Q.E⊖ values of some redox couples are given below. On the basis of these values choose the correct option.
E⊖ values: Br2/Br^- = +1.90; Ag^+/Ag(s) = +0.80; Cu^2+/Cu(s) = +0.34; I2(s)/I^- = +0.54

Showing the 12 most recent of 28 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.$X, Y$ and $Z$ represent three electrodes $Al^{3+}/Al$, $Cu^{2+}/Cu$ and $Ag^+/Ag$ with $E°$ values $-1.66$, $0.34$ and $0.80\ V$ respectively. The correct order of oxidising power of these three electrodes is (A) $X > Y > Z$ (B) $Z > Y > X$ (C) $X = Y = Z$ (D) $Y > Z > X$

   ›Reveal solutionSolution

   This tests reading standard reduction potentials as a measure of oxidising power; higher (more positive) $E°$ means stronger oxidising power, giving $Z>Y>X$.

   Concept and Intuition

   The standard reduction potential $E°$ of an electrode couple $M^{n+}/M$ measures how readily $M^{n+}$ is reduced to $M$. A more positive $E°$ means the ion has a greater tendency to be reduced — i.e., it is a stronger oxidising agent (it more readily takes electrons from something else, getting reduced itself). So ranking electrodes by oxidising power is the same as ranking them by $E°$ value, from most positive (strongest oxidiser) to most negative (weakest oxidiser / strongest reducing agent in its reduced form).

   Step-by-Step Solution

   1. Identify the labels: $X = Al^{3+}/Al$ ($E°=-1.66\ V$), $Y=Cu^{2+}/Cu$ ($E°=0.34\ V$), $Z=Ag^+/Ag$ ($E°=0.80\ V$).
   2. Rank by $E°$ value (most positive = strongest oxidising power): $Z\ (0.80) > Y\ (0.34) > X\ (-1.66)$.
   3. This directly gives the oxidising-power order: $Z > Y > X$.

   Common Mistakes

   • Confusing oxidising power with reducing power — a very negative $E°$ (like $Al^{3+}/Al$) means the metal Al is a strong reducing agent, not that the ion $Al^{3+}/Al$ couple has strong oxidising power.
   • Ranking by atomic number or reactivity series intuition instead of directly by the given $E°$ values.

   ✓Final answer

   The correct option is (B) — $Z > Y > X$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\Delta G^\circ$ (in kJ mol$^{-1}$) for the cell reaction given below is $2Al(s) + 3Cu^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Cu(s)$ (Given: $E^\circ_{Al^{3+}|Al} = -1.66$ V ; $E^\circ_{Cu^{2+}|Cu} = +0.34$ V ; F = 96500 C mol$^{-1}$) (A) -1158 (B) -579 (C) -386 (D) -772

   ›Reveal solutionSolution

   Standard electrochemistry problem: find $E^\circ_{cell}$ from the two standard reduction potentials, count the electrons transferred in the balanced equation, then apply $\Delta G^\circ = -nFE^\circ$. Result: $-1158\ kJ\,mol^{-1}$.

   Concept and Intuition

   The standard Gibbs free energy change of a cell reaction is related to its standard cell potential by

   $\Delta G^\circ = -nFE^\circ_{cell}$

   where $n$ is the number of moles of electrons transferred in the balanced overall reaction, and $F$ is the Faraday constant. Here, Al is oxidised (its half-reaction is reversed relative to the reduction potential given, so it becomes the anode), and $Cu^{2+}$ is reduced (cathode). The cell potential is always cathode potential minus anode potential (both taken as standard reduction potentials, without flipping signs manually):

   $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$

   Step-by-Step Solution

   1. Identify cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu$, $E^\circ = +0.34\ V$.
   2. Identify anode (oxidation, but use its reduction potential in the formula): $Al^{3+} + 3e^- \rightarrow Al$, $E^\circ = -1.66\ V$.
   3. $E^\circ_{cell} = 0.34 - (-1.66) = 2.00\ V$.
   4. Balance the overall reaction: $2Al \rightarrow 2Al^{3+}+6e^-$ and $3Cu^{2+}+6e^- \rightarrow 3Cu$, so total electrons transferred $n=6$.
   5. $\Delta G^\circ = -nFE^\circ = -(6)(96500\ C\,mol^{-1})(2.00\ V) = -1{,}158{,}000\ J\,mol^{-1} = -1158\ kJ\,mol^{-1}$.

