolyte are reduced at the cathode to deposit on the article, which is also true and correctly explains why the article is placed at the cathode.

Key Concept: Hazard Prevention

c) Rubber
[Solution Description]
Electrical insulators prevent the flow of electric current. Rubber is a common insulator used in electrical safety equipment such as gloves and mats because it does not conduct electricity.

Key Concept: Selective discharge theory

c) Oxygen gas ($O_2$)
[Solution Description]
Acidified water contains $H^+$ and $OH^-$ ions due to dissociation. At the anode (positive electrode), hydroxyl ions ($OH^-$) lose electrons preferentially and form oxygen gas ($O_2$) because oxygen has a higher tendency to be discharged compared to other anions in this setup. The reaction is: $4OH^- - 4e^- \rightarrow 2H_2O + O_2$.

Key Concept: Electrochemical Equivalence

b) $Z = \frac{E}{96500}$
[Solution Description]
The electrochemical equivalent ($Z$) is given by the formula $Z = \frac{E}{96500}$, where $E$ is the chemical equivalent weight and 96500 C is Faraday’s constant.

Key Concept: Electrochemical Series

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that sodium can displace copper from copper sulphate solution, which is true because sodium is more electropositive than copper and lies above it in the electrochemical series. The reason provided is also correct as it accurately explains why sodium can displace copper. Since both the assertion and reason are true, and the reason correctly explains the assertion, the correct answer is option a).

Key Concept: Electrochemical series, displacement reactions

a) Silver
[Solution Description]
Zinc ($Zn$) is higher in the electrochemical series compared to both copper ($Cu$) and silver ($Ag$). Therefore, zinc can displace both metals from their solutions. However, the ion with a lower position in the series (higher tendency to gain electrons) will be displaced first. Since $Ag^+$ is lower than $Cu^{2+}$ in the series, it will be reduced more easily and displaced first.
Step 1: Compare positions of $Ag^+$ and $Cu^{2+}$ in the electrochemical series.
Step 2: $Ag^+$ is lower than $Cu^{2+}$, so it has a higher tendency to gain electrons.
Step 3: Thus, $Ag^+$ will be reduced and displaced by zinc before $Cu^{2+}$.

Key Concept: Electroplating and Cathode Role

c) Metal ions reduce and deposit on it.
[Solution Description]
In electroplating, the article to be plated is made the cathode. Metal ions from the electrolytic solution are reduced and deposited on the cathode surface. For example, if silver plating is done, $Ag^+$ ions reduce to $Ag$ on the cathode:
$Ag^+ + e^- \rightarrow Ag$.

Key Concept: Insulation Resistance

c) To identify degradation of insulation material
[Solution Description]
Insulation resistance testing ensures the integrity of insulating materials, preventing leakage currents and potential short circuits or electric shocks.

Key Concept: Electrolysis of Molten Salts, General Principle

c) Copper ($Cu$)
[Solution Description]
Metals higher in the activity series (e.g., K, Na, Ca, Mg, Al) are extracted by electrolysis of their fused salts. Metals like Zn, Fe, Pb, Cu are extracted by reducing agents, while Pt and Au are extracted by thermal decomposition. Therefore, copper ($Cu$) is not extracted by electrolysis of its fused salt.

Key Concept: Purpose of Electroplating

c) To prevent rusting and improve aesthetics
[Solution Description]
Nickel electroplating prevents rusting by creating a protective layer over iron. It also enhances appearance, making the article more decorative.

Key Concept: Electrometallurgy, Activity series

b) Aluminum (Al)
[Solution Description]
To determine the metal X, we analyze the given data:
1. The metal is extracted by electrolysis of its fused chloride, indicating it is a reactive metal (higher in the activity series).
2. For 1 mole of X deposited, 3 moles of electrons are required. This implies the valency of X is 3 because:
$\text{Moles of electrons} = \text{Valency} \times \text{Moles of metal deposited}$
So, $3 = n \times 1$ → $n = 3$.
From the activity series, aluminum (Al) has a valency of 3 and is extracted by electrolysis of its fused chloride. Thus, X is likely Al.

Key Concept: Conductivity in Ionisation

d) Because it produces $H^+$ and $Cl^-$ ions in solution.
[Solution Description]
Aqueous hydrochloric acid conducts electricity because it ionises in water to produce mobile ions ($H^+$ and $Cl^-$). These free-moving ions are responsible for electrical conductivity. The ionisation equation is: $HCl \rightarrow H^+ + Cl^-$.

Key Concept: Electrolyte Concentration

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Assertion is correct because adding solvent dilutes the solution, reducing molarity. Reason correctly defines molarity but doesn't explain why it decreases with dilution - it only states its definition.

Key Concept: Polar Covalent Compounds, Ionization

c) HCl dissociates into ions, while sugar remains molecular.
[Solution Description]
HCl is a polar covalent compound that ionizes in water to form $H_3O^+$ and $Cl^-$ ions, which are responsible for electrical conduction. Sugar (a non-electrolyte) does not ionize in water; it remains as intact molecules and does not produce any free ions. Hence, only Solution A conducts electricity.
Reactions:
- For HCl: $HCl + H_2O \rightarrow H_3O^+ + Cl^-$.
- For sugar: No ionization occurs.
The presence of free ions in Solution A enables conductivity, whereas their absence in Solution B prevents it.

Key Concept: Electrolysis of Acidified Water

c) Hydroxide ions ($OH^-$)
[Solution Description]
In the electrolysis of acidified water, $OH^-$ and $SO_4^{2-}$ migrate to the anode. However, $OH^-$ is lower in the electrochemical series and is discharged first. The reaction is: $OH^- \rightarrow OH + e^-$, followed by $OH + OH \rightarrow H_2O + O$.

Key Concept: Electrolysis of acidified water

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
During the electrolysis of acidified water, the reaction at the cathode produces hydrogen gas ($H_2$) and the reaction at the anode produces oxygen gas ($O_2$). The stoichiometry shows that for every molecule of oxygen ($O_2$) produced, two molecules of hydrogen ($H_2$) are produced. According to Avogadro’s Law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Hence, the volume ratio of hydrogen to oxygen is 2:1. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Safety Measures, Non-Electrolytes

c) Ensure proper ventilation and wear protective gear
[Solution Description]
NaOH pellets are highly corrosive and release heat when dissolved in water (exothermic reaction). Proper precautions include wearing gloves and goggles to prevent chemical burns, adding NaOH slowly to water (not water to NaOH) to control heat generation, and working in a well-ventilated area.

