Q.Value of standard electrode potential for the oxidation of $Cl^-$ ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is $Cl^-$ oxidised at anode instead of water?

Concept and intuition:

Group 2 elements (alkaline earth metals) form salts with oxoacids like carbonic acid, sulphuric acid, and nitric acid. Their properties follow clear periodic trends due to increasing ionic size and decreasing lattice energy vs. hydration energy. For carbonates, the large carbonate ion makes lattice energy dominant, so they are insoluble. For sulphates, as the cation gets larger, hydration energy drops faster than lattice energy, so solubility decreases. Nitrates of group 2 are thermally unstable and decompose to oxide, nitrogen dioxide, and oxygen — a common property of many metal nitrates.

Step-by-step reasoning:

1. Statement I: "Their carbonates are generally insoluble in water"

   • Group 2 carbonates (e.g., MgCO₃, CaCO₃, SrCO₃, BaCO₃) have high lattice energies because the carbonate ion is large and doubly charged. The hydration energy of the small group 2 cations is not enough to overcome this lattice energy.
   • Result: All are insoluble (except BeCO₃, which is unstable and decomposes). So statement I is correct.

2. Statement II: "Solubility of their sulphates in water decreases from CaSO₄ to BaSO₄"

   • For sulphates, solubility depends on the balance between lattice energy and hydration energy. As we go down the group (Ca → Sr → Ba), the cation radius increases.
   • Lattice energy decreases slowly (since both ions are large), but hydration energy decreases more sharply (larger cation is less strongly hydrated). …

The key here is to recognise that aluminium is an amphoteric metal — it reacts with both acids and bases, but the products differ in each case. In acid, it forms a simple salt; in a strong base, it forms a complex aluminate. Both products are ionic and dissolve in water. The gas evolved in both reactions is hydrogen, so statements I and III are true, while II is false.

Let’s work through each reaction step by step.

1. Reaction with HCl (acid) Aluminium metal reacts with hydrochloric acid to give aluminium chloride and hydrogen gas. The balanced equation is:

Here, $X$ is $\mathrm{AlCl_3}$ and $A$ is $\mathrm{H_2}$.

Aluminium chloride is an ionic salt that dissolves readily in water (it is highly soluble). So statement II — “X is not soluble in water” — is false.

2. Reaction with water in presence of NaOH (base) Aluminium does not react with pure water at room temperature because of its protective oxide layer. However, in the presence of a strong base like NaOH, the oxide layer dissolves and aluminium reacts with water to form sodium aluminate and hydrogen gas. The balanced equation is:

Here, $Y$ is $\mathrm{NaAlO_2}$ (sodium aluminate) and $B$ is $\mathrm{H_2}$. …

The key idea is to compare the standard reduction potentials of all possible half‑reactions. In aqueous solution, water itself can be oxidised or reduced, and the species with the higher reduction potential (easier to reduce) wins at the cathode, while the species with the lower reduction potential (easier to oxidise) wins at the anode. For inert Pt electrodes, the electrode material does not participate.

1. Identify the ions and water present

   Aqueous CuSO₄ dissociates into Cu²⁺ and SO₄²⁻ ions. Water is also present (H₂O ⇌ H⁺ + OH⁻). At the cathode (reduction), possible reactions are:

   • Cu²⁺ + 2e⁻ → Cu(s) E° = +0.34 V
   • 2H₂O + 2e⁻ → H₂(g) + 2OH⁻ E° = –0.83 V (at pH 7) The more positive reduction potential (+0.34 V) means Cu²⁺ is much more easily reduced than water. So Cu metal deposits at the cathode.

2. At the anode (oxidation), consider the possibilities

   Possible oxidation reactions:

   • 2H₂O → O₂(g) + 4H⁺ + 4e⁻ E° = +1.23 V (reverse of O₂ reduction)
   • 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻ E° ≈ +2.01 V (very high)
   • 2H₂O → H₂O₂ + 2H⁺ + 2e⁻ E° = +1.78 V The easiest oxidation (lowest potential required) is water to O₂ at +1.23 V. Sulphate ions are extremely difficult to oxidise. So oxygen gas is evolved at the anode.

3. Check for any overpotential effects …

When an aqueous solution is electrolysed, there is a competition between the reduction of the metal cation and the reduction of water at the cathode. The species with the higher (less negative or more positive) standard reduction potential will be preferentially reduced.

