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.

COMEDK 2026Set 2026-A1 markMCQ
$\Delta \mathrm{G}^{\circ}$
›Reveal solutionSolution
The standard Gibbs free energy change is found from the cell potential using $\Delta G^\circ = -nFE^\circ_\text{cell}$ . Here $E^\circ_\text{cell} = +0.360\ \text{V}$ and $n=2$ , giving $\Delta G^\circ \approx -69.5\ \text{kJ}$ , so the correct option is (A).

The key idea is that the standard Gibbs free energy change for a redox reaction is directly related to the standard cell potential by $\Delta G^\circ = -nFE^\circ_\text{cell}$ . A positive cell potential means the reaction is spontaneous (negative $\Delta G^\circ$ ), and a negative cell potential means it is non-spontaneous (positive $\Delta G^\circ$ ). Here, we are given two half-cell reduction potentials; we must combine them correctly to get the cell potential for the reaction as written.

Identify the half-reactions and their standard reduction potentials.

Cadmium: $\mathrm{Cd^{2+} + 2e^- \rightarrow Cd(s)}$ , $E^\circ = -0.403\ \text{V}$

Zinc: $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn(s)}$ , $E^\circ = -0.763\ \text{V}$

Determine which half-reaction is oxidation and which is reduction in the given overall reaction.

The reaction is: $\mathrm{Cd^{2+}(aq) + Zn(s) \rightarrow Zn^{2+}(aq) + Cd(s)}$ .

$\mathrm{Zn(s)}$ loses electrons to become $\mathrm{Zn^{2+}}$ : this is oxidation.

$\mathrm{Cd^{2+}}$ gains electrons to become $\mathrm{Cd(s)}$ : this is reduction. So the cell is: $\mathrm{Zn(s) | Zn^{2+}(aq) || Cd^{2+}(aq) | Cd(s)}$ .

Calculate the standard cell potential $E^\circ_\text{cell}$ .

The standard cell potential is:

$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$
Here, the cathode (reduction) is $\mathrm{Cd^{2+}/Cd}$ and the anode (oxidation) is $\mathrm{Zn^{2+}/Zn}$ .
$E^\circ_\text{cell} = (-0.403\ \text{V}) - (-0.763\ \text{V}) = +0.360\ \text{V}$
Tip

A common mistake is to subtract the wrong way or to add the potentials. Remember: $E^\circ_\text{cell} = E^\circ_\text{reduction at cathode} - E^\circ_\text{reduction at anode}$ . Since both given values are reduction potentials, this formula works directly.

Determine the number of electrons transferred ( $n$ ). Both half-reactions involve 2 electrons: $\mathrm{Cd^{2+} + 2e^- \rightarrow Cd}$ and $\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}$ . So $n = 2$ . …
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COMEDK 2026Set 2026-A1 markMCQ
$\begin{aligned} &\text { Using the data given below, the strongest reducing agent is: }\\ &\begin{array}{ll} \mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7}^{\mathrm{o}}{ }^{2-} / \mathrm{Cr}^{3+}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\mathrm{o}}=1.51 \mathrm{~V} \\ \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\mathrm{O}}=1.36 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\mathrm{O}}=-0.74 \mathrm{~V} \end{array} \end{aligned}$
›Reveal solutionSolution
The strongest reducing agent is the species that is most easily oxidized, which corresponds to the most negative reduction potential. Among the given options, Cr(s) has the most negative reduction potential (−0.74 V), so it is the strongest reducing agent.

The key concept here is the relationship between reduction potential and reducing strength. A reducing agent is a species that donates electrons (gets oxidized). The more easily it donates electrons, the stronger it is as a reducing agent. In electrochemistry, the tendency to be oxidized is the opposite of the tendency to be reduced. So, a very negative reduction potential means the species is very hard to reduce — which means its oxidized form is very stable, and the reduced form (the species itself) is very eager to give up electrons. Thus, the more negative the standard reduction potential, the stronger the reducing agent.