Key Concept: Electrolytic Dissociation

b) Electrolytic dissociation into \$ Na^+ \$ and \$ Cl^- \$
[Solution Description]
Sodium chloride dissociates into ions in aqueous solution as shown in the reaction: \$ NaCl_{(aq)} \rightarrow Na^+ + Cl^- \$. The ionic solid breaks apart into free-moving ions when dissolved.

Key Concept: Electrolytic Dissociation

c) Lead bromide (\$PbBr_2\$)
[Solution Description]
Strong electrolytes dissociate almost completely in their molten or aqueous state. Examples include \$PbBr_2\$, \$CuCl_2\$, \$AgNO_3\$, \$Na_2CO_3\$, \$KHCO_3\$, and \$(CH_3COO)_2Pb\$.

Key Concept: Active Electrodes, Electrolysis of Copper Sulphate

b) Oxygen gas is evolved at the anode.
[Solution Description]
Initially, the reaction at the anode is oxidation of copper: $Cu \rightarrow Cu^{2+} + 2e^-$. However, as $Cu^{2+}$ ion concentration decreases in the solution due to prolonged electrolysis, the electrode potential for the oxidation of copper increases. At very low concentrations, the oxidation of water ($2H_2O \rightarrow O_2 + 4H^+ + 4e^-$) may become favorable instead, leading to oxygen gas evolution at the anode.

Key Concept: Anode mud composition

c) Silver and gold
[Solution Description]
Insoluble impurities like silver and gold settle down near the anode as anode mud because they do not dissolve in the electrolyte during the refining process.

Key Concept: Anode Function

a) Oxidation takes place at the anode.
[Solution Description]
The anode is the electrode connected to the positive terminal of the battery where oxidation occurs. It attracts anions and facilitates the release of electrons.

Key Concept: Mass-Proportionality, Charge Conservation

b) 63.5 grams
[Solution Description]
The reaction at the cathode for \$Cu^{2+}\$ is:
$Cu^{2+} + 2e^- \rightarrow Cu$
1 mole of \$Cu\$ requires 2 moles of electrons for deposition. Therefore, 2 moles of electrons will deposit 1 mole of \$Cu\$.
Mass of 1 mole of \$Cu\$ = 63.5 g.
Hence, 63.5 grams of copper will be deposited.

Key Concept: Air Exchange, Indoor Air Quality

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion (A) is true because proper ventilation helps dilute and remove indoor pollutants such as carbon dioxide, volatile organic compounds (VOCs), and particulate matter, thereby maintaining good indoor air quality. The Reason (R) is false because natural ventilation can also be achieved through strategies like stack effect (temperature-driven airflow) and mechanical means (e.g., exhaust fans), not just wind pressure and thermal buoyancy alone. Therefore, while the assertion highlights a correct fact about ventilation, the reason incorrectly simplifies the mechanisms of natural ventilation.

Key Concept: Electroplating Basics

c) Nickel
[Solution Description]
Chromium and nickel are commonly used metals for electroplating iron articles to prevent rusting because they form a protective layer on the surface.

Key Concept: Characteristics of Active Electrodes

d) They take part in the redox reaction
[Solution Description]
Active electrodes participate in the electrochemical reaction during electrolysis. They can be made of metals like copper, silver, or zinc. Unlike inert electrodes, they dissolve or deposit material during the process. Therefore, the correct statement is that active electrodes take part in the redox reaction.

Key Concept: Electrolysis of acidified water, charge calculations

d) 579000 C
[Solution Description]
For producing H$_2$, the reaction is $2H^+ + 2e^- \rightarrow H_2$. To produce 1 mole of H$_2$, 2 moles of electrons are needed. So for 3 moles of H$_2$, 6 moles of electrons are required.
1 mole of electrons = 96500 C (Faraday's constant). Therefore, total charge = 6 × 96500 = 579000 C.

Key Concept: Electrode Reactions, Precautions

d) Increases at both electrodes
[Solution Description]
The reactions at the electrodes are:
Cathode: $2H^+ + 2e^- \rightarrow H_2$
Anode: $4OH^- \rightarrow O_2 + 2H_2O + 4e^-$
Faraday's First Law states that the amount of substance produced is directly proportional to the current ($m = ZIt$). Thus, increasing the current increases the rate of gas production at both electrodes proportionally.

Key Concept: Anode reaction, Oxidation

c) Chlorine gas
[Solution Description]
During electrolysis of brine, $Cl^-$ ions migrate to the anode and undergo oxidation: $2Cl^- \rightarrow Cl_2 + 2e^-$. Oxygen evolution is less favorable under standard conditions due to overpotential, so chlorine gas ($Cl_2$) is the main product.

Key Concept: Inert Electrode Stability, Acidified Water Electrolysis

b) Copper corrodes in acidic medium
[Solution Description]
Platinum is preferred because it is chemically inert, meaning it does not react with the electrolyte or the products (oxygen and hydrogen). Copper, however, can oxidize or dissolve in acidic conditions, forming $Cu^{2+}$ ions, which would interfere with the intended reactions at the electrodes.
The reactions for acidified water electrolysis at inert platinum electrodes are:
- Cathode: $2H^+ + 2e^- \rightarrow H_2 \uparrow$.
- Anode: $4OH^- \rightarrow 2H_2O + O_2 \uparrow + 4e^-$.
Platinum ensures no side reactions occur.

Key Concept: Electrolysis of Water: Ions in Solution

b) $H^+$, $OH^-$, and $SO_4^{2-}$
[Solution Description]
In acidified water, $H_2SO_4$ dissociates into $2H^+ + SO_4^{2-}$ and water partially dissociates into $H^+ + OH^-$. Therefore, the ions present are $H^+$, $OH^-$, and $SO_4^{2-}$.

Key Concept: Reaction at electrodes

c) $Cu^{2+} + 2e^- \rightarrow Cu$
[Solution Description]
At the cathode, copper ions ($Cu^{2+}$) from the electrolyte gain electrons and get reduced to form pure copper metal. The reaction is:
$Cu^{2+} + 2e^- \rightarrow Cu$

Key Concept: First Law of Electrolysis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The Assertion states that mass deposited is proportional to quantity of electricity, which is true as per the First Law of Electrolysis. The Reason correctly provides the mathematical form of this law ($m = Z \times I \times t$) and thus explains the Assertion. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Oxidation Example

c) $Zn - 2e^- \rightarrow Zn^{2+}$
[Solution Description]
Oxidation involves the loss of electrons. Among the options, $Zn - 2e^- \rightarrow Zn^{2+}$ represents the oxidation of zinc at the anode by losing two electrons.

Key Concept: Electroplating Setup

c) It dissolves to supply $Ni^{2+}$ ions into the solution
[Solution Description]
The anode in nickel electroplating is made of pure nickel metal. During the process, it loses electrons to form $Ni^{2+}$ ions that replenish the electrolyte solution. This ensures continuous deposition of nickel on the cathode.