Concept: Electrolysis at the Cathode

Electrolysis is the process of using electrical energy to drive non-spontaneous chemical reactions. At the cathode, reduction occurs. In an aqueous solution containing a metal cation ($M^{n+}$) and water ($H_2O$), two possible reduction reactions can take place:

1. Reduction of the metal cation:

   $M^{n+}(aq) + ne^- \rightarrow M(s)$

2. Reduction of water:

   $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$

To determine which reaction occurs, we compare their standard reduction potentials ($E^\circ$). The reaction with the more positive (or less negative) $E^\circ$ will occur preferentially.

Let's consider the standard reduction potentials relevant to this problem:

• $K^+(aq) + e^- \rightarrow K(s)$ ; $E^\circ = -2.92 V$
• $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ ; $E^\circ = -0.83 V$ (at standard conditions, i.e., $[OH^-]=1M$)
  • At neutral pH (like in an initial aqueous solution), the potential for water reduction is approximately $-0.42 V$. Regardless, this value is significantly less negative than that for K$^+$.

Step-by-step Solution

1. Electrolysis of Aqueous KCl Solution

1. Identify species present: In an aqueous KCl solution, the species available are K$^+$ ions, Cl$^-$ ions, and H$_2$O molecules.
2. Consider reactions at the cathode (reduction):
   • Reduction of K$^+$ ions: $K^+(aq) + e^- \rightarrow K(s)$ ; $E^\circ = -2.92 V$
   • Reduction of water: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ ; $E^\circ = -0.83 V$ (or approx. $-0.42 V$ at neutral pH)
3. Compare reduction potentials: The reduction potential of water (approx. $-0.42 V$ to $-0.83 V$) is significantly higher (less negative) than that of K$^+$ ions ($-2.92 V$).
4. Determine the product: Since water is much easier to reduce than K$^+$ ions, water will be preferentially reduced at the cathode. The gas liberated at the cathode in the electrolysis of aqueous KCl is hydrogen (H$_2$).

2. Electrolysis of Aqueous K$_2$SO$_4$ Solution …

Concept & Intuition

Concentrated sulfuric acid is a strong oxidizing agent, especially when hot. Carbon (C) is a reducing agent. When they react, the carbon gets oxidized (its oxidation number increases) and the sulfuric acid gets reduced (sulfur’s oxidation number decreases). The key is to track the oxidation states to predict the stable products: carbon typically goes to CO₂ (not CO) under these strongly oxidizing conditions, and sulfuric acid is reduced to SO₂ (not SO₃, which is an even higher oxidation state and would require further oxidation, not reduction).

1. Identify the reactants and their roles

   • Carbon (C): oxidation state 0. It can be oxidized to +2 (in CO) or +4 (in CO₂).
   • Concentrated H₂SO₄: sulfur is at +6. It can be reduced to +4 (in SO₂) or to 0 (in S) or even –2 (in H₂S). With hot concentrated acid, the common reduction product is SO₂.

2. Write the half-reactions

   • Oxidation: C → CO₂ + 4e⁻ (carbon goes from 0 to +4)
   • Reduction: H₂SO₄ + 2e⁻ → SO₂ + 2H₂O (sulfur goes from +6 to +4) (Note: The water produced is already shown in the given equation.)

3. Balance the electrons

   • Multiply the reduction half-reaction by 2 to match the 4 electrons from oxidation:

• Combine with oxidation:

This matches the given skeleton: C + conc. H₂SO₄ → X + Y + H₂O, with X = CO₂ and Y = SO₂.

4. Check the other options
   • (A) CO, SO₃: CO would mean carbon only goes to +2, and SO₃ would mean sulfur stays at +6 (no reduction) — unlikely with a reducing agent present. …

The question asks you to identify the process of concentration of ore represented by the given chemical reaction. The key is to recognize what is happening chemically: solid silver metal (or silver in its ore) is being converted into a soluble complex ion, $[\text{Ag}(\text{CN})_2]^-$, using a cyanide solution. This is not a physical separation based on density or wettability — it is a chemical dissolution.

