Use the data given in Q.8 ($E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$) and find out the most stable ion in its reduced form. (i) $Cl^-$ (ii) $Cr^{3+}$ (iii) $Cr$ (iv) $Mn^{2+}$

Use the data given in Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1.36V; E^°MnO_4^-/Mn²⁺ = 1.51V; E^°Cr³⁺/C

NCERT Exemplar · Q11

E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V

(i)

Cl^-

(ii)

Cr^{3+}

(iii)

Cr

(iv)

Mn^{2+}

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Concept understanding — Standard Electrode Potentials

Standard Electrode Potentials: A Number for "How Badly It Wants Electrons"

Dip a zinc rod into a zinc-salt solution and a tiny tug-of-war begins at the surface:

metal atoms tend to dissolve as ions (leaving electrons behind on the rod) while ions

from the solution tend to deposit as metal (consuming electrons). The rod ends up with

a characteristic electrical potential relative to the solution — the electrode potential. It is a direct measure of the tendency of that redox couple to gain or

lose electrons.

The Core Idea

Different couples pull electrons with very different strengths. Copper's ion grabs

them readily; zinc's barely wants them. Put a number on each couple and you can

predict, before mixing anything, who will oxidise whom.

Two conventions make the numbers comparable:

Standard conditions. Every species at unit concentration (1 M), any gas at 1 atm, temperature 298 K. The potential measured then is the standard electrode potential, written $E^\ominus$ .
A common zero. Potentials can only be measured as differences, so one electrode is defined as the reference: the standard hydrogen electrode (SHE), $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ , is fixed at exactly 0.00 V. Every $E^\ominus$ is the voltage of a couple measured against it.

By convention the values are tabulated for the reduction direction:

\text{Oxidised form} + ne^- \rightarrow \text{Reduced form} \qquad E^\ominus \text{ (in volts, at 298 K)}

Reading the Table

The standard-potential table (Table 7.1 in the Class 11 chapter) runs from

$\text{F}_2/\text{F}^-$ at $+2.87$ V down to $\text{Li}^+/\text{Li}$ at $-3.05$ V.

Two rules unlock it:

More positive $E^\ominus$ → stronger oxidising agent (the oxidised form is hungrier for electrons). F₂ tops the table; that is why fluorine oxidises almost everything.
More negative $E^\ominus$ → stronger reducing agent (the reduced form gives electrons up most easily). Li, K, Ca, Na at the bottom are the great electron donors. A negative $E^\ominus$ means the couple is a stronger reducing agent than the H⁺/H₂ couple; a positive one, weaker.

Predicting Whether a Reaction Goes

For any proposed redox reaction, the species being reduced acts as the cathode couple

and the species being oxidised as the anode couple:

E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}

Worked feel: can Fe³⁺ oxidise iodide? $E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$ V is above $E^\ominus(\text{I}_2/\text{I}^-) = +0.54$ V, so

$E^\ominus_{\text{cell}} = +0.23$ V — yes. Can silver metal reduce Fe³⁺?

$0.77 - 0.80 = -0.03$ V — no.

This is also the logic of the activity series: a metal displaces, from solution, …

Why this formula?

Galvanic Corrosion: Why the Key Formulas Hold

Galvanic corrosion occurs when two dissimilar metals are electrically connected in the presence of an electrolyte. The key formula that governs this is the mixed potential theory, which leads to the galvanic current and corrosion rate expressions.

Let's build the reasoning step-by-step.

1. The Core Idea: Two Electrodes, One Circuit

When metals M₁ (more active, e.g., zinc) and M₂ (more noble, e.g., copper) are connected:

M₁ acts as the anode — it oxidizes (corrodes):

\text{M}_1 \rightarrow \text{M}_1^{n+} + n e^-

M₂ acts as the cathode — it reduces something (e.g., oxygen or H⁺):

\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^- \quad \text{(in neutral/alkaline)}

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad \text{(in acidic)}

The two metals are electrically connected (via a wire or direct contact), and the electrolyte completes the circuit. Electrons flow from M₁ to M₂.

2. The Mixed Potential: Why It Exists

Each metal, when alone in the electrolyte, has its own open-circuit potential (OCP) — the equilibrium potential for its half-reaction. For M₁, it's $E_{\text{corr},1}$ ; for M₂, it's $E_{\text{corr},2}$ .