Key Concept: Electrolytic refining of copper

c) They settle near the anode as anode mud
[Solution Description]
During the electrolytic refining of copper, the impure anode dissolves into the electrolyte as $Cu^{2+}$ ions. However, impurities like silver and gold are insoluble in the electrolyte and do not dissolve. Instead, they settle down near the anode as a sludge called anode mud. This anode mud is then collected and processed to recover these valuable metals.

Key Concept: Electrolysis of copper (II

b) 224 mL
[Solution Description]
From the cathode reaction $Cu^{2+} + 2e^- \rightarrow Cu$, 0.02 moles of Cu require:
$0.02 \times 2 = 0.04 \, \text{moles of electrons}$
At the platinum anode, the reaction is $4OH^- - 4e^- \rightarrow 2H_2O + O_2$. For 0.04 moles of electrons:
$\text{Moles of } O_2 = \frac{0.04}{4} = 0.01 \, \text{moles}$
At STP, 1 mole of gas occupies 22.4 L. Thus:
$\text{Volume of } O_2 = 0.01 \times 22.4 = 0.224 \, \text{L or 224 mL}$

Key Concept: Electrolysis, Conductors vs Non-conductors

c) Non-metals are liberated at the anode during electrolysis.
[Solution Description]
During electrolysis, conductors like metals allow the flow of electricity due to mobile electrons but do not undergo chemical change. Non-conductors (except graphite) do not conduct electricity. The reaction at electrodes involves reduction at the cathode (positive ions gain electrons) and oxidation at the anode (negative ions lose electrons).
Metals are electro-positive and liberated at the cathode, while non-metals are electro-negative and liberated at the anode. Alternating current (A.C.) does not cause chemical changes in electrolytes.
The correct statement is that only non-metals are liberated at the anode since they are electronegative elements.

Key Concept: Electrolysis of Molten Lead Bromide

b) Evolution of dark reddish-brown fumes of bromine
[Solution Description]
In the electrolysis of molten lead bromide, the anode reaction involves bromide ions (\$ Br^- \$) losing electrons to form bromine atoms, which then combine to form bromine gas (\$ Br_2 \$). The observations include dark reddish-brown fumes of bromine gas evolving at the anode.

Key Concept: Preferential discharge of ions, Electrode reactions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that during the electrolysis of aqueous $CuSO_4$ with platinum electrodes, $Cu^{2+}$ ions are reduced at the cathode instead of $H^+$ ions. This is correct because according to the electrochemical series, copper ($E^\circ = +0.34$ V) has a higher reduction potential than hydrogen ($E^\circ = 0.00$ V).
The Reason states that copper has a higher reduction potential than hydrogen, which is true and directly explains why $Cu^{2+}$ is preferentially reduced at the cathode. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Electrochemical Series

d) Sodium ($Na$)
[Solution Description]
The electrochemical series arranges metals in decreasing order of their tendency to lose electrons. Sodium ($Na$) loses electrons more easily than other metals like copper ($Cu$), zinc ($Zn$), and silver ($Ag$), as mentioned in the syllabus (6.6.1).

Key Concept: Second Law of Electrolysis, Molar Mass Calculation

c) 1.5:1
[Solution Description]
Let $M_X$ and $M_Y$ be the molar masses of X and Y.
For X ($n = 2$): $m_X = \frac{M_X \times Q_X}{96500 \times 2} \Rightarrow Q_X = \frac{2.7 \times 96500 \times 2}{M_X}$.
For Y ($n = 3$): $m_Y = \frac{M_Y \times Q_Y}{96500 \times 3} \Rightarrow Q_Y = \frac{5.6 \times 96500 \times 3}{M_Y}$.
Total charge: $Q_X + Q_Y = 9650 \, \text{C}$. Substitute and rearrange:
$\frac{5.4 \times 96500}{M_X} + \frac{16.8 \times 96500}{M_Y} = 9650$.
Simplify: $\frac{5.4}{M_X} + \frac{16.8}{M_Y} = 0.1$.
Assume $M_X = k M_Y$: $\frac{5.4}{k M_Y} + \frac{16.8}{M_Y} = 0.1$.
Solving gives $k \approx 1.5$, so the ratio $\frac{M_X}{M_Y} = 1.5$.

Key Concept: Current Efficiency, Faraday's Law

b) 3.55 g
[Solution Description]
First, calculate total charge passed using $Q = I \times t$ where $I = 1.5$ A and $t = 7200$ s (2 hours). So, $Q = 1.5 \times 7200 = 10800$ C. Using Faraday’s law, moles of Cu deposited are $\frac{Q}{nF} = \frac{10800}{2 \times 96500} \approx 0.05596$ mol. Mass of Cu deposited is $0.05596 \times 63.5 \approx 3.55$ g.

Key Concept: Nature of the Electrodes Used, Relative Position of Ions

c) Copper dissolves into the solution
[Solution Description]
Copper electrodes are active and participate in the reaction. At the anode, copper atoms lose electrons more readily than $OH^-$ or $SO_4^{2-}$ ions because copper is lower in the electrochemical series compared to these anions. Thus, the copper electrode dissolves, forming $Cu^{2+}$ ions.

Key Concept: Applications of Active Electrodes

c) Metal refining
[Solution Description]
Active electrodes are widely used in electroplating, where a thin layer of metal is deposited on another material. They are also used in refining impure metals by dissolving the impure metal at the anode and depositing pure metal at the cathode. Thus, one of their key applications is metal refining.

Key Concept: Relative Position of Ions in the Electrochemical Series, Concentration of Ions

c) $H^+$
[Solution Description]
The electrochemical series indicates that $H^+$ is lower than other common cations like $Na^+$ or $K^+$, making it easier to discharge. Additionally, since the concentration of $H^+$ is much higher, it further ensures its preferential discharge over any other cation present, even if they are lower in the series.

Key Concept: Reaction at cathode with copper electrodes

c) Reddish-brown copper is deposited
[Solution Description]
When copper electrodes are used in the electrolysis of copper sulphate solution, $Cu^{2+}$ ions are reduced at the cathode by gaining electrons, forming solid copper metal. The reaction is:
$Cu^{2+} + 2e^- \rightarrow Cu$
Hence, reddish-brown copper is deposited at the cathode.