   Common Mistakes

   • Using $n=2$ or $n=3$ (the individual half-reaction electron counts) instead of $n=6$, the electrons transferred in the fully balanced overall equation.
   • Sign errors: forgetting that $E^\circ_{cell}$ is cathode minus anode (both as reduction potentials), not adding the two potentials or flipping the wrong sign.
   • Forgetting to convert the final answer from joules to kilojoules.

   ✓Final answer

   The correct option is (A) — -1158.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Which of the following reactions is non-spontaneous? (A) $2F_2 + 2H_2O \rightarrow 4HF + O_2$ (B) $Cl_2 + H_2O \rightarrow HCl + HOCl$ (C) $Br_2 + H_2O \rightarrow HBr + HOBr$ (D) $2I_2 + 2H_2O \rightarrow 4HI + O_2$

   ›Reveal solutionSolution

   Comparing the standard reduction potentials of the halogens with that of $O_2/H_2O$ shows only $F_2$ (and, via ordinary hydrolysis, $Cl_2$/$Br_2$) can act spontaneously on water; iodine cannot oxidise water to $O_2$. The answer is (D).

   Concept and Intuition

   Whether a halogen reacts with water depends on its oxidising power (standard reduction potential) relative to the species it must oxidise. Fluorine has such an exceptionally high reduction potential that it can directly oxidise water all the way to $O_2$ gas, releasing HF — a highly spontaneous, even violent reaction. Chlorine and bromine are weaker oxidants and instead undergo a milder disproportionation-type hydrolysis, forming the hydrohalic acid and the hypohalous acid ($HOX$), which is also spontaneous (though the equilibrium lies less and less to the product side going down the group). Iodine is the weakest oxidant of these halogens; its reduction potential is too low to drive the oxidation of water to $O_2$, so a reaction analogous to fluorine's ($2I_2+2H_2O\rightarrow4HI+O_2$) does not occur spontaneously.

   Step-by-Step Solution

   1. (A) $2F_2+2H_2O\rightarrow4HF+O_2$: fluorine's reduction potential ($E° \approx 2.87$ V) vastly exceeds that of $O_2/H_2O$ ($E°\approx1.23$ V), so this reaction is strongly spontaneous — not the answer.
   2. (B) $Cl_2+H_2O\rightarrow HCl+HOCl$: this simple hydrolysis (disproportionation) reaction is spontaneous and is the actual observed reaction of chlorine with water — not the answer.
   3. (C) $Br_2+H_2O\rightarrow HBr+HOBr$: similarly a spontaneous hydrolysis reaction (to a smaller extent than chlorine, but still spontaneous) — not the answer.
   4. (D) $2I_2+2H_2O\rightarrow4HI+O_2$: this requires iodine to oxidise water to $O_2$, exactly like fluorine does — but iodine's reduction potential ($E°\approx0.54$ V) is far below that of $O_2/H_2O$ ($1.23$ V), so iodine simply cannot drive this reaction; it is thermodynamically non-spontaneous.

   Common Mistakes

   • Assuming all halogens behave identically toward water just because they're in the same group.
   • Not distinguishing between the "ordinary hydrolysis" reaction ($X_2+H_2O\rightarrow HX+HOX$, spontaneous down to Br) and the "water-oxidation-to-O₂" reaction (spontaneous only for F).

   ✓Final answer

   The correct option is (D) — $2I_2 + 2H_2O \rightarrow 4HI + O_2$.

   ANSWER: D

4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Identify the sets in which both the metals react with water? I. Be, Mg II. Li, Mg III. Na, K IV. K, Ca Correct answer is (A) II, IV only (B) II, III, IV only (C) III, IV only (D) I, II only

   ›Reveal solutionSolution

   Only Be fails to react with water; sets II, III and IV have both metals reacting → (B).

