Q.Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.What are X, Y in the following reactions? $2X_2(g) + 2H_2O(l) \rightarrow 4H^+(aq) + 4X^-(aq) + O_2(g)$ $Y_2(g) + H_2O(l) \rightarrow HY(aq) + HOY(aq)$ (A) X = F, Y = I (B) X = I, Y = Cl (C) X = F, Y = Cl (D) X = Cl, Y = I

   ›Reveal solutionSolution

   The first reaction is the vigorous oxidation of water by fluorine, so X = F. The second reaction is the disproportionation of chlorine in water, so Y = Cl. The correct option is (C).

   The key is to recognise the characteristic chemistry of the halogens with water. Not all halogens behave the same way — fluorine is so oxidising it rips oxygen from water, while chlorine, bromine and iodine undergo a different kind of reaction.

   1. First reaction: $2X_2(g) + 2H_2O(l) \rightarrow 4H^+(aq) + 4X^-(aq) + O_2(g)$ This is the oxidation of water to oxygen gas. For this to happen, the halogen must be a stronger oxidising agent than oxygen. Among the halogens, only fluorine ($F_2$) is powerful enough to do this. $F_2$ oxidises water according to:

   $2F_2(g) + 2H_2O(l) \rightarrow 4HF(aq) + O_2(g)$

   In aqueous solution, HF dissociates to $H^+$ and $F^-$, matching the given products exactly. No other halogen (Cl₂, Br₂, I₂) can displace oxygen from water — they instead dissolve to give acidic and hypohalous acid solutions.

   So X = F.

   2. Second reaction: $Y_2(g) + H_2O(l) \rightarrow HY(aq) + HOY(aq)$ This is a disproportionation reaction: the same element is both oxidised and reduced. Chlorine, bromine and iodine all do this in water, but fluorine does not — it never disproportionates because it has no positive oxidation state in water. For chlorine: $Cl_2(g) + H_2O(l) \rightarrow HCl(aq) + HOCl(aq)$ …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Consider the two redox reactions, I and II I. $2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 + \mathrm{I}_2 \rightarrow \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6 + 2\mathrm{NaI}$ II. $2\mathrm{KMnO}_4 + 10\mathrm{FeSO}_4 + 3\mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{MnSO}_4 + 5\mathrm{Fe}_2(\mathrm{SO}_4)_3 + 3\mathrm{H}_2\mathrm{O}$ The ratio of equivalent weights of $\mathrm{KMnO}_4$ and $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$ is (K = $39u$; Mn = $55u$; O = $16u$; Na = $23u$; S = $32u$) (A) $1:1$ (B) $2:1$ (C) $1:5$ (D) $5:1$

   ›Reveal solutionSolution

   The equivalent weight of a substance in a redox reaction is its molar mass divided by its n-factor (total change in oxidation state per molecule). For $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$, the n-factor is 1, and for $\mathrm{KMnO}_4$, it is 5. The ratio of their equivalent weights is $\boxed{1:5}$.

   The concept of equivalent weight is crucial in stoichiometry, especially for redox reactions. The equivalent weight of a substance is defined as its molar mass divided by its n-factor. For redox reactions, the n-factor represents the total change in the oxidation state of the atoms undergoing oxidation or reduction in one molecule of the substance. This factor tells us how many moles of electrons are gained or lost per mole of the substance.

   Here's how we determine the equivalent weights:

   1. Calculate the Molar Masses:

      First, we need the molar masses of $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$ and $\mathrm{KMnO}_4$.

      • For $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$: $M_{\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3} = (2 \times \text{Na}) + (2 \times \text{S}) + (3 \times \text{O})$ $M_{\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3} = (2 \times 23) + (2 \times 32) + (3 \times 16)$ $M_{\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3} = 46 + 64 + 48 = 158 \text{ u}$
      • For $\mathrm{KMnO}_4$: $M_{\mathrm{KMnO}_4} = (1 \times \text{K}) + (1 \times \text{Mn}) + (4 \times \text{O})$ $M_{\mathrm{KMnO}_4} = (1 \times 39) + (1 \times 55) + (4 \times 16)$ $M_{\mathrm{KMnO}_4} = 39 + 55 + 64 = 158 \text{ u}$

   2. Determine the n-factor for $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$ in Reaction I:

      The reaction is: $2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 + \mathrm{I}_2 \rightarrow \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6 + 2\mathrm{NaI}$

      We need to find the change in oxidation state for sulfur.