Concept and intuition:

In metallurgy, "concentration of ore" means separating the valuable mineral from the gangue (unwanted rock). There are several methods:

• Leaching: uses a chemical reagent to selectively dissolve the desired metal from the ore, leaving impurities behind. The metal is later recovered from the solution.
• Levigation: washing powdered ore with water to separate lighter gangue from heavier metal particles (used for native metals like gold).
• Froth floatation: uses air bubbles and a frothing agent to separate sulfide ores from gangue based on differences in wettability.
• Liquation: melting an ore to allow a low-melting-point metal to flow away from higher-melting impurities.

Here, the reaction shows silver being dissolved by cyanide ions in the presence of oxygen and water, forming a soluble complex. This is exactly the MacArthur-Forrest process for silver extraction — a classic example of leaching.

Step-by-step reasoning:

1. Identify the chemical action:

   The reaction is:

   $4\text{Ag}(s) + 8\text{CN}^-(aq) + 2\text{H}_2\text{O}(aq) + \text{O}_2(g) \rightarrow 4[\text{Ag}(\text{CN})_2]^-(aq) + 4\text{OH}^-(aq)$

   Solid silver is being oxidized and complexed by cyanide, turning it into a water-soluble ion. This is a chemical dissolution, not a physical separation.

2. Match to the concentration method:

   • Leaching involves using a solvent (here, cyanide solution) to selectively dissolve the metal. The silver goes into solution as $[\text{Ag}(\text{CN})_2]^-$, leaving gangue behind. This fits perfectly.
   • Levigation uses water flow to separate by density — no chemical reaction.
   • Froth floatation uses air bubbles and collectors — no dissolution.
   • Liquation uses melting — no aqueous chemistry.

3. Confirm with known industrial process: …

The core idea here is Faraday’s second law of electrolysis: when the same quantity of electricity passes through different electrolytes, the masses of substances deposited are directly proportional to their equivalent weights. Equivalent weight is atomic mass divided by valency. Since the charge and time are identical in both experiments, we can set up a direct proportion between the masses and the equivalent weights of silver and gold.

Let’s walk through it step by step.

1. Recall the relationship. For a given amount of charge $Q$, the mass $m$ deposited is:

where $E$ is the equivalent weight (mass per mole of electrons), and $F$ is Faraday’s constant. Since $Q$ and $F$ are the same for both experiments, we have:

2. Express equivalent weights. For silver (Ag), valency $n_{\text{Ag}} = 1$ (always monovalent in electrolysis). So:

For gold (Au), let its valency be $n$ (unknown). Then:

3. Set up the proportion. From $m \propto E$, we get:

Substitute the given masses:

4. Solve for $n$. Simplify the right side:

Multiply both sides by 197:

Calculate the left side: …

When ammonia ($\mathrm{NH_3}$) is added to solutions containing metal ions, two main types of reactions can occur, depending on the metal ion and the amount of ammonia added:

1. Precipitation of Metal Hydroxides: Ammonia acts as a weak base in water, producing a small concentration of hydroxide ions ($\mathrm{OH^-}$): $\mathrm{NH_3 (aq) + H_2O (l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq)}$ If the solubility product ($K_{sp}$) of the metal hydroxide is exceeded, a metal hydroxide precipitate will form.
2. Formation of Ammine Complexes: Many transition metal ions can act as Lewis acids and form stable coordination complexes with ammonia, which acts as a Lewis base (ligand). These complexes are often soluble and can have characteristic colors. If the ammine complex is sufficiently stable, it can dissolve a pre-formed metal hydroxide precipitate.

Let's analyze each reaction:

1. Reaction A: $\mathrm{FeCl_3 (aq) + NH_3 (aq)}$

• Concept: Iron(III) ions ($\mathrm{Fe^{3+}}$) react with hydroxide ions from ammonia solution to form iron(III) hydroxide.
• Explanation: $\mathrm{FeCl_3}$ in aqueous solution provides $\mathrm{Fe^{3+}}$ ions. When ammonia solution is added, it generates $\mathrm{OH^-}$ ions. These $\mathrm{OH^-}$ ions react with $\mathrm{Fe^{3+}}$ to form iron(III) hydroxide, $\mathrm{Fe(OH)_3}$, which is an insoluble precipitate. $\mathrm{Fe^{3+} (aq) + 3NH_3 (aq) + 3H_2O (l) \rightarrow Fe(OH)_3 (s) + 3NH_4^+ (aq)}$
• Product Color: Iron(III) hydroxide, $\mathrm{Fe(OH)_3}$, is a characteristic reddish-brown or brown precipitate.
• Match: This matches III) Brown ppt.