   Concept and Intuition

   Among the light s-block metals, reactivity with water rises down and across the alkali/alkaline-earth series. Beryllium is anomalous — it does not react with water even as steam (protective oxide, high hydration/ionisation energy). Magnesium reacts slowly with hot water/steam; alkali metals (Li, Na, K) react with water, and Ca reacts readily with cold water.

   Step-by-Step Solution

   1. Set I (Be, Mg): Be does not react with water → invalid.
   2. Set II (Li, Mg): Li reacts with water, Mg reacts with hot water/steam → valid.
   3. Set III (Na, K): both react vigorously with water → valid.
   4. Set IV (K, Ca): both react with water → valid.
   5. Valid sets = II, III, IV → option (B).

   Common Mistakes

   • Assuming Be behaves like other alkaline-earth metals — Be is the exception that does not react with water.
   • Overlooking that Mg does react, but only with hot water/steam.

   ✓Final answer

   The correct option is (B) — II, III, IV only.

   ANSWER: B

   Note

   This solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.

5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.For which of the following the $E^{\ominus}(\mathrm{M^{3+}/M^{2+}})$ is negative? (A) Mn (B) Co (C) Fe (D) Cr

   ›Reveal solutionSolution

   $\mathrm{Cr^{3+}/Cr^{2+}}$ has a negative standard reduction potential because $\mathrm{Cr^{2+}}$ is unstable/strongly reducing (readily loses an electron to reach the stable $d^3$ $\mathrm{Cr^{3+}}$), unlike Mn, Fe, Co which favour +2. The answer is (D) Cr.

   Concept and Intuition

   The sign and magnitude of $E^\ominus(M^{3+}/M^{2+})$ tells us the relative thermodynamic stability of the +2 vs +3 oxidation state for a transition metal:

   • A large positive $E^\ominus$ means $M^{3+}$ is a strong oxidising agent — i.e., $M^{2+}$ is the thermodynamically favoured/stable state (electron gain is favourable).
   • A negative $E^\ominus$ means the reverse: $M^{2+}$ readily loses an electron to become $M^{3+}$ — i.e., $M^{3+}$ is the stable state and $M^{2+}$ is a strong reducing agent.

   Extra stability of a particular $d^n$ configuration (half-filled $d^5$, or in this case the stability trend of $d^3$ for $\mathrm{Cr^{3+}}$) strongly influences these potentials. Known standard values (approx.): $\mathrm{Mn^{3+}/Mn^{2+}} = +1.57$ V, $\mathrm{Co^{3+}/Co^{2+}} = +1.82$ V, $\mathrm{Fe^{3+}/Fe^{2+}} = +0.77$ V, $\mathrm{Cr^{3+}/Cr^{2+}} = -0.41$ V.

   Step-by-Step Solution

   1. Recall/estimate the sign of $E^\ominus(M^{3+}/M^{2+})$ for each metal in the options.
   2. Mn: $\mathrm{Mn^{2+}}$ ($d^5$, half-filled, extra stable) is strongly favoured over $\mathrm{Mn^{3+}}$ ($d^4$) — so $E^\ominus$ is a large positive value (+1.57 V). Not the answer.
   3. Co: $\mathrm{Co^{2+}}$ ($d^7$) is far more stable than $\mathrm{Co^{3+}}$ ($d^6$, in aqueous simple ions) — $E^\ominus$ is strongly positive (+1.82 V). Not the answer.
   4. Fe: $\mathrm{Fe^{2+}}$ ($d^6$) to $\mathrm{Fe^{3+}}$ ($d^5$, half-filled and stable) is only mildly favourable to oxidise — $E^\ominus$ is positive but modest (+0.77 V). Not the answer.
   5. Cr: $\mathrm{Cr^{2+}}$ ($d^4$) readily loses an electron to reach the more stable $\mathrm{Cr^{3+}}$ ($d^3$, particularly stable half-filled $t_{2g}^3$ subshell in an octahedral field) — so $\mathrm{Cr^{2+}}$ is a strong reducing agent and $E^\ominus(\mathrm{Cr^{3+}/Cr^{2+}})$ is negative (≈ −0.41 V).
   6. Only Cr gives a negative value, matching option D.