Let’s work through the data step by step.

List the given half-reactions and their standard reduction potentials (E°)

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$ $E^\circ = +1.33\ \mathrm{V}$

$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$ $E^\circ = +1.51\ \mathrm{V}$

$\mathrm{Cl}_2 + 2e^- \rightarrow 2\mathrm{Cl}^-$ $E^\circ = +1.36\ \mathrm{V}$

$\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr}$ $E^\circ = -0.74\ \mathrm{V}$

Identify the species that are candidates as reducing agents

The question asks for the strongest reducing agent among the options:

(A) Cr (B) Mn²⁺ (C) Cr³⁺ (D) Cl⁻

These are the reduced forms of the couples given. For each, we look at the reduction potential of the couple where that species appears on the right (as the product of reduction).

For Cr(s): the couple is $\mathrm{Cr}^{3+}/\mathrm{Cr}$ with $E^\circ = -0.74\ \mathrm{V}$ .

For Mn²⁺: the couple is $\mathrm{MnO}_4^-/\mathrm{Mn}^{2+}$ with $E^\circ = +1.51\ \mathrm{V}$ .

For Cr³⁺: the couple is $\mathrm{Cr}_2\mathrm{O}_7^{2-}/\mathrm{Cr}^{3+}$ with $E^\circ = +1.33\ \mathrm{V}$ .

For Cl⁻: the couple is $\mathrm{Cl}_2/\mathrm{Cl}^-$ with $E^\circ = +1.36\ \mathrm{V}$ .

Compare reducing strength using the reduction potentials

The more negative the reduction potential, the stronger the reducing agent. Here:

Cr: $E^\circ = -0.74\ \mathrm{V}$ (negative → strong reducing agent) …
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COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$
›Reveal solutionSolution
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ . Since $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})$ isn't given directly, derive it from the two iron couples via $\Delta G^\circ = -nFE^\circ$ : it equals $3c-2b$ . Then $E^\circ_{\text{cell}} = a-(3c-2b) = (a+2b-3c)\,\mathrm{V}$ — option (A).

Concept

A galvanic cell's standard potential is $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ (both as reduction potentials). Here we must first build $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})$ from the given couples, and potentials are combined through Gibbs energy, not added directly.

Solution

Half-reactions. Reduction (cathode): $\mathrm{Ag^+ + e^- \to Ag}$ , $E^\circ = a$ . Oxidation (anode): $\mathrm{Fe^{2+} \to Fe^{3+} + e^-}$ , i.e. the $\mathrm{Fe^{3+}/Fe^{2+}}$ couple.

Derive $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})$ . With $\Delta G^\circ = -nFE^\circ$ :

$\mathrm{Fe^{3+}+3e^-\to Fe}:\ \Delta G_1 = -3Fc, \qquad \mathrm{Fe^{2+}+2e^-\to Fe}:\ \Delta G_2 = -2Fb.$
Subtracting gives $\mathrm{Fe^{3+}+e^-\to Fe^{2+}}$ : …
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KCET 2026Set D31 markMCQ
$^{2+}$
›Reveal solutionSolution

$E°$ values cannot be added directly when the electron counts differ; converting each half-reaction to $\Delta G°$ , combining, and converting back gives $E°_{cell} = -2.69$ V, so the disproportionation is non-spontaneous.

Step 1 — Set up the target reaction from the given half-reactions

We are given:
$Mn^{2+} + 2e^- \rightarrow Mn \qquad E°_1 = -1.18\ V \quad (n_1 = 2)$ $Mn^{3+} + e^- \rightarrow Mn^{2+} \qquad E°_2 = +1.51\ V \quad (n_2 = 1)$
We need $E°$ for $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ , obtained by adding the first equation to twice the reverse of the second equation:
$Mn^{2+} + 2e^- \rightarrow Mn$ $2 \times \left(Mn^{2+} \rightarrow Mn^{3+} + e^-\right)$
Adding gives $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$ , with the $2e^-$ cancelling — an overall $n = 2$ electron process.