      • In $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$: Let the oxidation state of S be $x$. $2(+1) + 2(x) + 3(-2) = 0$ $2 + 2x - 6 = 0$ $2x = 4 \Rightarrow x = +2$
      • In $\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6$: Let the oxidation state of S be $y$. $2(+1) + 4(y) + 6(-2) = 0$ $2 + 4y - 12 = 0$ $4y = 10 \Rightarrow y = +2.5$
      • The change in oxidation state per sulfur atom is $+2.5 - (+2) = +0.5$.
      • Since there are two sulfur atoms in one molecule of $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$, the total change in oxidation state (n-factor) for $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$ is $2 \times 0.5 = 1$.

      Equivalent Weight ($E$) = $\frac{\text{Molar Mass}}{\text{n-factor}}$

      • Equivalent weight of $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3$: $E_{\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3} = \frac{158}{1} = 158$

   3. Determine the n-factor for $\mathrm{KMnO}_4$ in Reaction II: …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The reaction of lithium with water is less vigorous than that of sodium. It is due to I. Most negative $E^\Theta$ value of $\mathrm{Li}^+|\mathrm{Li}$ II. Small size of Li III. Very high hydration energy of $\mathrm{Li}^+$ (A) I, III only (B) II, III only (C) I, II only (D) I, II, III

   ›Reveal solutionSolution

   The lower reactivity of lithium with water compared to sodium is due to its small size and very high hydration energy, not its electrode potential. The correct answer is (B) II, III only.

   The question asks why lithium reacts less vigorously with water than sodium, even though lithium is above sodium in Group 1. At first glance, you might think lithium should be more reactive — but the trend in reactivity down Group 1 is actually the opposite of what you’d expect from ionization energy alone. The key lies in the interplay between size, hydration energy, and the actual driving force of the reaction.

   The reaction of an alkali metal with water is:

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

   The vigour depends on how easily the metal loses an electron and how favourably the resulting ion is stabilized in water. Let’s examine each statement.

   1. Statement I: Most negative $E^\Theta$ value of $\mathrm{Li}^+|\mathrm{Li}$

      The standard electrode potential for $\mathrm{Li}^+ + e^- \rightarrow \mathrm{Li}$ is $-3.04\ \mathrm{V}$, which is indeed the most negative among alkali metals. A more negative $E^\Theta$ means a greater thermodynamic tendency to lose electrons and form ions. If this were the only factor, lithium should be the most reactive — but it isn’t. So this statement does not explain the lower vigour; in fact, it predicts the opposite. Statement I is incorrect for the given observation.

   2. Statement II: Small size of Li

      Lithium has the smallest atomic radius in Group 1. A small size means its valence electron is held more tightly by the nucleus (higher ionization energy), making it harder to remove compared to sodium. This directly reduces the initial step of electron transfer to water, slowing the reaction. So the small size contributes to lower reactivity. Statement II is correct.

   3. Statement III: Very high hydration energy of $\mathrm{Li}^+$ …

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Which of the following is an endothermic process? (A) $\mathrm{Xe}(g) + e^- \rightarrow \mathrm{Xe}^-(g)$ (B) $\mathrm{O}(g) + e^- \rightarrow \mathrm{O}^-(g)$ (C) $\mathrm{Cl}(g) + e^- \rightarrow \mathrm{Cl}^-(g)$ (D) $\mathrm{I}(g) + e^- \rightarrow \mathrm{I}^-(g)$

   ›Reveal solutionSolution

   The key idea is that adding an electron to a neutral atom releases energy (exothermic) when the atom has a high electron affinity, but requires energy (endothermic) when the added electron must enter a filled or half-filled subshell. Here, only Xe has a filled valence shell, so its electron gain is endothermic. The correct option is (A).

   The relevant concept is electron affinity — the energy change when a gaseous atom gains an electron. A negative electron affinity means energy is released (exothermic), while a positive value means energy must be absorbed (endothermic). Most atoms release energy when gaining an electron because the added electron experiences a net attractive force from the nucleus. However, if the atom already has a stable, filled valence shell (like a noble gas), the extra electron must enter a higher-energy orbital, making the process energetically unfavorable.

