Using the data given in Q.8 ($E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$) find out in which option the order of reducing power is correct. (i) $Cr^{3+} < Cl^- < Mn^{2+} < Cr$ (ii) $Mn^{2+} < Cl^- < Cr^{3+} < Cr$ (iii) $Cr^{3+} < Cl^- < Cr_2O_7^{2-} < MnO_4^-$ (iv) $Mn^{2+} < Cr^{3+} < Cl^- < Cr$

> [!TLDR] > Reducing power is the tendency to **lose electrons** — it is the **reverse** of the reduction half-reaction. A more **negative** (or less positive) reduction potential means a stronger reducing agent. Using the given $E^\circ$ values, the correct order of increasing reducing power is $Mn^{2+} < Cl^- < Cr^{3+} < Cr$, which matches option **(ii)**. The key to this question is understanding what "reducing power" actually means. A reducing agent is a species that **donates electrons** and gets oxidised itself. In electrochemistry, we measure the tendency of a species to **gain** electrons — that's the standard reduction potential $E^\circ$. So a strong reducing agent has a **low (or very negative) reduction potential** because it prefers to lose electrons rather than gain them. Let’s list the given half-reactions and their $E^\circ$ values: 1. $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$; $E^\circ = +1.33\ V$ 2. $Cl_2 + 2e^- \rightarrow 2Cl^-$; $E^\circ = +1.36\ V$ 3. $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$; $E^\circ = +1.51\ V$ 4. $Cr^{3+} + 3e^- \rightarrow Cr$; $E^\circ = -0.74\ V$ Now, the species we need to compare for reducing power are: $Cr^{3+}$, $Cl^-$, $Mn^{2+}$, and $Cr$. Notice that these are the **reduced forms** of the couples above (except $Cr^{3+}$ appears both as a product in reaction 1 and as a reactant in reaction 4 — we’ll handle that carefully). > [!WARNING] > A common mistake is to compare the $E^\circ$ values of the **oxidised forms** directly. Remember: reducing power belongs to the **reduced species** (the one on the right side of the reduction half-reaction). For example, $Cl^-$ is the reduced form of the $Cl_2/Cl^-$ couple, so its reducing power is related to the **reverse** of that reaction. Here’s the step-by-step reasoning: 1. **Identify the reduced species and their corresponding reduction potentials** - For $Cr^{3+}$: It is the reduced form of the $Cr_2O_7^{2-}/Cr^{3+}$ couple ($E^\c

Using 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⁺

NCERT Exemplar · Q10

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

(i)

Cr^{3+} < Cl^- < Mn^{2+} < Cr

(ii)

Mn^{2+} < Cl^- < Cr^{3+} < Cr

(iii)

Cr^{3+} < Cl^- < Cr_2O_7^{2-} < MnO_4^-

(iv)

Mn^{2+} < Cr^{3+} < Cl^- < Cr

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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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Reducing power is the tendency to lose electrons — it is the reverse of the reduction half-reaction. A more negative (or less positive) reduction potential means a stronger reducing agent. Using the given $E^\circ$ values, the correct order of increasing reducing power is $Mn^{2+} < Cl^- < Cr^{3+} < Cr$ , which matches option (ii).

The key to this question is understanding what "reducing power" actually means. A reducing agent is a species that donates electrons and gets oxidised itself. In electrochemistry, we measure the tendency of a species to gain electrons — that's the standard reduction potential $E^\circ$ . So a strong reducing agent has a low (or very negative) reduction potential because it prefers to lose electrons rather than gain them.

Let’s list the given half-reactions and their $E^\circ$ values:

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

Now, the species we need to compare for reducing power are: $Cr^{3+}$ , $Cl^-$ , $Mn^{2+}$ , and $Cr$ . Notice that these are the reduced forms of the couples above (except $Cr^{3+}$ appears both as a product in reaction 1 and as a reactant in reaction 4 — we’ll handle that carefully).

Watch out

A common mistake is to compare the $E^\circ$ values of the oxidised forms directly. Remember: reducing power belongs to the reduced species (the one on the right side of the reduction half-reaction). For example, $Cl^-$ is the reduced form of the $Cl_2/Cl^-$ couple, so its reducing power is related to the reverse of that reaction.