   Common Mistakes

   • Assuming all $M^{3+}/M^{2+}$ potentials are positive by default — the Cr case is the well-known textbook exception because $\mathrm{Cr^{3+}}$'s $d^3$ configuration is unusually stable.
   • Confusing this with $\mathrm{Mn^{3+}/Mn^{2+}}$, which is a large positive value (opposite sign) due to $\mathrm{Mn^{2+}}$'s $d^5$ stability.

   ✓Final answer

   The correct option is (D) — Cr.

   ANSWER: D

6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If $E^\circ_{\mathrm{Fe^{2+}/Fe}} = -0.441\ \mathrm{V}$ and $E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} = 0.771\ \mathrm{V}$, the standard emf of the cell reaction $\mathrm{Fe(s) + 2Fe^{3+}(aq) \longrightarrow 3Fe^{2+}(aq)}$ is (A) $-1.212\ \mathrm{V}$ (B) $+1.212\ \mathrm{V}$ (C) $-2.424\ \mathrm{V}$ (D) $+2.424\ \mathrm{V}$

   ›Reveal solutionSolution

   Combining the Fe$^{2+}$/Fe and Fe$^{3+}$/Fe$^{2+}$ half-cells for the disproportionation-type reaction Fe + 2Fe$^{3+}$ → 3Fe$^{2+}$ gives a standard cell potential of $+1.212$ V.

   Concept and Intuition

   Standard cell potential is computed as $E^\circ_{cell} = E^\circ_{cathode(reduction)} - E^\circ_{anode(reduction)}$, regardless of how many electrons each half-reaction involves — $E^\circ$ is an intensive quantity and is NOT multiplied when a half-reaction is scaled to balance electrons.

   Step-by-Step Solution

   1. Identify the two half reactions inside the overall reaction Fe(s) + 2Fe$^{3+}$(aq) → 3Fe$^{2+}$(aq):
      • Oxidation (anode): $\mathrm{Fe \rightarrow Fe^{2+} + 2e^-}$, using $E^\circ_{\mathrm{Fe^{2+}/Fe}} = -0.441$ V (as a reduction potential).
      • Reduction (cathode): $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$ (doubled to $\mathrm{2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}}$ for electron balance), using $E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}} = 0.771$ V — this value does NOT change when the equation is doubled.
   2. Apply $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.771 - (-0.441) = 1.212$ V.
   3. Since this is positive, the reaction as written is spontaneous, consistent with iron metal reducing Fe$^{3+}$ to Fe$^{2+}$.

   Common Mistakes

   • Doubling $E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}$ to "balance" the 2 electrons — standard potentials are never scaled with stoichiometric coefficients.
   • Sign errors when subtracting a negative anode potential (must become $+0.441$, not $-0.441$, in the addition).

   ✓Final answer

   The correct option is (B) — $+1.212\ \mathrm{V}$.

   ANSWER: B

7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Which one of the following reactions is not feasible ? (g = gas, l = liquid, s = solid, aq = aqueous) (A) $\mathrm{Cl_2(g) + 2KBr(aq) \longrightarrow 2KCl(g) + Br_2(l)}$ (B) $\mathrm{Cl_2(g) + 2KI(aq) \longrightarrow 2KCl(aq) + I_2(s)}$ (C) $\mathrm{Br_2(l) + 2KI(aq) \longrightarrow 2KBr(aq) + I_2(s)}$ (D) $\mathrm{I_2(s) + 2KBr(aq) \longrightarrow 2KI(aq) + Br_2(l)}$

   ›Reveal solutionSolution

   This tests the halogen displacement (reactivity) series; the infeasible reaction is I₂ + 2KBr → 2KI + Br₂, option (D).