Step 2 — Convert each step to $\Delta G°$ (never add $E°$ values directly)
$\Delta G°_1 = -n_1FE°_1 = -(2)F(-1.18) = +2.36F$
Reversing the second half-reaction flips the sign of $E°_2$ ; doubling it doubles $\Delta G°$ (since $\Delta G°$ , unlike $E°$ , is extensive):

$\Delta G°_{2,\ \text{reversed} \times 2} = -(2)F(-1.51) = +3.02F$ …
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KCET 2025Set D-41 markMCQ

\ce{Cl2_{(g)} + 2Br^-_{(aq)} \rightarrow 2Cl^-_{(aq)} + Br2_{(l)}}

List-I (Types of redox reactions)	List-II (Examples)
a. Combination reaction	i. $\ce{Cl2_{(g)} + 2Br^-_{(aq)} \rightarrow 2Cl^-_{(aq)} + Br2_{(l)}}$
b. Decomposition reaction	ii. $\ce{2H2O2_{(aq)} \rightarrow 2H2O_{(l)} + O2_{(g)}}$
c. Displacement reaction	iii. $\ce{CH4_{(g)} + 2O2_{(g)} \rightarrow CO2_{(g)} + 2H2O_{(l)}}$
d. Disproportionation reaction	iv. $\ce{2H2O_{(l)} \rightarrow 2H2_{(g)} + O2_{(g)}}$

›Reveal solution

Assign oxidation numbers in each of the four example reactions and name the redox type from what the numbers do — one substance formed (combination), one substance split (decomposition), one element pushing another out (displacement), or one element going both up and down (disproportionation).

Step 1 — Fix the four definitions

Combination redox: two substances combine to give one product, with an oxidation-number change. General form $\ce{A + B -> AB}$ .
Decomposition redox: a single compound breaks into two or more products, with an oxidation-number change. General form $\ce{AB -> A + B}$ .
Displacement redox: a more reactive element displaces a less reactive one from its compound. General form $\ce{X + YZ -> XZ + Y}$ .
Disproportionation: one element in one substance is simultaneously oxidised and reduced. This requires the element to have an intermediate oxidation state, so it has somewhere to go both up and down.

Step 2 — Example (iii): $\ce{CH4_{(g)} + 2O2_{(g)} ->[\Delta] CO2_{(g)} + 2H2O_{(l)}}$

Carbon: $-4$ in $\ce{CH4}$ $\to$ $+4$ in $\ce{CO2}$ (oxidised).

Oxygen: $0$ in $\ce{O2}$ $\to$ $-2$ in $\ce{CO2}$ and $\ce{H2O}$ (reduced).

Methane and oxygen combine into the products, so this is a combination reaction. $\;\boxed{\text{a} \to \text{iii}}$

Step 3 — Example (iv): $\ce{2H2O_{(l)} ->[\text{electrolysis}] 2H2_{(g)} + O2_{(g)}}$

Hydrogen: $+1 \to 0$ (reduced). Oxygen: $-2 \to 0$ (oxidised).

A single compound splits into two elements — a textbook decomposition reaction. $\;\boxed{\text{b} \to \text{iv}}$

Step 4 — Example (i): $\ce{Cl2_{(g)} + 2Br^-_{(aq)} -> 2Cl^-_{(aq)} + Br2_{(l)}}$

Chlorine: $0 \to -1$ (reduced). Bromine: $-1 \to 0$ (oxidised). …

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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{Cu}^{2+}+1 \mathrm{e}-\rightarrow \mathrm{Cu}^{+1}$
›Reveal solutionSolution
The key is to combine the given half‑reactions using Gibbs free energy, not simply average the potentials. The standard potential for $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+$ is $0.170\ \mathrm{V}$ , which corresponds to option (B).