Here’s the step-by-step reasoning:

Identify the reduced species and their corresponding reduction potentials
- For $Cr^{3+}$ : It is the reduced form of the $Cr_2O_7^{2-}/Cr^{3+}$ couple ( $E^\circ = +1.33\ V$ ). But $Cr^{3+}$ is also the oxidised form of the $Cr^{3+}/Cr$ couple ( $E^\circ = -0.74\ V$ ). Which one matters? When we talk about $Cr^{3+}$ as a reducing agent, we mean it can be oxidised to a higher state — that is, to $Cr_2O_7^{2-}$ . So the relevant half-reaction is the reverse of $Cr_2O_7^{2-} \rightarrow Cr^{3+}$ , i.e., $Cr^{3+} \rightarrow Cr_2O_7^{2-} + 3e^-$ . The potential for this oxidation is $-1.33\ V$ (reverse sign).
- For $Cl^-$ : Reduced form of $Cl_2/Cl^-$ couple ( $E^\circ = +1.36\ V$ ). Oxidation: $Cl^- \rightarrow \frac12 Cl_2 + e^-$ , potential = $-1.36\ V$ .
- For $Mn^{2+}$ : Reduced form of $MnO_4^-/Mn^{2+}$ couple ( $E^\circ = +1.51\ V$ ). Oxidation: $Mn^{2+} \rightarrow MnO_4^- + 5e^-$ , potential = $-1.51\ V$ .
- For $Cr$ (metallic chromium): Reduced form of $Cr^{3+}/Cr$ couple ( $E^\circ = -0.74\ V$ ). Oxidation: $Cr \rightarrow Cr^{3+} + 3e^-$ , potential = $+0.74\ V$ (reverse sign). …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing Reducing Power with Oxidising Power

The error: Students often think that a higher $E^\circ$ means stronger reducing power. In reality, higher $E^\circ$ = stronger oxidising agent (gets reduced easily). Reducing power is the opposite — a species with a more negative (lower) $E^\circ$ is a stronger reducing agent.

How to avoid: Always ask yourself: "Does this species get oxidised or reduced?"

If it gets reduced (higher $E^\circ$ ), it's an oxidising agent.
If it gets oxidised (lower $E^\circ$ ), it's a reducing agent.

Key rule: Reducing power $\propto$ lower (more negative) $E^\circ$ value.

Mistake 2: Using the Wrong Half-Cell Reaction

The error: For $Cr_2O_7^{2-}/Cr^{3+}$ ( $E^\circ = 1.33\ V$ ), students compare this directly with $Cr^{3+}/Cr$ ( $E^\circ = -0.74\ V$ ) without realising these are different couples. The reducing species in each case is different:

For $Cr_2O_7^{2-}/Cr^{3+}$ : the reduced form is $Cr^{3+}$ (which can act as a reducing agent if oxidised back to $Cr_2O_7^{2-}$ )
For $Cr^{3+}/Cr$ : the reduced form is $Cr$ metal

How to avoid: Identify the reduced form of each couple — that's the species whose reducing power you're comparing. List them clearly:

Couple	$E^\circ$ (V)	Reduced form (reducing agent)
$Cr_2O_7^{2-}/Cr^{3+}$	$+1.33$	$Cr^{3+}$
$Cl_2/Cl^-$	$+1.36$	$Cl^-$
$MnO_4^-/Mn^{2+}$	$+1.51$	$Mn^{2+}$
$Cr^{3+}/Cr$	$-0.74$	$Cr$

Mistake 3: Forgetting to Compare Only the Reduced Forms

The error: Option (iii) includes $Cr_2O_7^{2-}$ and $MnO_4^-$ — these are oxidised forms, not reducing agents. The question asks for reducing power order of the species listed, so you must compare only the reduced forms from the given data.

How to avoid:

The species in the options are: $Cr^{3+}$ , $Cl^-$ , $Mn^{2+}$ , $Cr$ , $Cr_2O_7^{2-}$ , $MnO_4^-$
From the data, the reduced forms are: $Cr^{3+}$ (from $+1.33\ V$ ), $Cl^-$ (from $+1.36\ V$ ), $Mn^{2+}$ (from $+1.51\ V$ ), and $Cr$ (from $-0.74\ V$ )
$Cr_2O_7^{2-}$ and $MnO_4^-$ are oxidised forms — they are not reducing agents here.

Mistake 4: Misordering Based on $E^\circ$ Values

The error: Arranging $E^\circ$ values from highest to lowest and assuming that's the reducing power order.

Correct approach:

Reducing power increases as $E^\circ$ decreases (becomes more negative).
Order of $E^\circ$ (reduced form): …

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←Previous questionQ9 — Use the data given in Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1…→Next questionQ11 — Use the data given in Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1…

Q.Using the data given in Q.8 ( $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$ ; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$ ; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$ ; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$ ) find out in which option the order of reducing power is correct.

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: Using Standard Electrode Potentials to Compare Reducing Power

Steps

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Reducing Power with Oxidising Power

Mistake 2: Using the Wrong Half-Cell Reaction

Mistake 3: Forgetting to Compare Only the Reduced Forms

Mistake 4: Misordering Based on $E^\circ$ Values

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