Key Concept: Electrolytic dissociation, Ion migration, Faraday's laws

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that during the electrolysis of molten $PbBr_2$, the mass of lead deposited at the cathode is proportional to the quantity of electricity passed. This is true because, according to Faraday's first law of electrolysis, the mass of a substance liberated or deposited at an electrode is directly proportional to the charge (quantity of electricity) passed through the electrolyte. The reason correctly explains the assertion by stating Faraday's first law, which forms the basis for the proportionality mentioned in the assertion.
Therefore, both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Electrolysis and Electrochemical Series

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Sodium ($Na$) has a higher tendency to lose electrons compared to copper ($Cu$), which is why it is placed above copper in the electrochemical series. The electrochemical series ranks metals based on their ability to lose electrons, with metals at the top losing electrons more readily. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Conductors and Non-conductors

c) Graphite
[Solution Description]
Graphite is a non-metal that conducts electricity due to its unique structure allowing electron mobility.

Key Concept: Electrolytic Cell, Stoichiometry

c) 448 mL
[Solution Description]
The reactions are:
At anode: $4OH^- \rightarrow 2H_2O + O_2 + 4e^-$
At cathode: $2H^+ + 2e^- \rightarrow H_2$
For every 1 mole of $O_2$ produced, 2 moles of $H_2$ are produced (from stoichiometry).
Given 224 mL of $O_2$ at STP:
1 mole of any gas occupies 22400 mL at STP.
So, moles of $O_2$ = $\frac{224}{22400} = 0.01 \, \text{moles}$
Moles of $H_2$ produced = 2 × moles of $O_2$ = 0.02 moles.
Volume of $H_2$ at STP = $0.02 \times 22400 = 448 \, \text{mL}$

Key Concept: Purpose of Electroplating

d) For decoration and protection against corrosion
[Solution Description]
Brass spoons are electroplated with silver primarily for decoration (to give them a shiny appearance) and protection (to prevent rusting and corrosion). It enhances both aesthetics and durability.

Key Concept: Reaction with Acids

a) Magnesium
[Solution Description]
Metals above hydrogen in the electrochemical series can displace hydrogen from acids. Among the options, magnesium (\text{Mg}) and zinc (\text{Zn}) are above hydrogen, while silver (\text{Ag}) and gold (\text{Au}) are below.
The reaction for magnesium is:
$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$
Thus, magnesium is the correct answer.

Key Concept: Electrolysis of Acidified Water

c) Oxygen
[Solution Description]
The reactions occurring at the anode during the electrolysis of acidified water are:
$OH^- \rightarrow OH + e^-$
$OH + OH \rightarrow H_2O + O$
$O + O \rightarrow O_2$
Therefore, oxygen gas ($O_2$) is evolved at the anode.

Key Concept: Electrolytic refining of copper, Purity

d) The refined copper contains residual impurities
[Solution Description]
Electrolytic refining produces copper that is 99.9\% pure, which is highly conductive. If the conductivity is lower than expected, it suggests that impurities are still present in the copper. These impurities could be due to incomplete refining or contamination during the process. Even small amounts of impurities can drastically reduce the conductivity of copper.

Key Concept: Concentration of Ions, Nature of the Electrodes Used

d) Chlorine gas
[Solution Description]
In a concentrated $NaCl$ solution, $Cl^-$ ions are present in much higher concentration than $OH^-$ ions. Despite $OH^-$ being lower in the electrochemical series, the high concentration of $Cl^-$ causes them to be discharged preferentially, forming chlorine gas ($Cl_2$) at the anode.

Key Concept: Non-electrolytes

c) Sugar solution
[Solution Description]
Non-electrolytes like sugar do not dissociate into ions in any state (solid, liquid, or aqueous). Hence, they do not conduct electricity under any conditions.

Key Concept: Safety and Precautions in Electrolysis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a fume hood must be used during the electrolysis of molten lead bromide ($PbBr_2$). This is correct because the process releases toxic bromine gas ($Br_2$), as mentioned in the Reason. The Reason accurately explains why a fume hood is necessary—to prevent inhalation of hazardous fumes. Thus, both statements are true, and the Reason correctly explains the Assertion.

Key Concept: Electrolysis of Acidified Water

b) Decreases
[Solution Description]
In the electrolysis of acidified water, \$H^+\$ ions are discharged at the cathode to form hydrogen gas. Since \$H^+\$ ions are consumed during this process, the concentration of \$H_2SO_4\$ decreases at the cathode as it dissociates to replenish \$H^+\$ ions.

Key Concept: Electrolysis of Molten Salts, Activity Series, Electrode Reactions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that graphite anode is used in the extraction of sodium from fused NaCl to prevent corrosion caused by chlorine gas. This is correct because chlorine gas is highly reactive and would corrode iron, forming iron chloride. The Reason explains why graphite is preferred: it is chemically inert to reactive gases like chlorine and oxygen. Since both the Assertion and Reason are true and the Reason correctly explains the Assertion, the correct answer is option (a).

Key Concept: Electrolytic cell, Molten electrolytes

d) Lead at cathode; bromine at anode
[Solution Description]
For the electrolysis of molten $PbBr_2$:
- At the cathode: $Pb^{2+}$ ions gain electrons to form lead metal:
$Pb^{2+} + 2e^- \rightarrow Pb(l)$
- At the anode: $Br^-$ ions lose electrons to form bromine gas:
$2Br^- \rightarrow Br_2(g) + 2e^-$
Therefore, lead is deposited at the cathode and bromine gas evolves at the anode.

Key Concept: Electrochemical Series and Concentration Effect

c) $SO_4^{2-}$
[Solution Description]
Although $OH^-$ is higher in the electrochemical series and generally harder to oxidize, if the concentration of $SO_4^{2-}$ is much higher, the high concentration can override the position in the electrochemical series. Thus, $SO_4^{2-}$ would be discharged at the anode due to its abundance.

Key Concept: Migration of Ions

c) Towards the cathode
[Solution Description]
Cations are positively charged ions. According to the characteristics of electrolysis, they migrate towards the cathode (negatively charged electrode).

Key Concept: Electrolytic refining of copper

b) It dissolves to provide $Cu^{2+}$ ions for the electrolyte while impurities settle as anode mud.
[Solution Description]
In the electrolytic refining of copper, the impure copper slab acts as the anode. When current is passed, copper atoms from the anode lose electrons and dissolve into the electrolyte as $Cu^{2+}$ ions. The reaction at the anode is:
$Cu - 2e^- \rightarrow Cu^{2+}$
The impurities like silver and gold settle as anode mud.

Key Concept: Cathode Material in Electrolysis

b) Iron
[Solution Description]
In the electrolytic extraction of reactive metals such as sodium, the cathode is typically made of iron (Fe) because it is a good conductor of electricity and does not react with the metal being deposited. Graphite is used as the anode.

Key Concept: Electrical Safety, Grounding

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true: Three-prong plugs are designed to include a grounding wire, which protects users by providing a low-resistance path for fault current. The reason is also true and correctly explains the assertion, as grounding indeed diverts dangerous currents away from the user, reducing shock risk.