   Concept and Intuition

   Among halogens, oxidizing power (and hence the ability to displace a halide from its salt) decreases down the group: $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$. A more powerful oxidizing halogen can displace a less powerful one from its halide salt, but the reverse cannot happen spontaneously.

   Step-by-Step Solution

   1. (A) $\mathrm{Cl_2 + 2KBr \to 2KCl + Br_2}$: Cl₂ is a stronger oxidizer than Br₂, so it displaces bromide — feasible.
   2. (B) $\mathrm{Cl_2 + 2KI \to 2KCl + I_2}$: Cl₂ is stronger than I₂, displaces iodide — feasible.
   3. (C) $\mathrm{Br_2 + 2KI \to 2KBr + I_2}$: Br₂ is stronger than I₂, displaces iodide — feasible.
   4. (D) $\mathrm{I_2 + 2KBr \to 2KI + Br_2}$: I₂ is WEAKER than Br₂ as an oxidizer, so it cannot displace bromide from KBr — NOT feasible.

   Common Mistakes

   • Forgetting the direction of the reactivity trend and assuming any halogen can displace any other.
   • Mixing up oxidizing power with reducing power of the halide ions.

   ✓Final answer

   The correct option is (D) — $\mathrm{I_2(s) + 2KBr(aq) \longrightarrow 2KI(aq) + Br_2(l)}$ is not feasible.

   ANSWER: D

8. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.The $E^{\ominus}_{M^+(aq)|M(s)}$ is highest with negative sign for the alkali metal 'x' and lowest with negative sign for the alkali metal 'y'. In flame test, the characteristic colours of $x$ and $y$ are respectively (A) Blue, Yellow (B) Yellow, Violet (C) Yellow, Crimson red (D) Crimson red, Yellow

   ›Reveal solutionSolution

   Li has the most negative $E^\ominus_{M^+/M}$ among alkali metals (anomalously, due to huge hydration enthalpy) and Na has the least negative. Their flame colours are crimson red (Li) and yellow (Na) respectively.

   Concept and Intuition

   Standard reduction potentials of alkali metals in water do not follow the simple ionisation-energy trend because hydration enthalpy also matters heavily. Lithium's very small ionic size gives it an unusually large hydration enthalpy, which overcompensates for its lower ionisation energy and makes $E^\ominus_{Li^+/Li}$ the most negative among the alkali metals — Li is thus the strongest reducing agent in aqueous solution, contrary to what its higher ionisation energy alone might suggest. Sodium, further down this thermodynamic cycle, ends up with the least negative potential among the common alkali metals.

   Step-by-Step Solution

   1. Typical standard reduction potentials (aqueous, 298 K): $\mathrm{Li^+/Li}\approx -3.05$ V, $\mathrm{Na^+/Na}\approx -2.71$ V, $\mathrm{K^+/K}\approx -2.93$ V, $\mathrm{Rb^+/Rb}\approx -2.93$ V, $\mathrm{Cs^+/Cs}\approx -2.92$ V.
   2. "Highest with negative sign" = most negative value = Li ($x=\text{Li}$).
   3. "Lowest with negative sign" = least negative (closest to zero) = Na ($y=\text{Na}$).
   4. Flame test colours (characteristic emission on excitation): Li → crimson red; Na → yellow (the well-known intense sodium D-line emission).
   5. So $x,y$ colours are crimson red, yellow — matching option (D).

   Common Mistakes

   • Assuming ionisation energy alone determines $E^\ominus$ trend and picking K (highest IE trend anomaly) instead of correctly identifying Li as most negative due to hydration enthalpy.
   • Swapping Li's and Na's flame colours (Li = crimson red, not Na; Na = yellow, not Li).

   ✓Final answer

   The correct option is (D) — Crimson red, Yellow.