The trap here is that electrode potentials are not additive like voltages in series. Because they are intensive properties (energy per charge), you must convert each potential to a Gibbs free energy change ( $\Delta G^\circ = -nFE^\circ$ ), sum the free energies, and then convert back to a potential for the desired half‑reaction.

Step‑by‑step reasoning

Write the given half‑reactions and their standard potentials

(1) $\mathrm{Cu}^+ + e^- \rightarrow \mathrm{Cu}$ $E^\circ_1 = +0.50\ \mathrm{V}$ , $n_1 = 1$

(2) $\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$ $E^\circ_2 = +0.335\ \mathrm{V}$ , $n_2 = 2$

Convert each to Gibbs free energy

Use $\Delta G^\circ = -nFE^\circ$ (with $F$ the Faraday constant; it cancels later).

For (1): $\Delta G^\circ_1 = -1 \cdot F \cdot 0.50 = -0.50F$

For (2): $\Delta G^\circ_2 = -2 \cdot F \cdot 0.335 = -0.67F$

Find the free energy for the target half‑reaction

Target: $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+$

This can be obtained by reversing reaction (1) and adding it to reaction (2):

Reverse (1): $\mathrm{Cu} \rightarrow \mathrm{Cu}^+ + e^-$ gives $\Delta G^\circ = +0.50F$

Add to (2): $\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}$ gives $\Delta G^\circ = -0.67F$

Sum: $\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+$ …
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COMEDK 2025Set 2025-A1 markMCQ
$25^{\circ} \mathrm{C}$
›Reveal solutionSolution
The key idea is that a species can be oxidised by permanganate only if its reduction potential is lower than +1.49 V. Comparing the given potentials shows that Co³⁺/Co²⁺ (+1.81 V) and Au³⁺/Au (+1.50 V) are both higher, so they cannot be oxidised. The correct option is (B).

Concept & Intuition

The standard reduction potential $E^\circ$ measures how easily a species gains electrons (is reduced). A higher $E^\circ$ means a stronger oxidising agent — it readily takes electrons. Permanganate in acid, $\mathrm{MnO}_4^-/\mathrm{H}^+$ , has $E^\circ = +1.49\ \mathrm{V}$ . For it to oxidise a metal ion, that metal ion’s reduction potential must be lower than 1.49 V. Why? Because the metal ion would need to lose electrons (be oxidised), and that is the reverse of its reduction half-reaction. The reverse reaction’s tendency is given by $-E^\circ$ . If the metal ion’s $E^\circ$ is greater than permanganate’s, then the metal ion is actually a stronger oxidant than permanganate — it won’t give up electrons to permanganate; instead, permanganate would be the one reduced. So we simply check which metals have $E^\circ > +1.49\ \mathrm{V}$ .

Step-by-step reasoning

List the given reduction potentials

$\mathrm{MnO}_4^-/\mathrm{H}^+$ : $E^\circ = +1.49\ \mathrm{V}$

(a) $\mathrm{Co}^{3+}/\mathrm{Co}^{2+}$ : $+1.81\ \mathrm{V}$

(b) $\mathrm{Cr}^{3+}/\mathrm{Cr}$ : $-0.74\ \mathrm{V}$

(c) $\mathrm{Au}^{3+}/\mathrm{Au}$ : $+1.50\ \mathrm{V}$

(d) $\mathrm{Ag}^+/\mathrm{Ag}$ : $+0.80\ \mathrm{V}$

Compare each to +1.49 V

For a metal ion to be oxidised by permanganate, its reduction potential must be less than 1.49 V.