Key Concept: First Law of Electrolysis

a) Quantity of electricity passed
[Solution Description]
Faraday's first law states that the mass ($m$) of a substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte. The mathematical form is $m = Z \cdot Q$, where $Z$ is the electrochemical equivalent of the substance.

Key Concept: Electroplating conditions

c) Applying low current for a longer duration
[Solution Description]
For uniform electroplating, a low current must be used for a longer duration. High current causes uneven deposition. Additionally, direct current (D.C.) must be used instead of alternating current (A.C.), as A.C. leads to ineffective coating due to alternating discharge and ionization.

Key Concept: Relative Position of Ions in Electrochemical Series

b) Hydrogen ion ($H^+$)
[Solution Description]
The ions present in the solution are $H^+$ from $H_2SO_4$ and $OH^-$ from water. The electrochemical series shows that $H^+$ is lower than $Na^+$ and higher than $Cu^{2+}$. Here, $H^+$ gets preferentially discharged over $OH^-$ at the cathode because it is lower in the electrochemical series.

Key Concept: Volume ratio of gases produced

b) 2:1
[Solution Description]
During electrolysis, 2 molecules of hydrogen ($H_2$) and 1 molecule of oxygen ($O_2$) are produced. According to Avogadro's Law, the volume ratio of gases corresponds to their molecular ratio. Hence, the volume ratio of hydrogen to oxygen is 2:1.

Key Concept: Quantity of Electricity

c) 96500 C
[Solution Description]
1 gram-equivalent of any substance requires 1 mole of electrons for deposition. Since 1 mole of electrons carries 96500 C of charge (Faraday's constant), the answer is 96500 C.

Key Concept: Current Density and Resistivity

c) $2.5 \times 10^7 \, \text{A/m}^2$
[Solution Description]
The relation between current density $J$, electric field $E$, and resistivity $\rho$ is:
$J = \frac{E}{\rho}$
Substituting the values:
$J = \frac{0.5}{2 \times 10^{-8}} = 2.5 \times 10^7 \, \text{A/m}^2$

Key Concept: Electrolytic Conduction

c) Direct current (D.C.)
[Solution Description]
Electrolysis involves chemical changes and requires a direct current (D.C.) because alternating current (A.C.) does not cause sustained ion migration or redox reactions.

Key Concept: Electroplating

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In electroplating, the article to be plated is placed at the cathode because the metal ions ($M^{n+}$) in the electrolyte gain electrons and get reduced to form a metallic layer on the cathode. The assertion correctly states the placement of the article, and the reason explains why it must be placed at the cathode (reduction occurs there). Thus, both are true, and the reason correctly explains the assertion.

Key Concept: Electrolysis of Molten Salts, Faraday's Law

a) 1.866 g
[Solution Description]
First, calculate the total charge passed using the formula $Q = I \times t$. Here, $I = 5$ A and $t = 30 \times 60 = 1800$ s. So, $Q = 5 \times 1800 = 9000$ C.
Next, find the number of moles of electrons using $n_e = Q / F$. Here, $F = 96500$ C/mol. So, $n_e = 9000 / 96500 \approx 0.0933$ mol of electrons.
For $Ca^{2+} + 2e^- \rightarrow Ca$, 2 moles of electrons are needed to deposit 1 mole of Ca. Therefore, moles of Ca deposited = $n_e / 2 = 0.0933 / 2 = 0.04665$ mol.
Finally, mass of Ca deposited = moles $\times$ molar mass = $0.04665 \times 40 \approx 1.866$ g.

Key Concept: Electrolysis of Molten Lead Bromide, Observations

c) Greyish-white deposit at the cathode and reddish-brown fumes at the anode.
[Solution Description]
At the cathode, greyish-white lead deposits are observed due to the reduction of $Pb^{2+}$ ions ($Pb^{2+} + 2e^- \rightarrow Pb$). At the anode, dark reddish-brown fumes of bromine gas are observed due to the oxidation of $Br^-$ ions ($Br^- \rightarrow Br + e^-$). These observations confirm the formation of lead and bromine as the products.

Key Concept: Anode Characteristics

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) is true because the anode is indeed the electrode where oxidation takes place, and it is connected to the positive terminal of the battery. The reason (R) is also true because oxidation involves the loss of electrons, and anions (negative ions) are attracted to the anode where they lose electrons. Moreover, the reason correctly explains why oxidation occurs at the anode, as the process of losing electrons (oxidation) is facilitated by the attraction of anions to the positively charged anode.

Key Concept: Electrolytic Dissociation, Crystal Structure

b) Its ions are fixed in the crystal lattice and cannot move
[Solution Description]
In solid state, sodium chloride has a fixed crystal lattice structure where $Na^+$ and $Cl^-$ ions are held strongly in position. These ions cannot move freely to conduct electricity. Only when melted or dissolved in water do the ions become mobile and can conduct electricity.

Key Concept: Electrometallurgy, Activity series

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Sodium and potassium are highly reactive metals placed at the top of the activity series. Due to their strong affinity for oxygen, they form stable oxides. The oxides of these metals ($Na_2O$, $K_2O$) have very high melting points, which makes electrolysis difficult as it requires a molten state. In contrast, their chlorides ($NaCl$, $KCl$) have comparatively lower melting points, so they can be easily melted and electrolyzed. Thus, the assertion is true. The reason correctly explains why chlorides are preferred over oxides for extraction via electrolysis, as lower melting points facilitate the process. Hence, both Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Anode mud formation

b) Gold and silver
[Solution Description]
During the electrolytic refining of copper, impurities such as gold and silver, which are less reactive than copper, do not dissolve in the electrolyte. Instead, they settle down near the anode as anode mud. Other soluble impurities remain dissolved in the solution.

Key Concept: Electrolysis of Molten Lead Bromide, Current Density

c) 2.51 L
[Solution Description]
Step 1: Calculate total current ($I$) using $J = \frac{I}{A}$.
$I = J \times A = 3 \, \text{A/cm}^2 \times 4 \, \text{cm}^2 = 12 \, \text{A}$
Step 2: Calculate total charge ($Q$) passed in 30 minutes using $Q = I \times t$.
$Q = 12 \, \text{A} \times 1800 \, \text{s} = 21600 \, \text{C}$
Step 3: Calculate moles of electrons ($n_e$) using $n_e = \frac{Q}{F}$.
$n_e = \frac{21600 \, \text{C}}{96500 \, \text{C/mol}} = 0.224 \, \text{mol}$
Step 4: For $\text{Br}_2$ evolution, $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$. Thus, moles of $\text{Br}_2$ ($n_{Br_2}$) is:
$n_{Br_2} = \frac{0.224 \, \text{mol}}{2} = 0.112 \, \text{mol}$
Step 5: Calculate volume of $\text{Br}_2$ at STP using $V = n_{Br_2} \times 22.4 \, \text{L/mol}$.
$V = 0.112 \, \text{mol} \times 22.4 \, \text{L/mol} = 2.51 \, \text{L}$

Key Concept: Nature of Electrodes

c) Oxygen
[Solution Description]
With inert platinum electrodes, $OH^-$ and $SO_4^{2-}$ migrate to the anode. $OH^-$ is discharged preferentially as it is lower in the electrochemical series, producing oxygen gas.

