Use the data of 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 oxidised species. (i) $Cr^{3+}$ (ii) $MnO_4^-$ (iii) $Cr_2O_7^{2-}$ (iv) $Mn^{2+}$

Use the data of Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1.36V; E^°MnO_4^-/Mn²⁺ = 1.51V; E^°Cr³⁺/Cr = -0

NCERT Exemplar · Q12

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

(i)

Cr^{3+}

(ii)

MnO_4^-

(iii)

Cr_2O_7^{2-}

(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 oxidised species is the one with the least tendency to get reduced (lowest reduction potential). Comparing the given $E^\circ$ values, $Cr^{3+}$ has the smallest reduction potential among the oxidised forms listed, making it the most stable. The correct option is (i).

The Core Idea: What Makes an Oxidised Species "Stable"?

In electrochemistry, an oxidised species is one that has lost electrons — it's in a higher oxidation state. Its "stability" in solution means it doesn't readily grab electrons back (i.e., it resists being reduced). The tendency to get reduced is measured by the standard reduction potential ( $E^\circ$ ). A more positive $E^\circ$ means a stronger oxidising agent — it wants to get reduced. A less positive (or more negative) $E^\circ$ means the species is happy staying oxidised; it's stable.

So, to find the most stable oxidised species among the options, we simply look for the one with the lowest (least positive) standard reduction potential for its reduction half-reaction.

Watch out

A common mistake is to confuse "oxidised species" with "oxidising agent". They are the same thing here — but the question asks which one is most stable, meaning which one is least eager to accept electrons. That's the one with the smallest $E^\circ$ .

Step-by-Step Reasoning

1. Identify the reduction half-reactions for each oxidised species given.

The data provides $E^\circ$ values for the following couples:

$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ ; $E^\circ = +1.33\ V$
$Cl_2 + 2e^- \rightarrow 2Cl^-$ ; $E^\circ = +1.36\ V$
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ ; $E^\circ = +1.51\ V$
$Cr^{3+} + 3e^- \rightarrow Cr$ ; $E^\circ = -0.74\ V$

The oxidised species in each couple are: $Cr_2O_7^{2-}$ , $Cl_2$ , $MnO_4^-$ , and $Cr^{3+}$ .

2. Compare their standard reduction potentials.

A higher $E^\circ$ means a stronger driving force for reduction. Let's list them in descending order:

Oxidised Species	$E^\circ$ (V)	Tendency to get reduced
$MnO_4^-$	+1.51	Highest (strongest oxidant)
$Cl_2$	+1.36	High
$Cr_2O_7^{2-}$	+1.33	Moderate
$Cr^{3+}$	-0.74	Lowest (weakest oxidant)

$Cr^{3+}$ has a negative reduction potential. This means the reverse reaction (oxidation of Cr to $Cr^{3+}$ ) is spontaneous under standard conditions. In other words, $Cr^{3+}$ has very little desire to accept electrons and turn back into Cr metal. It is content in its oxidised state. …

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✗ Mistake 1: Confusing “most stable oxidised species” with “strongest oxidising agent”

Why it happens:

Students see high positive $E^\circ$ values and think “strong oxidiser = stable”. But a strong oxidiser is actually unstable — it wants to get reduced.

How to avoid:

Strongest oxidising agent = highest $E^\circ$ (most positive) → most reactive, not most stable.
Most stable oxidised species = hardest to reduce → lowest tendency to gain electrons → lowest (or most negative) $E^\circ$ among the oxidised forms.

✓ For this question:

$E^\circ_{Cr^{3+}/Cr} = -0.74\ V$ is the lowest → $Cr^{3+}$ is the most stable oxidised species.

✗ Mistake 2: Comparing $E^\circ$ values of half-reactions that are not all “oxidised form → reduced form”

Why it happens:

The given data mixes different types of half-cells:

$Cr_2O_7^{2-}/Cr^{3+}$ (both are oxidised species, but $Cr_2O_7^{2-}$ is the higher oxidation state)
$Cl_2/Cl^-$ (oxidised = $Cl_2$ )
$MnO_4^-/Mn^{2+}$ (oxidised = $MnO_4^-$ )
$Cr^{3+}/Cr$ (oxidised = $Cr^{3+}$ )

Students sometimes compare $E^\circ$ values without checking which species is the oxidised form in each couple.

How to avoid:

Identify the oxidised species in each half-reaction (the one on the left of the slash in standard notation: $Ox/Red$ ).
For stability, compare the $E^\circ$ values only for the reduction of that oxidised species.

Here:

$Cr_2O_7^{2-}$ : $E^\circ = 1.33\ V$
$Cl_2$ : $E^\circ = 1.36\ V$
$MnO_4^-$ : $E^\circ = 1.51\ V$
$Cr^{3+}$ : $E^\circ = -0.74\ V$

The lowest $E^\circ$ belongs to $Cr^{3+}$ → most stable.

✗ Mistake 3: Forgetting that $Cr^{3+}$ is also an oxidised species

Why it happens: …

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←Previous questionQ11 — Use the data given in Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1…→Next questionQ13 — The quantity of charge required to obtain one mole of aluminium from A…

Q.Use the data of 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 oxidised species.

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

The Core Idea: What Makes an Oxidised Species "Stable"?

Step-by-Step Reasoning

Method: Comparison of Standard Reduction Potentials

Steps

✗ Mistake 1: Confusing “most stable oxidised species” with “strongest oxidising agent”

✗ Mistake 2: Comparing $E^\circ$ values of half-reactions that are not all “oxidised form → reduced form”

✗ Mistake 3: Forgetting that $Cr^{3+}$ is also an oxidised species

The Core Idea: What Makes an Oxidised Species "Stable"?

Step-by-Step Reasoning

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