Using the data given below find out the strongest reducing agent. $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$ (i) $Cl^-$ (ii) $Cr$ (iii) $Cr^{3+}$ (iv) $Mn^{2+}$

Using the data given below find out the strongest reducing agent. E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1.

NCERT Exemplar · Q8

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

(i)

Cl^-

(ii)

Cr

(iii)

Cr^{3+}

(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 strongest reducing agent is the species with the most negative (or least positive) standard reduction potential, because it is most easily oxidised. Comparing the given $E^\circ$ values, $Cr$ (with $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$ ) is the strongest reducing agent. The correct option is (ii).

The key idea here is simple but often twisted in exams: a reducing agent is something that gets oxidised — it donates electrons. So to find the strongest reducing agent, we need the species that is most willing to lose electrons. That willingness is measured by the reverse of the given reduction half-reaction.

Standard electrode potentials ( $E^\circ$ ) are always tabulated as reduction potentials. A more negative $E^\circ$ means the reduction is less favourable — which means the reverse reaction (oxidation) is more favourable. So the species with the lowest (most negative) reduction potential is the strongest reducing agent.

Let’s look at each option.

Option (i): $Cl^-$ The given potential is $E^\circ_{Cl_2/Cl^-} = +1.36\ V$ . This is for the reduction:

Cl_2 + 2e^- \rightarrow 2Cl^-

For $Cl^-$ to act as a reducing agent, it must be oxidised to $Cl_2$ :

2Cl^- \rightarrow Cl_2 + 2e^-

The potential for this oxidation is $-1.36\ V$ (reverse sign). That’s a large positive reduction potential for $Cl_2$ , meaning $Cl_2$ is a strong oxidising agent — so $Cl^-$ is a weak reducing agent. Not our answer.

Option (ii): $Cr$ The given potential is $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$ . This is for:

Cr^{3+} + 3e^- \rightarrow Cr

For $Cr$ to act as a reducing agent, it gets oxidised to $Cr^{3+}$ :

Cr \rightarrow Cr^{3+} + 3e^-

The oxidation potential is $+0.74\ V$ (reverse sign). A negative reduction potential means the metal is easily oxidised — that’s exactly what we want. This is a strong candidate.

Option (iii): $Cr^{3+}$

$Cr^{3+}$ is the oxidised form of chromium. To act as a reducing agent, it would need to be further oxidised (e.g., to $Cr_2O_7^{2-}$ ), but the given potential $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = +1.33\ V$ tells us that $Cr_2O_7^{2-}$ is a strong oxidiser — so $Cr^{3+}$ is a very weak reducing agent. Not our answer.
Option (iv): $Mn^{2+}$

The given potential $E^\circ_{MnO_4^-/Mn^{2+}} = +1.51\ V$ is for: …

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Method: Using Standard Reduction Potentials to Identify the Strongest Reducing Agent

Concept (Why this works)

A reducing agent is a species that gets oxidised (loses electrons) in a reaction.

The more negative (or less positive) the standard reduction potential $E^\circ$ , the greater the tendency of the species to get oxidised — i.e., the stronger it is as a reducing agent.

Key rule:

Lower (more negative) $E^\circ$ → stronger reducing agent.

Higher (more positive) $E^\circ$ → stronger oxidising agent.

Steps

Identify the reduced form of each half-cell — that is the species that can be oxidised (the potential reducing agent).

Half-reaction	$E^\circ$ (V)	Reduced form (potential reducing agent)
$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$	$+1.33$	$\text{Cr}^{3+}$
$\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$	$+1.36$	$\text{Cl}^-$
$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$	$+1.51$	$\text{Mn}^{2+}$

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Common Mistakes Students Make on This Concept

Mistake 1: Confusing "Strongest Reducing Agent" with "Highest $E^\circ$ "

The error: Students see the highest positive $E^\circ$ value (like $1.51\ V$ for $\text{MnO}_4^-/\text{Mn}^{2+}$ ) and pick that species as the strongest reducing agent.

Why it's wrong:

$E^\circ$ values given are reduction potentials — they tell you how easily a species gets reduced (gains electrons).
A reducing agent is a species that gets oxidised (loses electrons).
The strongest reducing agent is the one that is most easily oxidised — which corresponds to the most negative (or least positive) reduction potential.

How to avoid:

Always ask: "Which species would most readily lose electrons?"

For a reduction half-reaction: $\text{Oxidised form} + ne^- \rightarrow \text{Reduced form}$
The reduced form (right side) is the reducing agent.
The more negative the $E^\circ$ , the stronger the reducing agent.

✓ Correct approach:

Among the given options, the reduced forms are:

$\text{Cl}^-$ (from $\text{Cl}_2/\text{Cl}^-$ , $E^\circ = 1.36\ V$ )
$\text{Cr}$ (from $\text{Cr}^{3+}/\text{Cr}$ , $E^\circ = -0.74\ V$ )
$\text{Cr}^{3+}$ (from $\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}$ , $E^\circ = 1.33\ V$ )
$\text{Mn}^{2+}$ (from $\text{MnO}_4^-/\text{Mn}^{2+}$ , $E^\circ = 1.51\ V$ )

The most negative $E^\circ$ is $-0.74\ V$ for $\text{Cr}^{3+}/\text{Cr}$ , so Cr metal is the strongest reducing agent.

Answer: (ii) Cr

Mistake 2: Forgetting to Identify the Reduced Form

The error: Students look at the species on the left side of the half-reaction (like $\text{Cr}_2\text{O}_7^{2-}$ ) and treat it as the reducing agent.

Why it's wrong:

The half-reaction is written as a reduction:

$\text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7H_2O$

Left side = oxidising agent (gets reduced)
Right side = reducing agent (gets oxidised)

How to avoid:

Write the half-reaction in the reduction direction.
The reducing agent is always the product (right side).
For $\text{Cr}^{3+}/\text{Cr}$ : $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}$ → Cr is the reducing agent.

Mistake 3: Comparing $E^\circ$ Values Without Considering the Species

The error: Students compare $E^\circ$ values directly without checking which species is actually in the options.

Why it's wrong:

The question gives four $E^\circ$ values but asks about four specific species — some are oxidised forms, some are reduced forms.

How to avoid:

List the reduced form for each half-reaction.
Match them to the options:
- $\text{Cl}^-$ → from $\text{Cl}_2/\text{Cl}^-$
- $\text{Cr}$ → from $\text{Cr}^{3+}/\text{Cr}$
- $\text{Cr}^{3+}$ → from $\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}$
- $\text{Mn}^{2+}$ → from $\text{MnO}_4^-/\text{Mn}^{2+}$
Now compare $E^\circ$ $E^{\circ}$ of the corresponding reduction half-reactions:
- $\text{Cl}_2/\text{Cl}^-$ : $1.36\ V$
- $\text{Cr}^{3+}/\text{Cr}$ : $-0.74\ V$
- $\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}$ : $1.33\ V$
- $\text{MnO}_4^-/\text{Mn}^{2+}$ : $1.51\ V$

The most negative $E^\circ$ → strongest reducing agent → Cr

Mistake 4: Misinterpreting "Strongest" as "Most Reactive in General" …

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