   ANSWER: D

9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Match the following List-I (Transition metal, M): A) Ni B) Mn C) Fe D) Cr List-II ($E^{\ominus}_{M^{2+}/M}$): I) $-1.18$ II) $-0.91$ III) $-0.25$ IV) $-0.44$ The correct answer is (A) A-III, B-II, C-IV, D-I (B) A-III, B-IV, C-I, D-II (C) A-III, B-I, C-IV, D-II (D) A-I, B-IV, C-II, D-III

   ›Reveal solutionSolution

   This tests recall of standard reduction potentials $E^{\ominus}_{M^{2+}/M}$ for first-row transition metals. The answer is A-III, B-I, C-IV, D-II.

   Concept and Intuition

   The standard electrode potentials of $M^{2+}/M$ couples for 3d transition metals are largely governed by a combination of enthalpy of atomisation, ionisation enthalpy, and hydration enthalpy, and do not follow a simple monotonic trend across the series. These values are typically memorised from the standard NCERT table.

   Step-by-Step Solution

   1. Recall standard $E^{\ominus}_{M^{2+}/M}$ values (in volts): $\mathrm{Cr}^{2+}/\mathrm{Cr}=-0.91$, $\mathrm{Mn}^{2+}/\mathrm{Mn}=-1.18$, $\mathrm{Fe}^{2+}/\mathrm{Fe}=-0.44$, $\mathrm{Ni}^{2+}/\mathrm{Ni}=-0.25$.
   2. Match List-I to List-II: A) Ni $=-0.25=$ III; B) Mn $=-1.18=$ I; C) Fe $=-0.44=$ IV; D) Cr $=-0.91=$ II.
   3. This gives A-III, B-I, C-IV, D-II, which is option (C).

   Common Mistakes

   • Mixing up Mn (most negative, $-1.18$) with Cr (next most negative, $-0.91$) since both are strongly negative.
   • Confusing Fe and Ni values, which are numerically closer to zero.

   ✓Final answer

   The correct option is (C) — A-III, B-I, C-IV, D-II.

   ANSWER: C

10. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
    Q.The cell reaction of a cell is given below $\mathrm{2Cu^+ \rightarrow Cu + Cu^{2+}}$ What is $E^0_{cell}$ (in V)? (Given: $E^0_{\mathrm{Cu^{2+}/Cu^+}} = x$ V; $E^0_{\mathrm{Cu^+/Cu}} = y$ V) (A) $x - y$ (B) $y - x$ (C) $x + y$ (D) $-x - y$

    ›Reveal solutionSolution

    The key is to treat the given cell reaction as the sum of two half‑reactions, each with its own standard potential, and then combine them correctly — the result is $E^0_{\text{cell}} = y - x$, so the correct option is (B).

    ---

    Why this approach works

    Standard electrode potentials are intensive properties: they do not depend on how many electrons are transferred. When we combine half‑reactions to get a full cell reaction, we never multiply the potentials by coefficients — we simply add them (with the appropriate sign for the direction we use). The trick here is that the reaction $\mathrm{2Cu^+ \rightarrow Cu + Cu^{2+}}$ is a disproportionation: one $\mathrm{Cu^+}$ is reduced to $\mathrm{Cu}$ and the other is oxidised to $\mathrm{Cu^{2+}}$. So we need to identify which half‑reaction runs as reduction and which as oxidation, then combine their potentials.

    ---

    Step‑by‑step reasoning

    1. Identify the two half‑reactions hidden in the overall reaction The overall reaction is:

    $2\mathrm{Cu^+} \rightarrow \mathrm{Cu} + \mathrm{Cu^{2+}}$

    This can be split into:

    • Reduction half: $\mathrm{Cu^+} + e^- \rightarrow \mathrm{Cu}$ Its standard potential is given as $E^0_{\mathrm{Cu^+/Cu}} = y$ V.
    • Oxidation half: $\mathrm{Cu^+} \rightarrow \mathrm{Cu^{2+}} + e^-$ This is the reverse of $\mathrm{Cu^{2+}} + e^- \rightarrow \mathrm{Cu^+}$, whose potential is $x$ V. For the reverse reaction, the potential changes sign: $E^0_{\text{ox}} = -x$ V.

