E^°cell for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or m

NCERT Exemplar · Q44

E^\circ_{cell}

(a)

H^+(aq) + e^- \rightarrow \frac{1}{2}H_2(g)

;

E^\circ_{cell} = 0.00\ V

(b)

2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-

;

E^\circ_{cell} = 1.23\ V

(c)

2SO_4^{2-}(aq) \rightarrow S_2O_8^{2-}(aq) + 2e^-

;

E^\circ_{cell} = 1.96\ V

(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.

(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.

(iii) In dilute sulphuric acid solution, water will be oxidised at anode.

(iv) In dilute sulphuric acid solution,

SO_4^{2-}

ion will be oxidised to tetrathionate ion at anode.

Andhra Pradesh BieapMCQ· 1mImportance★★★★★

76% · 87/115 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

The key idea is that in electrolysis, the species with the lower reduction potential gets reduced at the cathode, and the species with the lower oxidation potential (i.e., the one that is hardest to oxidise) gets oxidised at the anode. For dilute $H_2SO_4$ , water oxidises at the anode ( $E^\circ = 1.23\ V$ ) before $SO_4^{2-}$ ( $E^\circ = 1.96\ V$ ), and $H^+$ reduces at the cathode. For concentrated $H_2SO_4$ , the effective concentration changes the competition — water oxidation becomes harder, so $SO_4^{2-}$ oxidation can occur. The correct options are (i) and (iii).

This is a classic electrolysis problem from electrochemistry. The given half-cell reactions are standard reduction potentials — but note that reaction (b) and (c) are written as oxidations in the problem statement. That’s a deliberate twist. Let’s first convert everything to a consistent language.

The core principle: In an electrolytic cell, the cathode is where reduction happens (gain of electrons), and the anode is where oxidation happens (loss of electrons). The cell is driven by an external voltage, so the reaction that occurs is not spontaneous — we force it. Which reaction actually takes place at each electrode depends on the competition among all species present.

For reduction at the cathode: the species with the higher (more positive) reduction potential gets reduced first — because it is easier to reduce.

For oxidation at the anode: the species with the lower (less positive) reduction potential (i.e., the one that is easiest to oxidise) gets oxidised first. Equivalently, look at the oxidation potentials (reverse of reduction potentials): the species with the higher oxidation potential gets oxidised first.

Let’s rewrite the given data as standard reduction potentials (all in one direction):

$2H^+ + 2e^- \rightarrow H_2$ ; $E^\circ = 0.00\ V$
$O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ ; $E^\circ = +1.23\ V$ (reverse of given)
$S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-}$ ; $E^\circ = +1.96\ V$ (reverse of given)

Now, in an aqueous solution of sulphuric acid ( $H_2SO_4$ ), the species present are: $H^+$ , $SO_4^{2-}$ , $H_2O$ , and also $OH^-$ (but in acidic solution, $OH^-$ concentration is negligible). At the cathode, possible reductions are:

$2H^+ + 2e^- \rightarrow H_2$ ( $E^\circ = 0.00\ V$ )
$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ( $E^\circ = -0.83\ V$ in neutral, but in acid it’s even less favourable)

Clearly, $H^+$ reduction has a much higher reduction potential (0.00 V) than water reduction (−0.83 V). So hydrogen ions are reduced at the cathode in both dilute and concentrated acid. That makes option (i) correct.

At the anode, possible oxidations are:

$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ (reverse of reaction 2); $E^\circ_{ox} = -1.23\ V$ (since reduction potential is +1.23 V, oxidation potential is −1.23 V)
$2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$ (reverse of reaction 3); $E^\circ_{ox} = -1.96\ V$

The more positive the oxidation potential, the easier the oxidation. Here, −1.23 V is greater than −1.96 V, so water oxidation is easier than sulphate oxidation. Therefore, in dilute sulphuric acid, water gets oxidised at the anode, producing oxygen gas. That makes option (iii) correct and option (iv) incorrect. …

Start your 14-day free trial to unlock the full solution →

Method: Electrode Potential Comparison for Electrolysis

This method uses standard reduction potentials to predict which species gets oxidised (at anode) and which gets reduced (at cathode) during electrolysis.

Key rule:

Cathode (reduction): The species with the higher (more positive) reduction potential gets reduced.
Anode (oxidation): The species with the lower (less positive) reduction potential gets oxidised (reverse the sign for oxidation potential).

Step-by-step solution

Step 1: Write all half-reactions as reductions with their $E^\circ$ values

Reduction half-reaction	$E^\circ$ (V)
$2H^+ + 2e^- \rightarrow H_2$	$0.00$
$O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$	$+1.23$
$S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-}$	$+1.96$

Step 2: Identify possible reactions at each electrode in dilute $H_2SO_4$

At cathode (reduction):

$H^+$ reduction: $E^\circ = 0.00\ V$
$H_2O$ reduction: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ; $E^\circ = -0.83\ V$ (not given, but known)
Higher $E^\circ$ wins: $H^+$ reduction ( $0.00\ V$ ) > water reduction ( $-0.83\ V$ )
✓ Hydrogen is reduced at cathode → Option (i) is correct.