2. Reaction B: $\mathrm{AgCl (aq) + NH_3 (aq)}$

• Concept: Silver chloride is an insoluble precipitate that dissolves in ammonia due to the formation of a stable ammine complex.
• Explanation: $\mathrm{AgCl}$ is silver chloride, which is a white precipitate and is sparingly soluble in water. When ammonia solution is added, the $\mathrm{Ag^+}$ ions from the dissolved $\mathrm{AgCl}$ react with ammonia molecules to form a stable, soluble complex ion, diamminesilver(I) ion, $\mathrm{[Ag(NH_3)_2]^+}$. This formation shifts the equilibrium of $\mathrm{AgCl}$ dissolution, causing the precipitate to dissolve. $\mathrm{AgCl (s) + 2NH_3 (aq) \rightarrow [Ag(NH_3)_2]^+ (aq) + Cl^- (aq)}$
• Product Color: The diamminesilver(I) complex, $\mathrm{[Ag(NH_3)_2]^+}$, is colorless in solution.
• Match: This matches IV) Colourless.

3. Reaction C: $\mathrm{Cu^{2+} (aq) + NH_3 (aq)}$

• Concept: Copper(II) ions initially form a hydroxide precipitate with ammonia, but in excess ammonia, they form a characteristic deep blue ammine complex. …

The reactions between ammonia ($\mathrm{NH_3}$) and chlorine ($\mathrm{Cl_2}$) are classic examples of how the stoichiometry (relative amounts of reactants) dictates the products, especially in redox reactions. Ammonia contains nitrogen in its lowest possible oxidation state ($-3$), making it a reducing agent. Chlorine is a strong oxidizing agent.

The key concept here is understanding how the limiting reagent influences the extent of oxidation of nitrogen and the fate of the hydrogen atoms from ammonia.

Reaction 1: $\mathrm{NH_3}$(excess)$+\mathrm{Cl_2} \longrightarrow P+Q$

1. Identify Reactants and their Roles:

   • Ammonia ($\mathrm{NH_3}$): Nitrogen is in the $-3$ oxidation state. It will be oxidized. Since it's in excess, it will also react with any acidic products formed.
   • Chlorine ($\mathrm{Cl_2}$): Chlorine is in the $0$ oxidation state. It will be reduced.

2. Initial Redox Reaction:

   When ammonia reacts with chlorine, nitrogen is oxidized, and chlorine is reduced. The most common oxidation product of nitrogen when ammonia acts as a reducing agent is nitrogen gas ($\mathrm{N_2}$), where nitrogen is in the $0$ oxidation state. Chlorine is reduced to chloride ions, forming hydrogen chloride ($\mathrm{HCl}$).

   The unbalanced reaction is: $\mathrm{NH_3} + \mathrm{Cl_2} \longrightarrow \mathrm{N_2} + \mathrm{HCl}$

   Balancing this gives:

3. Effect of Excess Ammonia: Since ammonia is in excess, it is a basic compound and will react with the acidic hydrogen chloride ($\mathrm{HCl}$) produced in the reaction. This acid-base reaction forms ammonium chloride ($\mathrm{NH_4Cl}$).

4. Overall Reaction and Products: To get the overall reaction, we combine the redox reaction with the acid-base reaction. For every $6$ moles of $\mathrm{HCl}$ produced, $6$ moles of $\mathrm{NH_3}$ will react with it.

Adding these two equations:

Therefore, the major products $P+Q$ are $\mathrm{N_2}$ and $\mathrm{NH_4Cl}$.

Reaction 2: $\mathrm{NH_3}+\mathrm{Cl_2}$(excess)$\longrightarrow R+S$

1. Identify Reactants and their Roles:

   • Ammonia ($\mathrm{NH_3}$): Nitrogen is in the $-3$ oxidation state. It will be oxidized. Since it is the limiting reagent, it will be completely consumed.
   • Chlorine ($\mathrm{Cl_2}$): Chlorine is in the $0$ oxidation state. It will be reduced. Since it's in excess, it acts as a strong oxidizing agent and will drive the oxidation of nitrogen to a higher state.