When connected, the system cannot stay at two different potentials. The entire metal couple must reach a single potential — the mixed potential $E_{\text{mix}}$ .

$E_{\text{mix}}$ lies between $E_{\text{corr},1}$ and $E_{\text{corr},2}$ .
At $E_{\text{mix}}$ , the total anodic current from M₁ equals the total cathodic current from M₂ (charge conservation):

I_{\text{anode}} = I_{\text{cathode}}

This is the fundamental equation of galvanic corrosion.

3. Deriving the Galvanic Current

Assume each electrode follows Butler-Volmer kinetics (for activation-controlled reactions). For the anode (M₁), the anodic current density $i_a$ at potential $E$ is:

i_a = i_{0,1} \exp\left( \frac{\alpha_a F (E - E_{0,1})}{RT} \right)

For the cathode (M₂), the cathodic current density $i_c$ is:

i_c = i_{0,2} \exp\left( -\frac{\alpha_c F (E - E_{0,2})}{RT} \right)

Where:

$i_{0,1}, i_{0,2}$ = exchange current densities
$\alpha_a, \alpha_c$ = transfer coefficients (typically ~0.5)
$F$ = Faraday constant
$R$ = gas constant
$T$ = temperature
$E_{0,1}, E_{0,2}$ = standard reduction potentials

At the mixed potential $E_{\text{mix}}$ :

I_{\text{galvanic}} = A_1 \cdot i_a(E_{\text{mix}}) = A_2 \cdot i_c(E_{\text{mix}})

Where $A_1$ and $A_2$ are the surface areas of the anode and cathode.

Why this holds: The net current from the anode must exactly balance the net current consumed at the cathode — otherwise, charge would accumulate, which is impossible in a steady-state circuit.

4. The Corrosion Rate Formula

The corrosion rate (mass loss per time) of the anode is given by Faraday's law:

\text{Corrosion rate} = \frac{I_{\text{galvanic}} \cdot M}{n \cdot F \cdot \rho}

Where:

$M$ = molar mass of the anode metal
$n$ = number of electrons transferred per atom
$\rho$ = density of the metal
$F$ = Faraday constant (96,485 C/mol)

Why this holds: Each mole of metal oxidized releases $n$ moles of electrons. The total charge passed $Q = I_{\text{galvanic}} \cdot t$ corresponds to moles of metal lost:

\text{moles lost} = \frac{Q}{nF} = \frac{I_{\text{galvanic}} \cdot t}{nF}

Multiply by $M/\rho$ to get volume or thickness loss.

5. The Area Effect: Why It Matters

From the mixed potential equation:

A_1 \cdot i_a(E_{\text{mix}}) = A_2 \cdot i_c(E_{\text{mix}})

If the cathode area $A_2$ is large relative to the anode area $A_1$ , then $i_a(E_{\text{mix}})$ must be large to balance the current. This means:

Small anode + large cathode → severe galvanic corrosion (high current density on the anode). …

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The most stable reduced species is the one with the least negative (or most positive) reduction potential for its half‑reaction, because that indicates the strongest tendency to stay reduced. Comparing the given potentials shows that $Mn^{2+}$ is the most stable reduced ion.

The question asks: among $Cl^-$ , $Cr^{3+}$ , $Cr$ (metal), and $Mn^{2+}$ , which is the most stable in its reduced form?

“Stable in reduced form” means the species has little tendency to get oxidised back — it prefers to stay as it is. In electrochemistry, that tendency is measured by the standard reduction potential $E^\circ$ of the corresponding half‑reaction.

E^\circ_{\text{red}} \text{ (more positive)} \implies \text{stronger oxidising agent (oxidised form)}

But careful: we want the reduced form to be stable. That means the reduced form should have a low tendency to get oxidised. The tendency to get oxidised is the reverse of the reduction potential. So we look at the reduction potential of the oxidised form — if that potential is very positive, the oxidised form is a strong oxidiser and the reduced form is weak (unstable as a reductant). Conversely, if the reduction potential is very negative, the reduced form is a strong reductant (easily oxidised, hence unstable in its reduced state).

Therefore, the most stable reduced form corresponds to the most positive reduction potential of the oxidised species. Let’s list the given data properly.

