The increasing order of reduction of alkyl halides – 1st PUC Chemistry

NCERT Exemplar · Q3

Q. The increasing order of reduction of alkyl halides with zinc and dilute HCl is

(i) R-Cl < R-I < R-Br

(ii) R-Cl < R-Br < R-I

(iii) R-I < R-Br < R-Cl

(iv) R-Br < R-I < R-Cl

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Concept understanding — Galvanic Corrosion

Galvanic Corrosion: From Intuition to Precision

Imagine you have two different metals — say, a copper pipe and an iron nail — and you connect them with a wire, then dip both into a bucket of salt water. If you come back a few hours later, the iron nail will be badly rusted, while the copper pipe will look almost untouched. Why?

The answer is galvanic corrosion. It is the accelerated corrosion of one metal when it is in electrical contact with a different metal in the presence of an electrolyte (like water with dissolved salts).

The Intuition: A "Battery" That Eats Metal

Think of a simple battery: you have two different metals (electrodes) and a chemical solution (electrolyte). One metal wants to give away electrons (it gets eaten away), and the other wants to accept them (it stays protected). That is exactly what happens in galvanic corrosion.

The more reactive metal (the one that "wants" to corrode) becomes the anode. It loses electrons and dissolves into the electrolyte — that is the corrosion you see.
The less reactive metal becomes the cathode. It does not corrode; instead, it accepts electrons from the anode, often causing the electrolyte near it to become alkaline or to produce hydrogen gas.

The key point: the two metals do not need to be physically touching. They just need electrical contact (through a wire or direct contact) and a continuous electrolyte (water, soil, concrete, etc.) to complete the circuit.

The Precise Statement

Important

Galvanic corrosion is the electrochemical process in which a more active metal (the anode) corrodes preferentially when electrically coupled to a less active metal (the cathode) in the presence of an electrolyte. The driving force is the difference in their electrode potentials.

The Galvanic Series: The "Who Eats Whom" Chart

Not all metal pairs corrode equally. The galvanic series ranks metals and alloys by their tendency to corrode in seawater (a common electrolyte). The more negative (active) a metal is, the more likely it is to be the anode and corrode.

Here is a simplified version of the series (from most active/anodic to most noble/cathodic):

Metal / Alloy	Relative Activity
Magnesium	Most active (anodic)
Zinc
Aluminium
Cadmium
Mild steel / Iron
Stainless steel (active)
Tin
Lead
Copper
Nickel
Stainless steel (passive)
Silver
Titanium
Gold / Platinum	Most noble (cathodic)

Watch out

A common mistake: students think the larger metal always corrodes. In reality, it is the more active metal that corrodes, regardless of size. However, the area ratio matters enormously — a small anode coupled to a large cathode corrodes very fast (like a tiny iron rivet holding a huge copper plate).

The Three Conditions for Galvanic Corrosion

For galvanic corrosion to occur, all three must be present:

Two dissimilar metals (or the same metal in different environments, e.g., a steel pipe in soil vs. in air).
Electrical contact between them (direct physical contact or through a wire).
An electrolyte bridging them (water, moisture, soil, concrete, etc.).

Remove any one, and galvanic corrosion stops.

Real-World Examples

The Statue of Liberty: The copper skin was originally separated from the iron framework by asbestos cloth. When the cloth degraded, the iron (anode) corroded rapidly because it was coupled to the huge copper (cathode) surface.
Plumbing: Connecting a copper pipe directly to a galvanized steel pipe causes the steel to corrode near the joint.
Marine environments: Aluminium boat hulls with bronze propellers — the aluminium corrodes unless protected by sacrificial anodes (zinc blocks).

How to Prevent It

Avoid dissimilar metal contact where possible.
Insulate the two metals with a non-conductive gasket or coating.
Use a sacrificial anode — attach a more active metal (like zinc) that corrodes instead of the structure you want to protect.
Coat both metals with paint or sealant, but be careful: if the coating on the anode is damaged, corrosion concentrates at the defect.

Note

The same principle is used deliberately in cathodic protection — for example, zinc blocks are bolted to ship hulls or underground pipelines. The zinc corrodes sacrificially, protecting the steel.

The Bottom Line

Galvanic corrosion is not magic — it is simply a galvanic cell (a battery) where the anode metal is the "fuel" that gets consumed. The greater the difference in the galvanic series between the two metals, the stronger the driving force, and the faster the corrosion.