Key Concept: Electrolysis basics, Cations and Anions migration, Faraday's laws

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that during electrolysis of molten sodium chloride ($NaCl$), the mass of sodium deposited at the cathode is directly proportional to the quantity of electricity passed. This is correct because Faraday’s first law of electrolysis states that the amount of chemical change (mass deposited or gas evolved) is directly proportional to the quantity of electricity passed.
The Reason states that according to Faraday’s first law of electrolysis, the mass of a substance liberated at an electrode is proportional to the quantity of electricity passing through the electrolyte. This statement is also true and correctly explains the Assertion, as the proportionality described in the Assertion is a direct application of Faraday’s first law.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Electrode Material, Ion Migration

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion is true because platinum electrodes are indeed inert and do not react with the electrolyte or products. The Reason is also true as during electrolysis of molten $PbBr_2$, $Pb^{2+}$ ions migrate to the cathode (reduction: $Pb^{2+} + 2e^- \rightarrow Pb$) and $Br^-$ ions migrate to the anode (oxidation: $2Br^- \rightarrow Br_2 + 2e^-$). However, the Reason does not explain why platinum electrodes are preferred, as it describes ion migration and electrode reactions, not electrode material selection. Hence, both are true but the Reason is not the correct explanation of the Assertion.

Key Concept: Electrolysis

d) Copper
[Solution Description]
In the electrochemical series, copper is below hydrogen, meaning its cations gain electrons more easily than hydrogen ions during electrolysis. Therefore, copper is deposited at the cathode instead of hydrogen, as per the example in the syllabus.

Key Concept: Electrolysis Products

b) Oxygen gas ($O_2$)
[Solution Description]
In the electrolysis of aqueous $CuSO_4$ with a platinum anode, $OH^-$ ions discharge preferentially over $SO_4^{2-}$ because $OH^-$ is lower in the electrochemical series. The reaction is: $4OH^- - 4e^- \rightarrow 2H_2O + O_2$. Thus, oxygen gas ($O_2$) is evolved.

Key Concept: First Law of Electrolysis, Multiple Elements

b) 3.175 g
[Solution Description]
Step 1: Write the reduction reaction for copper.
$Cu^{2+} + 2e^- \rightarrow Cu$
Step 2: Calculate the moles of electrons involved.
$\text{Moles of electrons} = \frac{Q}{F} = \frac{9650}{96500} = 0.1 \text{ moles}$
Step 3: Determine the moles of copper deposited.
$\text{Moles of Cu} = \frac{\text{Moles of electrons}}{2} = \frac{0.1}{2} = 0.05 \text{ moles}$
Step 4: Calculate the mass of copper deposited.
$\text{Mass of Cu} = 0.05 \times 63.5 = 3.175 \text{ g}$

Key Concept: Electroplating

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that electroplating is used to deposit a thin layer of metal on another metallic article using electricity, which is true as per the syllabus. The reason states that the article to be electroplated is always placed at the cathode, which is also true and directly explains why electroplating works, as the metal ions are attracted to the cathode where they gain electrons and deposit as atoms.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the mass of copper deposited during electrolysis is directly proportional to the current passed. This is true because Faraday's first law states that mass ($m$) is proportional to the quantity of electricity ($Q$), and since $Q = I \times t$, mass is also proportional to current ($I$) for a fixed time ($t$). The reason correctly explains the assertion as it connects mass deposition to the fundamental relationship $m \propto Q$ given by Faraday's first law.

Key Concept: Relative Position in Electrochemical Series

c) Zinc ($Zn$)
[Solution Description]
Metals above hydrogen in the electrochemical series can displace hydrogen from acids, while metals below cannot. Among the given options:
- Copper ($Cu$) is below hydrogen and cannot displace it.
- Silver ($Ag$) is also below hydrogen.
- Zinc ($Zn$) is above hydrogen and can displace it.
- Gold ($Au$) is far below hydrogen.
Thus, zinc ($Zn$) is the correct answer.

Key Concept: Electroplating anode reaction

b) Silver atoms from the anode lose electrons and dissolve into the electrolyte
[Solution Description]
In electroplating, the anode is made of pure silver. The silver atoms lose electrons and oxidize to form silver ions ($Ag^+$), which go into the electrolyte solution. The reaction is $Ag - e^- \rightarrow Ag^+$. The anions ($CN^-$ and $OH^-$) do not discharge here.

Key Concept: Electrolyte Composition, Impurities

c) Collected as anode mud
[Solution Description]
During electrorefining, soluble impurities dissolve in the electrolyte, while insoluble impurities like gold settle as anode mud near the anode because they do not participate in the electrochemical reactions. Thus, gold remains at the anode.

Key Concept: Electroplating process, Cathode and Anode reactions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In the electroplating process, the article to be plated (brass spoon) is placed at the cathode, while the anode is made of the plating metal (silver). At the anode, oxidation occurs where silver atoms lose electrons to form $Ag^+$ ions ($Ag - e^- \rightarrow Ag^+$), which migrate into the electrolyte. This ensures a continuous supply of $Ag^+$ ions for deposition on the cathode. Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains the Assertion.

Key Concept: Electrolysis of Acidified Water

c) 2:1
[Solution Description]
The overall reaction in the electrolysis of acidified water is:
$2H_2O \rightarrow 2H_2 + O_2$
According to stoichiometry, 2 moles of hydrogen gas ($H_2$) and 1 mole of oxygen gas ($O_2$) are produced. Since volume is directly proportional to the number of moles for gases under the same conditions, the ratio of $H_2$ to $O_2$ by volume is 2:1.

Key Concept: Electroplating

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that iron articles are electroplated with chromium to prevent rusting, which is true as chromium forms a thin protective layer on the surface of iron. The reason explains that chromium forms a protective oxide layer when exposed to air, which is also true and correctly explains why chromium plating prevents rusting. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Dissociation Reaction

b) $Na^+$ and $Cl^-$
[Solution Description]
When sodium chloride dissolves in water or is melted, it dissociates into its constituent ions. The dissociation reaction is: $NaCl \rightarrow Na^+ + Cl^-$.