For $Cl^-$ : The half‑reaction is

Cl_2 + 2e^- \rightarrow 2Cl^- \quad E^\circ = +1.36\ \text{V}

The reduced form is $Cl^-$ . The potential is quite positive, so $Cl_2$ is a strong oxidiser and $Cl^-$ is a weak reductant — fairly stable.

For $Cr^{3+}$ : There are two relevant half‑reactions:

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad E^\circ = +1.33\ \text{V}

and

Cr^{3+} + 3e^- \rightarrow Cr \quad E^\circ = -0.74\ \text{V}

The reduced form $Cr^{3+}$ appears on the right of the first reaction and on the left of the second. To judge its stability, we need the potential for the half‑reaction that produces $Cr^{3+}$ from a higher oxidation state — that’s the $+1.33\ \text{V}$ one. But also, $Cr^{3+}$ can be further reduced to $Cr$ metal at $-0.74\ \text{V}$ , which tells us $Cr^{3+}$ is a weak oxidiser (hard to reduce). However, the question asks about stability of the reduced form — here $Cr^{3+}$ is the reduced form relative to dichromate. Its $E^\circ = +1.33\ \text{V}$ is slightly less positive than $Cl_2/Cl^-$ .

For $Cr$ (metal): The half‑reaction is $Cr^{3+} + 3e^- \rightarrow Cr \quad E^\circ = -0.74\ \text{V}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing "most stable reduced form" with "strongest reducing agent"

The error: Students often think the species with the most negative $E^\circ$ is the most stable reduced form. They pick Cr ( $-0.74\ V$ ) because it's the most negative.

Why it's wrong:

A more negative $E^\circ$ means the reduced form is a stronger reducing agent (easily gets oxidised) — meaning it is less stable in its reduced state.
A more positive $E^\circ$ means the reduced form is harder to oxidise — hence more stable.

How to avoid:

Remember: Higher $E^\circ$ → Reduced form is more stable (it "wants" to stay reduced).
Think of it like a tug-of-war: a high positive $E^\circ$ means the oxidised form pulls electrons strongly, so the reduced form is content and stable.

Mistake 2: Comparing potentials of half-cells with different numbers of electrons

The error: Students directly compare $E^\circ$ values without noticing that some half-reactions involve different $n$ (number of electrons). For example:

$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ ( $n=6$ )
$Cl_2 + 2e^- \rightarrow 2Cl^-$ ( $n=2$ )
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ ( $n=5$ )
$Cr^{3+} + 3e^- \rightarrow Cr$ ( $n=3$ )

Why it's wrong:

$E^\circ$ is an intensive property — it does not depend on $n$ . So comparing $E^\circ$ directly is actually correct here.
However, some students mistakenly try to "scale" $E^\circ$ by multiplying with $n$ (like $\Delta G^\circ = -nFE^\circ$ ). That would be wrong for stability comparison.

How to avoid:

For stability of the reduced form, compare $E^\circ$ values directly — not $\Delta G^\circ$ or $n \times E^\circ$ .
Only use $\Delta G^\circ$ if the question asks for spontaneity or thermodynamic favourability of a specific reaction.

Mistake 3: Forgetting to check which species is the "reduced form"

The error: Students look at the given $E^\circ$ values and pick the oxidised form (like $MnO_4^-$ or $Cr_2O_7^{2-}$ ) instead of the reduced form.

Why it's wrong:

The question asks for the most stable ion in its reduced form.
For each half-cell, the reduced form is:
- $Cl^-$ (from $Cl_2/Cl^-$ )
- $Cr^{3+}$ (from $Cr_2O_7^{2-}/Cr^{3+}$ )
- $Mn^{2+}$ (from $MnO_4^-/Mn^{2+}$ )
- $Cr$ (from $Cr^{3+}/Cr$ )

How to avoid:

Always write the half-reaction in the reduction direction:

\text{Oxidised form} + ne^- \rightarrow \text{Reduced form}

The species on the right is the reduced form — that's what you compare.

Mistake 4: Misinterpreting the sign of $E^\circ$ for $Cr^{3+}/Cr$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 28 on this concept.

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$X, Y$
›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

Identify the labels: $X = Al^{3+}/Al$ ( $E°=-1.66\ V$ ), $Y=Cu^{2+}/Cu$ ( $E°=0.34\ V$ ), $Z=Ag^+/Ag$ ( $E°=0.80\ V$ ).