Galvanic corrosion is discussed in the NCERT/CBSE Class 12 Chemistry chapter on Electrochemistry, and ‘galvanic corrosion vs rusting’ or ‘sacrificial anode protection’ are frequently searched important-question topics for board exams and JEE Main. Understanding this electrochemical-cell-based explanation of corrosion is also useful for application-based NEET and CET chemistry questions.

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).
Large anode + small cathode → mild corrosion.

Why this holds: The current density on the anode is $i_a = I_{\text{galvanic}} / A_1$ . For a fixed $I_{\text{galvanic}}$ , smaller $A_1$ gives higher $i_a$ , which accelerates corrosion.

6. The Driving Force: Potential Difference

The driving force for galvanic corrosion is the difference in open-circuit potentials:

\Delta E = E_{\text{corr},2} - E_{\text{corr},1}

A larger $\Delta E$ generally leads to a larger $I_{\text{galvanic}}$ , but the exact relationship depends on the polarization behavior (Tafel slopes) of both electrodes.

Why this holds: The mixed potential $E_{\text{mix}}$ is determined by the intersection of the anodic and cathodic polarization curves. A larger separation between the two curves shifts the intersection to a higher current.

Summary of Key Takeaways

Concept	Formula	Why It Holds
Mixed potential	$I_{\text{anode}} = I_{\text{cathode}}$	Charge conservation in a closed circuit
Galvanic current	$A_1 i_a(E_{\text{mix}}) = A_2 i_c(E_{\text{mix}})$	Butler-Volmer kinetics + area balance
Corrosion rate	$\frac{I_{\text{galvanic}} M}{n F \rho}$	Faraday's law of electrolysis
Area effect	Small anode → high $i_a$	Current density inversely proportional to area
Driving force	$\Delta E = E_{\text{corr},2} - E_{\text{corr},1}$	Larger potential difference → larger current (generally)

Exam tip: Always start with the mixed potential condition — it's the foundation. Then apply Faraday's law for the rate. Never forget the area ratio — it's the most common trick in exam problems.

The reduction of alkyl halides by Zn/dil. HCl follows the ease of C–X bond cleavage: weaker bonds react faster. Since bond strength decreases I > Br > Cl, the increasing order of reduction (slowest to fastest) is R–I < R–Br < R–Cl.

Why bond strength governs reduction rate

When zinc and dilute HCl reduce an alkyl halide to an alkane, the first and rate-determining step is breaking the carbon–halogen bond. Zinc donates electrons (acts as a reducing agent), and the halogen must leave as a halide ion:

\ce{R-X + Zn + HCl -> R-H + ZnCl_2 + HX}

The ease of reduction depends on how readily the C–X bond breaks. A weaker bond snaps more easily, so the reaction proceeds faster. Conversely, a stronger bond resists cleavage, making reduction slower.

The C–X bond strengths follow the order:

\text{C–I} < \text{C–Br} < \text{C–Cl}

Iodine is the largest halogen; its valence electrons are far from the nucleus and poorly overlap with carbon's orbital, yielding a weak bond. Chlorine is smallest, with tight overlap and a strong bond. Bromine sits in between.

Important

Weaker bond = faster reduction. The halide that holds on most weakly (R–I) reacts fastest; the one that grips tightest (R–Cl) reacts slowest.

Resolving the question's wording

The question asks for the increasing order of reduction. In chemistry, "increasing order" of a rate or reactivity means arranging from slowest to fastest (least reactive to most reactive).

Because R–I has the weakest bond, it reduces most readily (fastest). R–Cl, with the strongest bond, reduces least readily (slowest). Therefore:

Increasing order of reduction (slowest → fastest):

\text{R–Cl} < \text{R–Br} < \text{R–I}

Reading left to right: R–Cl is the slowest (least reduced), R–I is the fastest (most reduced).

Watch out

Do not confuse "increasing order of reduction" with "increasing bond strength." The two run in opposite directions. Stronger bonds reduce more slowly.

Matching the options

(i) R–Cl < R–I < R–Br: incorrect order; places R–I in the middle.
(ii) R–Cl < R–Br < R–I: matches our reasoning—slowest to fastest.
(iii) R–I < R–Br < R–Cl: inverted; this would be decreasing order of reduction (fastest to slowest).
(iv) R–Br < R–I < R–Cl: incorrect order.