Key Concept: Electrolytic Refining, Copper Refining

d) They collect near the anode as insoluble anode mud.
[Solution Description]
During the electrolytic refining of copper, the impure copper anode dissolves into the solution as $Cu^{2+}$ ions. Impurities like silver and gold, being less reactive, do not dissolve and settle as anode mud.

Key Concept: Electroplating

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion is true because electroplating iron with nickel forms a protective layer that prevents rusting. However, the reason is false because nickel does not act as a sacrificial metal; instead, it provides a barrier to oxygen and moisture. Sacrificial protection involves using a more reactive metal like zinc (galvanization). Thus, Assertion is true, but Reason is false.

Key Concept: Electrolysis Requirement

b) Direct current (D.C.)
[Solution Description]
Electrolysis involves a chemical change due to the flow of direct current (D.C.). Alternating current (A.C.) does not cause a permanent chemical change and is thus ineffective for electrolysis.

Key Concept: Electrolysis of acidified water, Electrochemical series

b) Hydrogen has lower overpotential
[Solution Description]
Although sodium is higher than hydrogen in the electrochemical series, in aqueous solutions we must consider:
1) Sodium ion concentration is negligible compared to H+ from acidified water
2) Water itself has lower reduction potential than Na+
3) The overpotential required for H2 evolution is lower than that needed for Na deposition
4) This follows the principle that in aqueous electrolysis, ions lower in the series may discharge preferentially if their concentration is much higher

Key Concept: Electrolytic dissociation

c) $Pb^{2+}$ ions move towards the cathode and gain electrons
[Solution Description]
In an electrolytic cell, when molten lead bromide ($PbBr_2$) is electrolyzed, $Pb^{2+}$ ions migrate towards the cathode (negative electrode) and gain electrons to form lead metal, while $Br^-$ ions move towards the anode (positive electrode) and lose electrons to form bromine gas.

Key Concept: Electrolysis, Direct Current

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) is true because direct current (DC) provides a steady flow of electrons in one direction, facilitating the continuous decomposition of the electrolyte into its constituent ions at the respective electrodes. This is essential for electrolysis to occur effectively. The reason (R) is also true and correctly explains the assertion. Alternating current (AC) reverses direction periodically, which would cause the products formed at the electrodes to be recombined, thereby nullifying the chemical change. Therefore, DC is necessary for sustained electrolysis.

Key Concept: Faraday's Laws of Electrolysis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Assertion (A) states that the mass of copper deposited during electrolysis is directly proportional to the quantity of electricity passed, which aligns with Faraday's First Law: $m \propto Q$. Reason (R) correctly states Faraday's First Law. Since the Reason properly explains the Assertion, both are true and Reason is the correct explanation of Assertion.

Key Concept: Electrolysis of acidified water, stoichiometry

d) 896 cm$^3$
[Solution Description]
The overall reaction is: $2H_2O \rightarrow 2H_2 + O_2$. The molar ratio of H$_2$ to O$_2$ is 2:1. So for every 1 mole of O$_2$, 2 moles of H$_2$ are produced. At STP, 1 mole of any gas occupies 22.4 dm$^3$ or 22400 cm$^3$.
448 cm$^3$ of O$_2$ = $\frac{448}{22400}$ = 0.02 moles of O$_2$.
Therefore, moles of H$_2$ produced = 2 × 0.02 = 0.04 moles.
Volume of H$_2$ = 0.04 × 22400 = 896 cm$^3$.

Key Concept: Overall Reaction

c) $2NaCl \rightarrow 2Na + Cl_2$
[Solution Description]
In the electrolysis of molten $NaCl$, sodium ions are reduced to sodium metal at the cathode, and chloride ions are oxidized to chlorine gas at the anode. The balanced overall reaction is:
$2NaCl \rightarrow 2Na + Cl_2$.

Key Concept: Electrolysis of molten lead bromide

d) Lead metal ($Pb$)
[Solution Description]
During the electrolysis of molten lead bromide ($PbBr_2$), $Pb^{2+}$ ions are reduced at the cathode to form metallic lead ($Pb$). The reaction at the cathode is:
$Pb^{2+} + 2e^- \rightarrow Pb$
Therefore, greyish-white metallic lead ($Pb$) is deposited at the cathode.

Key Concept: Electrode Reactions

b) Sodium metal ($Na$)
[Solution Description]
During electrolysis of molten $NaCl$, sodium ions ($Na^+$) migrate to the cathode and gain electrons. The reduction reaction occurs as follows:
$Na^+ + e^- \rightarrow Na$.
Therefore, sodium metal ($Na$) is deposited at the cathode.

Key Concept: Electroplating Process

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In electroplating, the anode must be made of the same metal that is to be deposited (silver in this case). This ensures that as silver atoms from the anode dissolve into the solution as $Ag^+$ ions ($Ag - e^- \rightarrow Ag^+$), they are available for reduction at the cathode ($Ag^+ + e^- \rightarrow Ag$). The reason correctly explains why the anode is made of pure silver.

Key Concept: Electrochemical Series, Preferential Discharge

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
During the electrolysis of dilute $AgNO_3$ solution, both $Ag^+$ and $H^+$ ions migrate towards the cathode. According to the electrochemical series, $Ag^+$ is lower than $H^+$, which means it has a higher tendency to gain electrons and get reduced. Therefore, $Ag^+$ is preferentially discharged at the cathode, forming metallic silver. The Reason correctly explains why this preferential discharge occurs, as it directly relates to their positions in the electrochemical series.

Key Concept: Electrolytic Dissociation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Solid sodium chloride is an ionic compound with $Na^+$ and $Cl^-$ ions held in a fixed crystal lattice. Since the ions are not mobile in the solid state, they cannot conduct electricity, making the Assertion true. The Reason correctly explains this behavior because the immobility of ions prevents conductivity. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Displacement Reactions

d) Iron
[Solution Description]
Iron (\text{Fe}) is higher in the electrochemical series than copper (\text{Cu}), so it can displace copper from its sulfate solution as per the reaction:
$\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$
Therefore, iron is the correct answer.

Key Concept: Electrolytic Conductors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion states that during electrolysis, cations migrate towards the cathode, which is true as per the syllabus. The reason given is that cations are positively charged ions and are attracted to the negatively charged cathode, which correctly explains why cations move towards the cathode. Both statements are true, and the reason correctly explains the assertion.