    2. Combine the half‑reaction potentials The standard cell potential is the sum of the reduction potential of the cathode and the oxidation potential of the anode:

    $E^0_{\text{cell}} = E^0_{\text{red}} + E^0_{\text{ox}}$

    Here:

    • Cathode (reduction): $\mathrm{Cu^+} + e^- \rightarrow \mathrm{Cu}$, $E^0_{\text{red}} = y$
    • Anode (oxidation): $\mathrm{Cu^+} \rightarrow \mathrm{Cu^{2+}} + e^-$, $E^0_{\text{ox}} = -x$

    Therefore:

    $E^0_{\text{cell}} = y + (-x) = y - x$

    3. Check the sign convention A positive $E^0_{\text{cell}}$ would mean the reaction is spontaneous as written. Since $y$ is typically more positive than $x$ for copper (because $\mathrm{Cu^+}$ is unstable in water and tends to disproportionate), $y - x$ is indeed positive — consistent with the known spontaneous disproportionation of $\mathrm{Cu^+}$.

    Watch out

    A common mistake is to treat the reaction as two identical $\mathrm{Cu^+}$ ions and try to average or double the potentials. Remember: potentials are not multiplied by stoichiometric coefficients. The number of electrons cancels out when you add the half‑reactions, but the potentials themselves are simply added with the correct sign.

    Tip

    If you ever forget the sign rule: write the two half‑reactions as reductions with their given potentials, then reverse the one that must be oxidation and flip its sign. Then add. That always works.

    ---

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.The standard reduction potentials of $2H^+/H_2$, $Cu^{2+}/Cu$, $Zn^{2+}/Zn$ and $NO_3^-, H^+/NO$ are 0.0, +0.34, -0.76 and 0.97 V respectively. Observe the following reactions I. $\mathrm{Zn + HCl \rightarrow}$ II. $\mathrm{Cu + HCl \rightarrow}$ III. $\mathrm{Cu + HNO_3 \rightarrow}$ Which reactions does not liberate $H_2(g)$? (A) II, III only (B) I, II only (C) I, III only (D) I, II, III

    ›Reveal solutionSolution

    H2 is liberated only when the metal's reduction potential is below 0 V (more easily oxidized than H2); Zn (−0.76 V) liberates H2 with HCl, but Cu (+0.34 V) cannot liberate H2 with either HCl or HNO3.

    Concept and Intuition

    A metal displaces H2 from a dilute acid only if it is a stronger reducing agent than hydrogen, i.e., its standard reduction potential is more negative than that of 2H⁺/H2 (0.0 V). Cu, with a positive reduction potential, cannot reduce H⁺ to H2 under any of these acids — with HNO3 specifically, the acid itself acts as an oxidizer (reduced to NO) rather than being a source of H2.

    Step-by-Step Solution

    1. Zn²⁺/Zn = −0.76 V is more negative than 2H⁺/H2 = 0.0 V, so Zn CAN reduce H⁺ to H2 — reaction I liberates H2.
    2. Cu²⁺/Cu = +0.34 V is more positive than 0.0 V, so Cu CANNOT reduce H⁺ to H2 with HCl — reaction II does NOT liberate H2.
    3. With HNO3, the oxidizing species is NO3⁻/H⁺ → NO (E° = 0.97 V), which is even more strongly oxidizing than H⁺; Cu reacts with HNO3 by reducing NO3⁻ to NO (or NO2), not by liberating H2 — reaction III does NOT liberate H2.
    4. So the reactions that do NOT liberate H2 are II and III, matching option (A).

    Common Mistakes

    • Assuming any metal + acid automatically gives H2 — this fails whenever the metal's reduction potential is positive (less reactive than H2).
    • Forgetting that HNO3 is a special oxidizing acid: it reacts with metals via redox to give nitrogen oxides, not H2, even with reactive metals in most cases.

    ✓Final answer

    The correct option is (A) — reactions II (Cu+HCl) and III (Cu+HNO3) do not liberate H2.