✓Final answer

The correct option is (ii) : R-Cl < R-Br < R-I.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set D-41 markMCQ
$\mathrm{Cmol}^{-1}$
›Reveal solutionSolution
The charge needed per mole of metal is just $n \times F$ , where $n$ is the cation's charge and $F = 96500\,\mathrm{C\,mol^{-1}}$ — so simply multiply $96500$ by $1, 2, 3, 4$ .

Step 1 — The governing law.

The cathodic reduction is
$\mathrm{M}^{n+} + n e^- \rightarrow \mathrm{M}$
One mole of $\mathrm{M}^{n+}$ therefore consumes exactly $n$ moles of electrons. Faraday's constant is the charge on one mole of electrons,
$F = 96500\ \mathrm{C\,mol^{-1}}$
so the quantity of electricity required per mole of metal deposited is
$Q = nF$
Step 2 — Compute for each cation.

(a) $\mathrm{Ag^+}$ , $n = 1$ : $Q = 1 \times 96500 = 96500\ \mathrm{C\,mol^{-1}}$ → (iii)

(b) $\mathrm{Mg^{2+}}$ , $n = 2$ : $Q = 2 \times 96500 = 193000\ \mathrm{C\,mol^{-1}}$ → (iv)

(c) $\mathrm{Al^{3+}}$ , $n = 3$ : $Q = 3 \times 96500 = 289500\ \mathrm{C\,mol^{-1}}$ → (ii)

(d) $\mathrm{Ti^{4+}}$ , $n = 4$ : $Q = 4 \times 96500 = 386000\ \mathrm{C\,mol^{-1}}$ → (i)

Step 3 — Assemble the matching.
$\mathrm{a\!-\!iii},\quad \mathrm{b\!-\!iv},\quad \mathrm{c\!-\!ii},\quad \mathrm{d\!-\!i}$
This is exactly option (B). Note the neat check: the four List-II values are just $96500$ multiplied by $1, 4, 3, 2$ in the order printed, and the charge required rises monotonically with the cation's charge — $\mathrm{Ag^+}$ (cheapest) to $\mathrm{Ti^{4+}}$ (dearest).

✓Final answer
$The correct option is (B) — a - iii, b - iv, c - ii, d - i.$

ANSWER: B

KCET 2024Set B-21 markMCQ

\approx 5\%

›Reveal solution

Recall the composition of Misch metal: ~95% lanthanoid + ~5% iron, with traces of S, C, Ca and Al.

Step 1 — What Misch metal is.

Misch metal is a commercially important lanthanoid alloy. Its composition is:

~95% lanthanoid metal (mostly cerium, with lanthanum, neodymium and praseodymium),
~5% iron,
traces of S, C, Ca and Al.

Step 2 — Identify the transition element.

The question asks specifically for the transition element present at the ~5% level. Screening the options against the $d$ -block:

Option	Element	Block
(A)	Mg	$s$ -block (alkaline earth) — not a transition metal
(B)	Fe	$3d$ transition series ✓
(C)	Zn	$3d$ series, but $d^{10}$ in all its compounds — and not the 5% component
(D)	Co	$3d$ series, but not a component of Misch metal

Only Fe is both a transition element and the ~5% component of Misch metal, so it satisfies the question completely.

Step 3 — Why it is worth remembering.

Misch metal's headline use is in cigarette lighter flints and tracer bullets: it is pyrophoric — scraping it throws off sparks. It is also added to steel as a scavenger, because the lanthanoids mop up sulphur and oxygen.

✓Final answer

The correct option is (B) — Fe.

ANSWER: B

COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{E}^{\circ}$
›Reveal solutionSolution
To prevent iron corrosion by coating, the coating metal must be more easily oxidized (more negative reduction potential) than iron, so it acts as a sacrificial anode. Only element A ( $E^\circ = -2.37\ \mathrm{V}$ ) satisfies this, making option (D) correct.