2. Redox Reaction with Excess Chlorine:

   When chlorine is in excess, it is a powerful oxidizing agent. It oxidizes the nitrogen in ammonia to a higher oxidation state than $0$. In this case, it replaces all hydrogen atoms in ammonia with chlorine atoms, forming nitrogen trichloride ($\mathrm{NCl_3}$), where nitrogen is in the $+3$ oxidation state. The hydrogen atoms from ammonia combine with chlorine to form hydrogen chloride ($\mathrm{HCl}$).

   The unbalanced reaction is: $\mathrm{NH_3} + \mathrm{Cl_2} \longrightarrow \mathrm{NCl_3} + \mathrm{HCl}$

   Balancing this gives: …

The concept first: where does ammonia actually come from?

There are exactly two routine routes on the syllabus:

1. Hydrolysis of a nitrogen compound whose nitrogen is already in the $-3$ state (urea, nitrides, cyanamide).
2. Displacement from an ammonium salt by a stronger base — because $NH_4^{+}$ is a weak acid, a strong base deprotonates it:

What does not give ammonia is an oxidation by $HNO_3$, where nitrogen goes down only as far as an ammonium ion in solution (or not at all).

Step-by-step through each reaction

1. $NH_2CONH_2 + 2H_2O \rightarrow$ (urea)

Urea hydrolyses (readily with the enzyme urease, or on boiling):

Ammonium carbonate is unstable and breaks down:

Ammonia is produced. ✓

2. $2NH_4Cl + Ca(OH)_2 \rightarrow$

This is the classic laboratory preparation of ammonia:

The strong base $Ca(OH)_2$ deprotonates $NH_4^{+}$. Ammonia is produced. ✓

3. $4Zn + 10HNO_3(\text{dil.}) \rightarrow$

With very dilute nitric acid, zinc reduces the nitrate all the way to the $-3$ oxidation state — but in the acidic medium the product is captured as the salt: …

The problem involves electrolysis, where a quantity of electricity causes the deposition of a certain mass of a substance. The key to solving this lies in Faraday's Laws of Electrolysis, particularly the second law, which deals with the deposition of different substances by the same quantity of electricity.

Concept: Faraday's Laws and Equivalent Weight

Faraday's First Law of Electrolysis states that the mass ($m$) of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte. Mathematically, $m \propto Q$.

Faraday's Second Law of Electrolysis states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited or liberated are directly proportional to their equivalent weights.

The equivalent weight ($E$) of an element is its atomic weight ($M$) divided by its valency ($n$).

Combining these ideas, if the same quantity of electricity is passed, then the number of gram equivalents deposited for each substance will be the same. The number of gram equivalents is given by:

$\text{Number of gram equivalents} = \frac{\text{Mass deposited}}{\text{Equivalent weight}}$

Therefore, for substances A, B, and C, if the same quantity of electricity is used:

$\frac{\text{Mass of A}}{\text{Equivalent weight of A}} = \frac{\text{Mass of B}}{\text{Equivalent weight of B}} = \frac{\text{Mass of C}}{\text{Equivalent weight of C}}$

Substituting $E = M/n$ into this relationship, we get:

$\frac{m_A}{M_A/n_A} = \frac{m_B}{M_B/n_B} = \frac{m_C}{M_C/n_C}$

This simplifies to:

$\frac{m_A n_A}{M_A} = \frac{m_B n_B}{M_B} = \frac{m_C n_C}{M_C}$

We can use this relationship to find the unknown valencies.

Here's how to solve the problem step-by-step:

1. Identify the given information:

   • Atomic weight of A ($M_A$) = 8
   • Mass of A deposited ($m_A$) = 2.4 g
   • Atomic weight of B ($M_B$) = 18
   • Mass of B deposited ($m_B$) = 1.8 g
   • Atomic weight of C ($M_C$) = 50
   • Mass of C deposited ($m_C$) = 7.5 g
   • The same quantity of electricity was used for all three.

2. Apply the principle of equal gram equivalents:

   Since the same quantity of electricity was passed, the number of gram equivalents deposited for A, B, and C must be equal.

   $\frac{m_A n_A}{M_A} = \frac{m_B n_B}{M_B} = \frac{m_C n_C}{M_C}$

3. Substitute the given values into the equation:

   For A: $\frac{2.4 \times n_A}{8} = 0.3 n_A$ …