   Let’s examine each option:

   1. Option (A): $\mathrm{Xe}(g) + e^- \rightarrow \mathrm{Xe}^-(g)$

      Xenon is a noble gas with a completely filled $5p^6$ subshell. Adding an electron forces it into the next available orbital ($6s$), which is much higher in energy and shielded from the nucleus. This requires energy input — the process is endothermic. Electron affinity of Xe is positive.

   2. Option (B): $\mathrm{O}(g) + e^- \rightarrow \mathrm{O}^-(g)$

      Oxygen has the configuration $2s^2 2p^4$. Adding an electron completes the half-filled $2p$ subshell to $2p^5$, which is energetically favorable. This releases energy (exothermic). Oxygen’s first electron affinity is negative.

   3. Option (C): $\mathrm{Cl}(g) + e^- \rightarrow \mathrm{Cl}^-(g)$ …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Match the following List-1 (Transition metal) List-2 ($E^{\ominus}_{\mathrm{M^{2+}|M}}$ in V) A. Cr I. $-1.18$ B. Co II. $-0.28$ C. Mn III. $-0.44$ D. Fe IV. $-0.90$ The correct answer is (A) A – IV, B – I, C – II, D – III (B) A – I, B – II, C – IV, D – III (C) A – III, B – IV, C – II, D – I (D) A – IV, B – II, C – I, D – III

   ›Reveal solutionSolution

   The standard reduction potentials for $\mathrm{M^{2+}|M}$ electrodes follow periodic trends and exceptions due to electronic configurations. The correct mapping is A–IV, B–II, C–I, D–III.

   The question tests your recall of standard reduction potentials for first-row transition metals. These values are not random — they reflect the stability of the $\mathrm{M^{2+}}$ ion relative to the metal. The key is to remember that $\mathrm{Mn^{2+}}$ has a half-filled d⁵ configuration (extra stable), making it harder to reduce, so its $E^\ominus$ is more negative. Conversely, $\mathrm{Co^{2+}}$ is relatively easier to reduce.

   Let’s match them step by step.

   1. Cr ($\mathbf{E^\ominus = -0.90\ V}$)

      Chromium has an anomalous electronic configuration: $\mathrm{Cr}$ is $3d^5 4s^1$, and $\mathrm{Cr^{2+}}$ is $3d^4$. The $d^4$ configuration is less stable, so $\mathrm{Cr^{2+}}$ is fairly easily reduced (less negative potential). The value $-0.90\ \mathrm{V}$ is the most negative among the given options except Mn. So A matches IV.

   2. Co ($\mathbf{E^\ominus = -0.28\ V}$)

      Cobalt’s $\mathrm{Co^{2+}}$ is $3d^7$, which is relatively stable. Among the four, $-0.28\ \mathrm{V}$ is the least negative (closest to zero), meaning $\mathrm{Co^{2+}}$ is the easiest to reduce. So B matches II.

   3. Mn ($\mathbf{E^\ominus = -1.18\ V}$) …

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6. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.100 mL of 0.05 M Cu$^{2+}$ aqueous solution is added to 1 L of 0.1 M KI solution. The number of moles of I$_2$ and CuI$_2$ formed are respectively (A) $2.5 \times 10^{-3},\ 5 \times 10^{-3}$ (B) $5 \times 10^{-3},\ 5 \times 10^{-3}$ (C) $5 \times 10^{-3},\ 2.5 \times 10^{-3}$ (D) $2.5 \times 10^{-3},\ 2.5 \times 10^{-3}$

   ›Reveal solutionSolution

   The reaction between Cu²⁺ and I⁻ produces I₂ and CuI (not CuI₂). Using stoichiometry and identifying the limiting reagent, the moles of I₂ and CuI formed are both $2.5 \times 10^{-3}$, so the correct option is (D).