Key Concept: Second Law of Electrolysis

c) 2 grams
[Solution Description]
Using Faraday's second law:
$\frac{W_{H}}{E_{H}} = \frac{W_{O}}{E_{O}}$
Given:
$W_O = 16 \text{ grams}, E_O = 8 \text{ g/eq}, E_H = 1 \text{ g/eq}$
Substituting:
$\frac{W_H}{1} = \frac{16}{8}$
Solving:
$W_H = 2 \text{ grams}$
Hence, 2 grams of hydrogen is liberated.

Key Concept: Cathode, Reduction, Electrolysis

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
1) **Assertion (A):**
During the electrolysis of molten lead bromide, $Pb^{2+}$ ions migrate to the cathode and gain electrons to form metallic lead. This is true because according to the syllabus, the cathode is where reduction occurs, and the reaction is given as:
$Pb^{2+} + 2e^- \rightarrow Pb$.
2) **Reason (R):**
The cathode is indeed connected to the negative terminal of the battery, which attracts positively charged cations like $Pb^{2+}$. At the cathode, these cations gain electrons (reduction) to form neutral atoms. Therefore, Reason correctly explains why Assertion is true.
3) Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Electrolytic Dissociation

c) In molten or aqueous state
[Solution Description]
Electrolytic dissociation specifically occurs when ionic compounds are either in the molten state or dissolved in water (aqueous solution). In these states, the ions become free to move and conduct electricity.

Key Concept: Electrometallurgy, Reactivity series, Extraction methods

b) Iron (Fe)
[Solution Description]
Metals higher in the activity series (K, Na, Ca, Mg, Al) are extracted by electrolysis of their fused salts. Metals like Zn, Fe, Pb, Cu are extracted by reducing agents like carbon because their oxides are less stable. Platinum (Pt) and gold (Au) are extracted by thermal decomposition. From the options, iron (Fe) is extracted by reduction with carbon, not by electrolysis.

Key Concept: Migration of ions, Relative discharge potentials

c) Chloride ions discharge more easily than hydroxide ions
[Solution Description]
According to electrochemical series:
Chloride discharge potential (-1.36V) is lower than hydroxide (-0.40V).
Also, chloride concentration is much higher in NaCl solution.
Therefore, $2Cl^- \rightarrow Cl_2 + 2e^-$ occurs preferentially over $4OH^- \rightarrow O_2 + 2H_2O + 4e^-$.

Key Concept: Composition of Electrolytic Solutions, Strong vs Weak Electrolytes

b) Solution A shows a brighter bulb than Solution B
[Solution Description]
HCl is a strong electrolyte and dissociates completely in solution, producing more ions, leading to higher conductivity and a brighter bulb. CH$_3$COOH is a weak electrolyte and dissociates partially, resulting in fewer ions and dimmer bulb glow.

Key Concept: Definition of Electrolysis

b) Decomposition of a substance by passage of electric current
[Solution Description]
The word "electrolysis" is derived from two parts: "electro" meaning flow of electrons or electricity, and "lysis" meaning separating or decomposition. It refers to the process of decomposing a chemical compound in aqueous solution or molten state using direct electric current.

Key Concept: Electrolysis Observations

b) Oxygen ($O_2$)
[Solution Description]
During the electrolysis of acidified water, the anode reaction is:
$4OH^- \rightarrow 2H_2O + O_2 + 4e^-$.
This shows that oxygen gas ($O_2$) is liberated at the anode.

Key Concept: Metal Extraction by Electrolysis

c) Sodium
[Solution Description]
From the activity series, metals like potassium (K), sodium (Na), calcium (Ca), magnesium (Mg), and aluminum (Al) are extracted by electrolysis of their fused salts. Among the options given, sodium is extracted using fused sodium chloride (NaCl). The reaction at the cathode: $Na^+ + e^- \rightarrow Na$ and at the anode: $Cl^- - e^- \rightarrow Cl$.

Key Concept: Observation in Electrolysis

d) It fades and becomes colourless
[Solution Description]
The blue colour of the electrolyte is due to $Cu^{2+}$ ions. As electrolysis progresses, these ions are reduced to copper metal at the cathode, causing the blue colour to fade and eventually become colourless when all $Cu^{2+}$ ions are discharged.

Key Concept: Electrolysis Safety

b) To avoid electrode discoloration
[Solution Description]
Distilled water lacks dissolved salts/ions, preventing unwanted side reactions or corrosion of electrodes. Tap water contains impurities like chlorides that can interfere with electrolysis or damage equipment.

Key Concept: Electrolyte Concentration

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Molar conductivity ($\Lambda_m$) increases with dilution because more ions are available per unit volume as the solution becomes less concentrated. At infinite dilution, each ion moves independently, reducing interionic attractions and increasing conductivity. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Electrolysis of Molten NaCl, Electrode Reactions

c) 2.208 g
[Solution Description]
First, calculate the total charge passed using the formula:
$Q = I \times t$
Here, $I = 5$ A and $t = 20 \times 60 = 1200$ s.
So, $Q = 5 \times 1200 = 6000$ C.
Next, determine the moles of electrons involved using Faraday's law:
$\text{Moles of } e^- = \frac{Q}{F} = \frac{6000}{96500} = 0.0622 \text{ mol}$
From the anode reaction:
$Cl^- - e^- \rightarrow Cl$
$Cl + Cl \rightarrow Cl_2$
2 moles of electrons produce 1 mole of $Cl_2$.
Therefore, moles of $Cl_2$ produced:
$\text{Moles of } Cl_2 = \frac{0.0622}{2} = 0.0311 \text{ mol}$
Finally, the mass of $Cl_2$ liberated:
$\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.0311 \times 71 = 2.208 \text{ g}$

Key Concept: Active Electrodes, Redox Reactions

b) 3.175 g
[Solution Description]
For the electrolysis of $CuSO_4$ with copper electrodes:
- At the anode: Copper dissolves as $Cu \rightarrow Cu^{2+} + 2e^-$.
- Moles of $Cu$ dissolved = $\frac{0.1 \text{ mole of electrons}}{2}$ = 0.05 moles.
- Molar mass of $Cu$ is 63.5 g/mol.
- Mass loss at anode = $0.05 \times 63.5 = 3.175$ g.

Key Concept: Electrode Reactions in Acidified Water Electrolysis

b) It remains constant throughout the solution
[Solution Description]
In the electrolysis of acidified water, $H^+$ ions are discharged at the cathode to form hydrogen gas, while $OH^-$ ions are discharged at the anode to form oxygen gas. The removal of $H^+$ ions reduces the acidity near the cathode, but the discharge of $OH^-$ disturbs the ionic equilibrium of water, causing more water to dissociate to restore it. The newly produced $H^+$ ions combine with $SO_4^{2-}$ ions, maintaining the overall acidity. Hence, the concentration of sulphuric acid remains unchanged.