    ANSWER: A

12. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.Consider the following standard electrode potentials ($E^0$ in volts) in aqueous solution:

    Element	$M^{3+}/M$	$M^+/M$ ;
    Al	$-1.66$	$+0.55$ ;
    Tl	$+1.26$	$-0.34$.
    Based on this data, which of the following statements is correct?
    (A) $\mathrm{Tl^{3+}}$ is more stable than $\mathrm{Al^{3+}}$
    (B) $\mathrm{Tl^{+}}$ is more stable than $\mathrm{Al^{3+}}$
    (C) $\mathrm{Al^{+}}$ is more stable than $\mathrm{Al^{3+}}$
    (D) $\mathrm{Tl^{+}}$ is more stable than $\mathrm{Al^{+}}$

    ›Reveal solutionSolution

    Combining the two given electrode potentials for each metal (via $\Delta G = -nFE^0$) shows Al's +1 state is unstable (disproportionates to Al$^{3+}$) while Tl's +1 state is the stable one — the classic inert-pair-effect result, so $Tl^+$ is more stable than $Al^+$.

    Concept and Intuition

    Whether an intermediate oxidation state ($M^+$) is stable or disproportionates depends on the relative reducing/oxidizing strength of the two half-reactions that flank it. We can combine $E^0(M^{3+}/M)$ and $E^0(M^+/M)$ using free energies (which are additive, unlike potentials) to get $E^0(M^{3+}/M^+)$, and that tells us directly whether $M^+$ wants to disproportionate into $M$ and $M^{3+}$.

    Step-by-Step Solution

    1. For Al: $E^0(Al^{3+}/Al) = -1.66$ V ($n=3$), $E^0(Al^+/Al) = +0.55$ V ($n=1$).
    2. $\Delta G^0(Al^{3+}\to Al) = -3F(-1.66) = +4.98F$; $\Delta G^0(Al^+\to Al) = -1F(0.55) = -0.55F$.
    3. $\Delta G^0(Al^{3+}\to Al^+) = \Delta G^0(Al^{3+}\to Al) - \Delta G^0(Al^+\to Al) = 4.98F - (-0.55F) = 5.53F$ (for a 2-electron step), so $E^0(Al^{3+}/Al^+) = -5.53F/2F \approx -2.77$ V — very negative, meaning $Al^{3+}$ strongly resists being reduced to $Al^+$; equivalently, $Al^+$ is a strong enough reducing agent to be oxidized to $Al^{3+}$ spontaneously (disproportionates, i.e. $Al^+$ is unstable).
    4. For Tl: $E^0(Tl^{3+}/Tl) = +1.26$ V, $E^0(Tl^+/Tl) = -0.34$ V.
    5. $\Delta G^0(Tl^{3+}\to Tl) = -3F(1.26) = -3.78F$; $\Delta G^0(Tl^+\to Tl) = -1F(-0.34) = +0.34F$.
    6. $\Delta G^0(Tl^{3+}\to Tl^+) = -3.78F - 0.34F = -4.12F$ (2-electron step), so $E^0(Tl^{3+}/Tl^+) = +4.12F/2F \approx +2.06$ V — strongly positive, meaning $Tl^{3+}$ is readily reduced to $Tl^+$: $Tl^+$ is the thermodynamically favoured, stable state and does not disproportionate.
    7. Conclusion: $Al^+$ is unstable (it disproportionates to $Al + Al^{3+}$), while $Tl^+$ is stable. Comparing the two +1 ions directly, $Tl^+$ is therefore more stable than $Al^+$ — the textbook inert-pair-effect result (heavier p-block elements like Tl prefer the lower oxidation state).

    Common Mistakes

    • Trying to compare $Tl^+$ directly against $Al^{3+}$ (different oxidation states of different elements) without a valid thermodynamic link — the data only lets us rigorously compare same-oxidation-state ions (i.e. $M^+$ vs $M^+$, or the disproportionation tendency within each element).
    • Forgetting that potentials are NOT additive — only free energies ($nFE^0$) can be combined via Hess's law.

    ✓Final answer

    The correct option is (D) — $Tl^{+}$ is more stable than $Al^{+}$.

    ANSWER: D