   Concept & Intuition

   This is a redox reaction between copper(II) ions and iodide ions. Cu²⁺ is reduced to Cu⁺, and I⁻ is oxidized to I₂. However, Cu⁺ is unstable in water and immediately precipitates as white CuI (copper(I) iodide) because I⁻ is present in excess. The common mistake is to think CuI₂ forms — but copper(II) iodide does not exist under these conditions; Cu²⁺ oxidizes I⁻ and the resulting Cu⁺ binds with I⁻ to form CuI. The balanced equation is:

   $2\,\text{Cu}^{2+} + 4\,\text{I}^- \rightarrow 2\,\text{CuI} \downarrow + \text{I}_2$

   So for every 2 moles of Cu²⁺, we get 1 mole of I₂ and 2 moles of CuI. We just need to find the limiting reagent and compute the moles.

   Step-by-step solution

   1. Calculate initial moles of each reactant
      • Cu²⁺: volume = 100 mL = 0.100 L, concentration = 0.05 M

   $n_{\text{Cu}^{2+}} = 0.100 \times 0.05 = 0.005 \text{ mol} = 5 \times 10^{-3} \text{ mol}$

   • I⁻: volume = 1 L, concentration = 0.1 M

   $n_{\text{I}^-} = 1 \times 0.1 = 0.1 \text{ mol}$

   2. Determine the stoichiometric requirement From the balanced equation:

   $2\,\text{Cu}^{2+} + 4\,\text{I}^- \rightarrow 2\,\text{CuI} + \text{I}_2$

   For 0.005 mol Cu²⁺, the required I⁻ is:

   $\text{Required I}^- = 0.005 \times \frac{4}{2} = 0.01 \text{ mol}$

   We have 0.1 mol I⁻, which is far more than needed. So Cu²⁺ is the limiting reagent.

   3. Calculate moles of products
      • From the equation, 2 mol Cu²⁺ produce 1 mol I₂.

   $n_{\text{I}_2} = \frac{0.005}{2} = 0.0025 \text{ mol} = 2.5 \times 10^{-3} \text{ mol}$

   • Also, 2 mol Cu²⁺ produce 2 mol CuI (1:1 mole ratio between Cu²⁺ and CuI). $n_{\text{CuI}} = 0.005 \text{ mol} = 5 \times 10^{-3} \text{ mol}$ …

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7. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Two reactions I and II are given below I.

   [!FORMULA] $\begin{array}{c} \chemfig{*6([:270]-=-(-OH)-=-=)} + SOCl_2 \rightarrow \chemfig{*6([:270]-=-(-Cl)-=-=)} + SO_2 + HCl \end{array}$

   II.

   [!FORMULA] $\begin{array}{c} \chemfig{*6([:270]-=-(-Cl)-(-NO_2)-=-=)} + H_2O \xrightarrow{300\,K} \chemfig{*6([:270]-=-(-OH)-(-NO_2)-=-=)} + HCl \end{array}$

   Correct statement regarding the reactions is (A) Both reactions I and II are feasible (B) Both reactions I and II are not feasible (C) Reaction I feasible, but Reaction II not feasible (D) Reaction I not feasible, but Reaction II feasible

   ›Reveal solutionSolution

   The key idea is that reaction I (phenol to chlorobenzene via SOCl₂) is not feasible because the C–O bond in phenol has partial double-bond character, making direct nucleophilic substitution impossible, while reaction II (chlorobenzene to phenol with aqueous NaOH) is not feasible under the given mild conditions (300 K) because the C–Cl bond in chlorobenzene is stabilized by resonance. Thus both reactions are not feasible, so option (B) is correct.

   ---

   Concept and Intuition

   The problem tests your understanding of nucleophilic aromatic substitution and the reactivity of phenol derivatives. The key is recognizing that:

   • Phenol has a strong C–O bond due to resonance between the oxygen lone pair and the aromatic ring. This makes the –OH group a poor leaving group; you cannot simply replace it with Cl using SOCl₂ (which works for aliphatic alcohols).
   • Chlorobenzene has a C–Cl bond that is also resonance-stabilized (the chlorine lone pairs delocalize into the ring). This makes the chlorine atom less susceptible to nucleophilic displacement under mild conditions. For hydrolysis to occur, you typically need harsh conditions (high temperature, strong base, or a catalyst).

   Let’s examine each reaction step by step.

   ---

   Step-by-Step Reasoning

   1. Reaction I: Phenol + SOCl₂ → Chlorobenzene

   • What is attempted?

     Replacing the –OH group of phenol with –Cl using thionyl chloride (SOCl₂). This is a classic method for converting aliphatic alcohols to alkyl chlorides.