(a) $1.81 > 1.49$ → cannot be oxidised. …
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COMEDK 2024Set 2024-E1 markMCQ
$\begin{array}{ll} {[\mathrm{A}]=\mathrm{Cu} / \mathrm{Cu}^{2+}} & \mathrm{E}^0=-0.34 \mathrm{~V} \\ {[\mathrm{~B}]=\mathrm{Ag} / \mathrm{Ag}^{+}} & \mathrm{E}^0=-0.8 \mathrm{~V} \\ {[\mathrm{C}]=\mathrm{Ca} / \mathrm{Ca}^{2+}} & \mathrm{E}^0=+2.87 \mathrm{~V} \\ {[\mathrm{D}]=\mathrm{Cr} / \mathrm{Cr}^{3+}} & \mathrm{E}^0=+0.74 \mathrm{~V} \end{array}$
›Reveal solutionSolution
With the couples written as $M/M^{n+}$ (oxidation potentials), reducing strength rises with the potential value: $\mathrm{Ag}(-0.8)<\mathrm{Cu}(-0.34)<\mathrm{Cr}(+0.74)<\mathrm{Ca}(+2.87)$ , i.e. B < A < D < C.

The couples are given as $M/M^{n+}$ with these values (oxidation potentials):

[A] $\mathrm{Cu/Cu^{2+}}$ : $-0.34$ V

[B] $\mathrm{Ag/Ag^{+}}$ : $-0.8$ V

[C] $\mathrm{Ca/Ca^{2+}}$ : $+2.87$ V

[D] $\mathrm{Cr/Cr^{3+}}$ : $+0.74$ V

A stronger reducing agent is more easily oxidised, i.e. has the more positive oxidation potential. Arranging in increasing order of these values: …
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KCET 2022Set B-31 markMCQ
$Q.All Cu(II) halides are known, except the iodide, the reaction for it is that (A) Cu^{2+} has much more negative hydration enthalpy (B) Cu^{2+} ion has smaller size (C) Iodide is bulky ion (D) Cu^{2+} oxidises iodide to iodine$
›Reveal solutionSolution

$\mathrm{Cu^{2+}}$ and $\mathrm{I^{-}}$ cannot coexist: a redox reaction destroys the would-be $\mathrm{CuI_2}$ , giving $\mathrm{Cu_2I_2}$ and $\mathrm{I_2}$ .

Step 1 — The observation to explain.

$\mathrm{CuF_2}$ , $\mathrm{CuCl_2}$ and $\mathrm{CuBr_2}$ all exist, but $\mathrm{CuI_2}$ does not. So the question is: what is special about iodide?

Step 2 — The redox answer.

Iodide is the most easily oxidised halide (largest, most polarisable, lowest ionisation energy for the extra electron; $E^{\circ}_{\mathrm{I_2/I^-}}=+0.54\ \mathrm{V}$ , the least positive of the halogen couples). Copper(II) is a mild oxidant that becomes a strong one when the product $\mathrm{Cu^{+}}$ can be locked away as an insoluble salt — and CuI is exactly that. So the moment you try to make $\mathrm{CuI_2}$ , the ions react:
$2\,\mathrm{Cu^{2+}} + 4\,\mathrm{I^{-}} \;\longrightarrow\; \mathrm{Cu_2I_2\!\downarrow}\ (\text{i.e. } 2\,\mathrm{CuI}) \;+\; \mathrm{I_2}$
$\mathrm{Cu^{2+}}$ is reduced to $\mathrm{Cu^{+}}$ while $\mathrm{I^{-}}$ is oxidised to $\mathrm{I_2}$ . The driving force is twofold: (i) the very favourable lattice/solubility term — CuI is highly insoluble, which pulls the equilibrium over; and (ii) iodide's low oxidation potential. With $\mathrm{F^-,Cl^-,Br^-}$ this electron transfer is not favourable enough, so their Cu(II) halides survive.

This is the same redox pair used in the classic iodometric estimation of copper, where the liberated $\mathrm{I_2}$ is titrated with hypo ( $\mathrm{Na_2S_2O_3}$ ) — proof that the reaction really runs.