At anode (oxidation):

Reverse the given reduction potentials to get oxidation potentials:

$H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ; oxidation potential = $-1.23\ V$
$2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$ ; oxidation potential = $-1.96\ V$
Higher (less negative) oxidation potential wins: $-1.23\ V > -1.96\ V$
✓ Water gets oxidised at anode → Option (iii) is correct. …

Start your 14-day free trial to unlock the full solution →

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Reduction Potential with Oxidation Potential

The error: Students treat the given $E^\circ_{cell}$ values as if they are all reduction potentials. Reaction (b) and (c) are written as oxidation half-reactions, but their $E^\circ$ values are still given as positive numbers. This leads to wrong comparisons.

How to avoid:

Always check the direction of the arrow. If electrons are on the right side, it's an oxidation half-reaction.
For oxidation, the actual potential is the negative of the given value when comparing with reduction potentials.
Correct approach: Convert all to reduction potentials:
- (a) $H^+ + e^- \rightarrow \frac12 H_2$ ; $E^\circ_{red} = 0.00\ V$
- (b) $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$ ; $E^\circ_{red} = +1.23\ V$
- (c) $S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-}$ ; $E^\circ_{red} = +1.96\ V$

Mistake 2: Forgetting the Effect of Concentration on Electrode Potential

The error: Students assume the same reaction occurs at both dilute and concentrated $H_2SO_4$ , ignoring that concentration changes the actual potential via the Nernst equation.

How to avoid:

Remember: Higher concentration of $H^+$ makes $H^+$ reduction easier (more positive potential).
In dilute $H_2SO_4$ , $[H^+]$ is low → $H^+$ reduction potential is less than 0.00 V.
In concentrated $H_2SO_4$ , $[H^+]$ is high → $H^+$ reduction potential is greater than 0.00 V.
Key insight: At the anode, the species with the lowest oxidation potential (most negative or least positive) gets oxidised first.

Mistake 3: Misidentifying Which Species Gets Oxidised at the Anode

The error: Students think $SO_4^{2-}$ oxidation (option D) happens in dilute acid because its $E^\circ$ is high, without comparing with water oxidation.

How to avoid:

At the anode, oxidation occurs. Compare oxidation potentials (reverse of reduction potentials):
- Water oxidation: $E^\circ_{ox} = -1.23\ V$
- $SO_4^{2-}$ oxidation: $E^\circ_{ox} = -1.96\ V$
More positive oxidation potential means easier oxidation.
$-1.23 > -1.96$ , so water oxidises more easily than $SO_4^{2-}$ .
Correct conclusion: In dilute acid, water oxidation occurs at anode → option (iii) is correct, not (iv).

Mistake 4: Assuming Option (ii) Must Be Correct Just Because SO4^2- Oxidation Becomes Possible in Concentrated Acid

The error: Students reason that since $SO_4^{2-}$ oxidation becomes competitive in concentrated acid, option (ii) — which claims water is oxidised in concentrated acid — must be the correct description of what happens there.

Why it's wrong:

In concentrated $H_2SO_4$ , the activity of free water is drastically reduced while $[SO_4^{2-}]$ is very high. This shifts the competition away from water and toward sulphate — it is $SO_4^{2-}$ that gets oxidised to $S_2O_8^{2-}$ (peroxodisulphate) at the anode in concentrated acid, not water. Option (ii) states the opposite of this (that water is oxidised in concentrated acid), so option (ii) is actually incorrect, not correct.

How to avoid:

For each scenario (dilute vs concentrated), determine both half-reactions and check exactly what species each option names:
- Dilute $H_2SO_4$ :
  - Cathode: $H^+$ reduction (since $E^\circ_{red} \approx 0$ and no better option) …

Start your 14-day free trial to unlock the full solution →

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): $Z\ (0.80) > Y\ (0.34) > X\ (-1.66)$ . …
Start your 14-day free trial to see the full solution →
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 $n=6$ . …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 ( $y=\text{Na}$ ). …
Start your 14-day free trial to see the full solution →
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): $\mathrm{Cr}^{2+}/\mathrm{Cr}=-0.91$ , $\mathrm{Mn}^{2+}/\mathrm{Mn}=-1.18$ , $\mathrm{Fe}^{2+}/\mathrm{Fe}=-0.44$ , $\mathrm{Ni}^{2+}/\mathrm{Ni}=-0.25$ . …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →

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. …

Start your 14-day free trial to see the full solution →

←Previous questionQ43 — The positive value of the standard electrode potential of Cu²⁺/Cu indi…→Next questionQ45 — E^°Cell = 1.1V for Daniel cell. Which of the following expressions are…

Q.Ecell∘E^\circ_{cell}Ecell∘​ for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or more than two options may be correct.)

🎓Unlock everything free for 14 days

Q. $E^\circ_{cell}$ for some half cell reactions are given below. On the basis of these mark the correct answer. (Two or more than two options may be correct.)

Unlock everything free for 14 days