   • Why does it fail for phenol?

     In phenol, the oxygen’s lone pairs are conjugated with the aromatic π-system, giving the C–O bond partial double-bond character. This makes the C–O bond very strong and the –OH group a poor leaving group. SOCl₂ works by first converting –OH into a better leaving group (e.g., –OSOCl), but in phenol, the oxygen is too tightly bound to the ring for this to happen effectively.

     Moreover, the mechanism for aliphatic alcohols involves an SN₂ or SN₁ pathway, but the aromatic ring does not undergo such substitution easily.

   • Conclusion: Reaction I is not feasible under normal conditions.

   Watch out

   A common mistake is to assume that because SOCl₂ works for alcohols, it works for phenol. But phenol is not an alcohol — it’s an aromatic compound with a very different C–O bond.

   ---

   2. Reaction II: Chlorobenzene + H₂O at 300 K → Phenol

   • What is attempted?

     Hydrolysis of chlorobenzene to phenol using water at 300 K (about 27°C).

   • Why does it fail? …

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.Identify the correct statements from the following

   (i) Ti (IV) is more stable than Ti(III) and Ti(II)

   (ii) Among 3d – series elements (from Z = 22 to 29) only copper has positive reduction potential $(\mathrm{M}^{2+}/\mathrm{M})$

   (iii) Both Sc and Zn exhibit +1 oxidation state (A) i, ii only (B) i, iii only (C) ii, iii only (D) i, ii, iii

   ›Reveal solutionSolution

   The question tests stability of Ti oxidation states, reduction potentials in the 3d series, and the common oxidation states of Sc and Zn. Only statements (i) and (ii) are correct, so the answer is option (A).

   Concept & Intuition

   This problem combines three separate ideas from transition metal chemistry:

   • Stability of oxidation states depends on electronic configuration and the tendency to lose electrons. For titanium, the +4 state (empty d‑orbitals) is particularly stable.
   • Reduction potential measures how easily a metal ion gains electrons. In the 3d series, most M²⁺/M couples have negative potentials; copper is the exception because of its high ionization energy and low hydration enthalpy.
   • Oxidation states of Sc and Zn are governed by their d‑electron counts: Sc typically loses three electrons (Sc³⁺), and Zn loses two (Zn²⁺); +1 is not common for either.

   ---

   Step‑by‑step reasoning

   1. Statement (i): Ti(IV) is more stable than Ti(III) and Ti(II)

      • Titanium has the electronic configuration $[Ar]\,3d^2 4s^2$.
      • Ti(IV) has a noble‑gas configuration $[Ar]$, which is highly stable.
      • Ti(III) ($3d^1$) and Ti(II) ($3d^2$) are less stable because they can be further oxidized to the empty‑d‑shell state.
      • In aqueous solution, Ti²⁺ is a strong reducing agent and Ti³⁺ is also easily oxidized.
      • Conclusion: Statement (i) is true.

   2. Statement (ii): Among 3d‑series elements (Z = 22 to 29) only copper has positive reduction potential (M²⁺/M)

      • The standard reduction potentials for M²⁺/M in the 3d series (from Ti to Zn) are mostly negative.
      • For example: $$$$

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9. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Identify the correct set (A) SiO$_2$ – ionic solid – conductor – high melting point (B) SiO$_2$ – network solid – conductor – low melting point (C) NaCl – ionic solid – insulator in solid state – high melting point (D) Fe – ionic solid – conductor – low melting point

   ›Reveal solutionSolution

   The key is to match each substance with its correct classification (bonding type, electrical conductivity, and melting point). Only option (C) correctly describes NaCl as an ionic solid, an insulator in the solid state, and having a high melting point.

   Concept & Intuition

   This question tests your ability to connect a material’s internal bonding structure to its macroscopic properties. Every solid falls into one of four broad categories: ionic, metallic, covalent network, or molecular. Each category has a signature set of properties:

   • Ionic solids (e.g., NaCl) are made of ions held by strong electrostatic forces. They are brittle, have high melting points, and do not conduct electricity as solids (ions are locked in place), but do when molten or dissolved.
   • Covalent network solids (e.g., SiO₂ as quartz) have atoms bonded in a continuous 3D network by covalent bonds. They are very hard, have extremely high melting points, and are insulators (no free electrons or mobile ions).
   • Metallic solids (e.g., Fe) have a “sea” of delocalized electrons. They are malleable, conduct electricity well, and have variable melting points (Fe’s is high, but many metals have low melting points).
   • Molecular solids (e.g., ice) are held by weak intermolecular forces; they are soft, insulators, and have low melting points.