Rank by $E°$ value (most positive = strongest oxidising power): $Z\ (0.80) > Y\ (0.34) > X\ (-1.66)$ . …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$\Delta G^\circ$
›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

Identify cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu$ , $E^\circ = +0.34\ V$ .

Identify anode (oxidation, but use its reduction potential in the formula): $Al^{3+} + 3e^- \rightarrow Al$ , $E^\circ = -1.66\ V$ .

$E^\circ_{cell} = 0.34 - (-1.66) = 2.00\ V$ .

Balance the overall reaction: $2Al \rightarrow 2Al^{3+}+6e^-$ and $3Cu^{2+}+6e^- \rightarrow 3Cu$ , so total electrons transferred $n=6$ . …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$2F_2 + 2H_2O \rightarrow 4HF + O_2$
›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

(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.

(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. …
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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

Set I (Be, Mg): Be does not react with water → invalid.

Set II (Li, Mg): Li reacts with water, Mg reacts with hot water/steam → valid.

Set III (Na, K): both react vigorously with water → valid.

Set IV (K, Ca): both react with water → valid.

Valid sets = II, III, IV → option (B).

Common Mistakes …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$E^{\ominus}(\mathrm{M^{3+}/M^{2+}})$
›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

Recall/estimate the sign of $E^\ominus(M^{3+}/M^{2+})$ for each metal in the options.

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.

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. …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$E^\circ_{\mathrm{Fe^{2+}/Fe}} = -0.441\ \mathrm{V}$
›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

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.

Apply $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.771 - (-0.441) = 1.212$ V. …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\mathrm{Cl_2(g) + 2KBr(aq) \longrightarrow 2KCl(g) + Br_2(l)}$
›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

(A) $\mathrm{Cl_2 + 2KBr \to 2KCl + Br_2}$ : Cl₂ is a stronger oxidizer than Br₂, so it displaces bromide — feasible.

(B) $\mathrm{Cl_2 + 2KI \to 2KCl + I_2}$ : Cl₂ is stronger than I₂, displaces iodide — feasible.

(C) $\mathrm{Br_2 + 2KI \to 2KBr + I_2}$ : Br₂ is stronger than I₂, displaces iodide — feasible. …
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AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$E^{\ominus}_{M^+(aq)|M(s)}$
›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

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.

"Highest with negative sign" = most negative value = Li ( $x=\text{Li}$ ).

"Lowest with negative sign" = least negative (closest to zero) = Na ( $y=\text{Na}$ ). …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$E^{\ominus}_{M^{2+}/M}$
›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

Recall standard $E^{\ominus}_{M^{2+}/M}$ values (in volts): $\mathrm{Cr}^{2+}/\mathrm{Cr}=-0.91$ , $\mathrm{Mn}^{2+}/\mathrm{Mn}=-1.18$ , $\mathrm{Fe}^{2+}/\mathrm{Fe}=-0.44$ , $\mathrm{Ni}^{2+}/\mathrm{Ni}=-0.25$ . …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$\mathrm{2Cu^+ \rightarrow Cu + Cu^{2+}}$
›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

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.

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

Check the sign convention …
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AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$2H^+/H_2$
›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

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.

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.

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. …
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AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ

E^0

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 solution

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

For Al: $E^0(Al^{3+}/Al) = -1.66$ V ( $n=3$ ), $E^0(Al^+/Al) = +0.55$ V ( $n=1$ ).
$\Delta G^0(Al^{3+}\to Al) = -3F(-1.66) = +4.98F$ ; $\Delta G^0(Al^+\to Al) = -1F(0.55) = -0.55F$ .
$\Delta G^0(Al^{3+}\to Al^+) = \Delta G^0(Al^{3+}\to Al) - \Delta G^0(Al^+\to Al) = 4.98F - (-0.55F) = 5.53F$ (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).
For Tl: $E^0(Tl^{3+}/Tl) = +1.26$ V, $E^0(Tl^+/Tl) = -0.34$ V.
$\Delta G^0(Tl^{3+}\to Tl) = -3F(1.26) = -3.78F$ ; $\Delta G^0(Tl^+\to Tl) = -1F(-0.34) = +0.34F$ .
$\Delta G^0(Tl^{3+}\to Tl^+) = -3.78F - 0.34F = -4.12F$ (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. …

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