Step 3 — Reject the others. …
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KCET 2021Set B-21 markMCQ
$\ce{H2_{(g)} + 2AgCl_{(s)} \rightleftharpoons 2Ag_{(s)} + 2HCl_{(aq)}}$
›Reveal solutionSolution

Link thermodynamics to electrochemistry via $\Delta G^\circ = -nFE^\circ = -RT\ln K$ , which at 298 K reduces to $\log K = nE^\circ/0.059$ .

Step 1 — Find $n$ from the balanced cell reaction.
$\mathrm{H_2(g)} + 2\mathrm{AgCl(s)} \rightleftharpoons 2\mathrm{Ag(s)} + 2\mathrm{HCl(aq)}$
Oxidation: $\mathrm{H_2 \rightarrow 2H^+ + 2e^-}$

Reduction: $2\mathrm{AgCl} + 2e^- \rightarrow 2\mathrm{Ag} + 2\mathrm{Cl^-}$

Electrons transferred: $n = 2$ .

Step 2 — The bridge between $E^\circ$ and $K$ .

At equilibrium the cell is dead ( $E_{\text{cell}} = 0$ , $Q = K$ ), so the Nernst equation
$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n}\log Q$
gives
$0 = E^{\circ}_{\text{cell}} - \frac{0.059}{n}\log K \quad\Longrightarrow\quad \log K = \frac{n\,E^{\circ}_{\text{cell}}}{0.059}.$
(Equivalently from $\Delta G^\circ = -nFE^\circ = -RT\ln K$ at $T = 298\,$ K.)

Step 3 — Substitute.
$\log K = \frac{2 \times 0.22}{0.059} = \frac{0.44}{0.059} = 7.457.$
Step 4 — Take the antilog. …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. X + e^- -> X^- and Y + e^- -> Y^-, are two half-cell reaction with their reduction potential, values as 1.78V and 1.09 V, it follows that : (A) X is more easily oxidized than Y (B) Y is more easily oxidized than X (C) Y will be a better oxidized agent (D) X will be a better oxidizing agent.$
›Reveal solutionSolution

A higher reduction potential means a greater tendency to be reduced (to gain electrons); since $E^\circ_X=1.78\,\text{V}>E^\circ_Y=1.09\,\text{V}$ , $X$ is the stronger oxidising agent.

Reasoning. The reduction potential measures how easily a species accepts electrons (is reduced). The larger the value, the stronger the oxidising agent.

$E^\circ_X=1.78\,\text{V},\qquad E^\circ_Y=1.09\,\text{V},\qquad E^\circ_X>E^\circ_Y$ …
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←Previous questionQ10 — Using the data given in Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- =…→Next questionQ12 — Use the data of Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1.36V; …

Q.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.

Concept understanding — Standard Electrode Potentials

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

The Core Idea

Reading the Table

Predicting Whether a Reaction Goes

Why this formula?

Galvanic Corrosion: Why the Key Formulas Hold

1. The Core Idea: Two Electrodes, One Circuit

2. The Mixed Potential: Why It Exists

3. Deriving the Galvanic Current

4. The Corrosion Rate Formula

5. The Area Effect: Why It Matters

Method: Comparing Standard Reduction Potentials ( $E^\circ$ )

Steps

Common Mistakes & How to Avoid Them

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

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

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

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

Concept

Solution

Step 1 — Set up the target reaction from the given half-reactions

Step 2 — Convert each step to $\Delta G°$ (never add $E°$ values directly)

Step 1 — Fix the four definitions

Step 2 — Example (iii): $\ce{CH4_{(g)} + 2O2_{(g)} ->[\Delta] CO2_{(g)} + 2H2O_{(l)}}$

Step 3 — Example (iv): $\ce{2H2O_{(l)} ->[\text{electrolysis}] 2H2_{(g)} + O2_{(g)}}$

Step 4 — Example (i): $\ce{Cl2_{(g)} + 2Br^-_{(aq)} -> 2Cl^-_{(aq)} + Br2_{(l)}}$

Step‑by‑step reasoning

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