   The trick is to spot which option gets all three descriptors right for the given substance.

   Step-by-step reasoning

   1. Examine option (A): SiO₂ – ionic solid – conductor – high melting point

      • SiO₂ is not ionic; it’s a covalent network solid (each Si bonds to four O atoms in a tetrahedral lattice).
      • As a network solid, it is an insulator, not a conductor.
      • The “high melting point” part is correct (~1600 °C), but the first two descriptors are wrong. → Eliminate (A).

   2. Examine option (B): SiO₂ – network solid – conductor – low melting point

      • “Network solid” is correct for SiO₂.
      • But network solids are insulators, not conductors.
      • Also, SiO₂ has a high melting point, not low. → Eliminate (B).

   3. Examine option (C): NaCl – ionic solid – insulator in solid state – high melting point

      • NaCl is indeed an ionic solid (Na⁺ and Cl⁻ ions). …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.The standard electrode potentials $E^\circ$ (V) for Li$^+$/Li, Na$^+$/Na respectively are (A) $-3.04,\ -2.714$ (B) $-2.714,\ -3.04$ (C) $-3.04,\ -3.04$ (D) $-2.714,\ -2.714$

    ›Reveal solutionSolution

    The standard electrode potentials for Li⁺/Li and Na⁺/Na are −3.04 V and −2.714 V respectively, so the correct order is option (A).

    The key here is remembering the actual numerical values from the electrochemical series. Many students get confused because lithium is above sodium in the periodic table, but its reduction potential is more negative (i.e., it is a stronger reducing agent). The standard reduction potentials are experimentally measured values, not something you deduce from periodic trends alone — though the trend does help you check consistency.

    1. Recall the standard reduction potentials For the half‑reaction

    $\text{M}^+ + e^- \rightarrow \text{M}$

    the standard electrode potentials (at 25 °C, 1 M, 1 atm) are:

    • Li⁺/Li: $E^\circ = -3.04\ \text{V}$
    • Na⁺/Na: $E^\circ = -2.714\ \text{V}$

    2. Interpret the sign and magnitude

       A more negative value means the metal is more difficult to reduce (i.e., it is a stronger reducing agent). Lithium’s potential is more negative than sodium’s, so lithium is the stronger reductant — consistent with its position in the reactivity series.

    3. Match to the options …

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11. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.Two statements are given below Statement I: Molten NaCl is electrolysed using Pt electrodes. Cl$_2$ is liberated at anode Statement II: Aqueous CuSO$_4$ is electrolysed using Pt electrodes. O$_2$ is liberated at cathode The correct answer is (A) Both statement I and II are correct (B) Both statement I and II are not correct (C) Statement I is correct but statement II is not correct (D) Statement I is not correct but statement II is correct

    ›Reveal solutionSolution

    Molten $\ce{NaCl}$ really does give $\ce{Cl2}$ at the anode, but electrolysis of aqueous $\ce{CuSO4}$ deposits $\ce{Cu}$ at the cathode and gives $\ce{O2}$ at the anode. So statement I is right, II is wrong — option (C).

    The concept first. Electrolysis is not about "which ion moves where" alone; it is about which species is easiest to reduce at the cathode and easiest to oxidise at the anode. Two rules never bend:

    • Cathode = reduction (electrons pushed in), anode = oxidation (electrons pulled out).
    • In an aqueous solution, water itself competes with the solute ions, and the species with the more favourable discharge potential wins.

    Step 1 — Molten NaCl (Statement I).

    The melt contains only $\ce{Na+}$ and $\ce{Cl-}$; there is no water to compete.

    Cathode: $\ce{Na+ + e- -> Na}$ (molten sodium).

    Anode: $\ce{2Cl^- -> Cl2 + 2e^-}$.

    This is exactly the Down's process, and chlorine is indeed the anode product. Statement I is correct.

    Step 2 — Aqueous CuSO4 with Pt electrodes (Statement II).

    Species present: $\ce{Cu^{2+}}$, $\ce{SO4^{2-}}$, $\ce{H2O}$ (and traces of $\ce{H+}$, $\ce{OH-}$). …

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12. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.Which of the following orders is correct for the property given? (A) Cr < Mn < Fe - standard electrode potential value for $M^{3+}/M^{2+}$ (B) Cr$^{2+}$ < Mn$^{2+}$ < Fe$^{2+}$ - magnetic moments (C) VO$_2^+$ < Cr$_2$O$_7^{2-}$ < MnO$_4^-$ - oxidizing power (D) Ti < V < Cr - first ionization enthalpy

    ›Reveal solutionSolution

    The question tests periodic trends in transition-metal properties. By analyzing each option against known chemical principles, only option (C) — the order of oxidizing power $\mathrm{VO}_2^+ < \mathrm{Cr}_2\mathrm{O}_7^{2-} < \mathrm{MnO}_4^-$ — is correct.

    Concept and Intuition

    Transition metals show subtle but systematic trends across a period. For standard electrode potentials, magnetic moments, oxidizing power, and ionization enthalpy, the key is to recall how effective nuclear charge, electronic configuration, and stability of oxidation states change from left to right. A common mistake is to assume monotonic trends where exceptions exist (e.g., Cr and Mn in ionization enthalpy). Here, we check each option carefully.

    ---

    1. Option (A): Standard electrode potential for $M^{3+}/M^{2+}$

       The standard reduction potential $E^\circ$ for $M^{3+} + e^- \rightarrow M^{2+}$ does not increase smoothly from Cr to Fe.

       • Cr: $E^\circ(\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}) = -0.41\ \mathrm{V}$
       • Mn: $E^\circ(\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}) = +1.51\ \mathrm{V}$ (very high because $\mathrm{Mn}^{2+}$ is exceptionally stable, half-filled d⁵)
       • Fe: $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77\ \mathrm{V}$ So the order is Cr < Fe < Mn, not Cr < Mn < Fe. Hence (A) is false.

    2. Option (B): Magnetic moments of $\mathrm{Cr}^{2+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}$

       Magnetic moment $\mu = \sqrt{n(n+2)}$ Bohr magnetons, where $n$ = number of unpaired electrons.

       • $\mathrm{Cr}^{2+}$: $[\mathrm{Ar}]3d^4$ → 4 unpaired (high-spin) → $\mu \approx 4.90\ \mathrm{BM}$
       • $\mathrm{Mn}^{2+}$: $[\mathrm{Ar}]3d^5$ → 5 unpaired → $\mu \approx 5.92\ \mathrm{BM}$
       • $\mathrm{Fe}^{2+}$: $[\mathrm{Ar}]3d^6$ → 4 unpaired (high-spin) → $\mu \approx 4.90\ \mathrm{BM}$ So the order is $\mathrm{Mn}^{2+} > \mathrm{Cr}^{2+} = \mathrm{Fe}^{2+}$, not $\mathrm{Cr}^{2+} < \mathrm{Mn}^{2+} < \mathrm{Fe}^{2+}$. Hence (B) is false.

    3. Option (C): Oxidizing power of $\mathrm{VO}_2^+, \mathrm{Cr}_2\mathrm{O}_7^{2-}, \mathrm{MnO}_4^-$

       Oxidizing power in oxoanions increases with the oxidation state of the central metal and with the element’s position in the periodic table (higher electronegativity, more stable lower oxidation state).

       • V in $\mathrm{VO}_2^+$: oxidation state +5
       • Cr in $\mathrm{Cr}_2\mathrm{O}_7^{2-}$: oxidation state +6
       • Mn in $\mathrm{MnO}_4^-$: oxidation state +7 Across the period, the tendency to accept electrons (oxidizing power) increases: $\mathrm{MnO}_4^-$ is the strongest, $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ intermediate, $\mathrm{VO}_2^+$ weakest. Standard reduction potentials confirm:
         • $\mathrm{VO}_2^+ + 2H^+ + e^- \rightarrow \mathrm{VO}^{2+} + H_2O$: $E^\circ \approx +1.00\ \mathrm{V}$ …

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