Concept & Intuition

Corrosion of iron (rusting) is an electrochemical process where iron is oxidized:
$\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+} + 2e^- \quad (E^\circ = -0.44\ \mathrm{V})$
If we coat iron with another metal, we want that metal to oxidize instead of iron — it “sacrifices” itself. This happens if the coating metal has a more negative reduction potential than iron, meaning it is a stronger reducing agent (more easily oxidized). A metal with a less negative or positive reduction potential would actually cause iron to corrode faster (it would act as a cathode, forcing iron to be the anode).

Step-by-step reasoning

Identify iron’s tendency to oxidize

Iron’s reduction potential is $E^\circ_{\mathrm{Fe}^{2+}/\mathrm{Fe}} = -0.44\ \mathrm{V}$ . The more negative this value, the easier it is for the metal to lose electrons (oxidize). So iron itself is moderately prone to oxidation.

Compare each candidate metal’s reduction potential

Element A: $E^\circ_{\mathrm{A}^{2+}/\mathrm{A}} = -2.37\ \mathrm{V}$ — much more negative than iron.

Element B: $E^\circ_{\mathrm{B}^{2+}/\mathrm{B}} = -0.15\ \mathrm{V}$ — less negative than iron.

Element C: $E^\circ_{\mathrm{C}^{2+}/\mathrm{C}} = +0.34\ \mathrm{V}$ — positive, meaning it is very hard to oxidize.

Determine which metal will oxidize preferentially

For a coating to protect iron, the coating metal must have a more negative reduction potential than iron. That way, when an electrochemical cell forms (e.g., with moisture and oxygen), the coating metal becomes the anode and corrodes, while iron remains the cathode and is protected.

A ( $-2.37\ \mathrm{V}$ ) is more negative than Fe ( $-0.44\ \mathrm{V}$ ) → A will protect iron.

B ( $-0.15\ \mathrm{V}$ ) is less negative than Fe → B is harder to oxidize than iron, so iron would corrode first.

C ( $+0.34\ \mathrm{V}$ ) is positive → C is very noble; it would actually accelerate iron corrosion (iron becomes the sacrificial anode).

Apply to the multiple-choice options

Only element A qualifies. Options:

(A) Element C only — incorrect.

(B) Elements B and C — incorrect.

(C) Element B only — incorrect.

(D) Element A only — correct.

Watch out

A common mistake is thinking that any metal with a less negative potential than iron will protect it. In reality, only metals with a more negative potential (like zinc, magnesium, or here element A) act as sacrificial anodes. Metals like tin or copper (with less negative or positive potentials) actually promote rusting if the coating is scratched.

Tip

Think of the reduction potential as a “ladder”: the more negative the value, the higher the metal sits on the reactivity series. A higher (more reactive) metal will sacrifice itself for a lower (less reactive) metal. Iron at $-0.44\ \mathrm{V}$ is protected only by metals above it (more negative), like A at $-2.37\ \mathrm{V}$ .

✓Final answer
$The correct option is (D) . ANSWER: D$
COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{O}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+4 \mathrm{e} \rightarrow 4 \mathrm{OH}^{-}$
›Reveal solutionSolution
The cathodic (reduction) reaction in rusting is $O_2+4H^+ +4e^-\rightarrow 2H_2O$ with $E^0=+1.23\,V$ — statement (B).

Electrochemical mechanism of rusting (NCERT):

Anode (oxidation): $Fe(s)\rightarrow Fe^{2+}+2e^-$ , $E^0_{(Fe^{2+}/Fe)}=-0.44\,V$ .

Cathode (reduction): in the presence of $H^+$ (from dissolved $CO_2$ /water) and atmospheric oxygen, $O_2(g)+4H^+(aq)+4e^-\rightarrow 2H_2O(l)$ , $E^0=+1.23\,V$ .

Overall: $2Fe+O_2+4H^+\rightarrow 2Fe^{2+}+2H_2O$ ; the $Fe^{2+}$ is then further oxidised to hydrated $Fe_2O_3$ (rust).

Checking the options: (A) gives the neutral/basic oxygen-reduction half-reaction, not the one used in the standard rusting description; (C) wrongly makes water oxidation the anodic reaction (the anode is iron dissolution); (D) gives a wrong overall reaction with an incorrect negative $E^0$ (corrosion is spontaneous, net $E^0>0$ ). Only (B) is correct.

✓Final answer
$O_2(g)+4H^+(aq)+4e\rightarrow 2H_2O(l);\;E^0=+1.23\,V$

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Q.The increasing order of reduction of alkyl halides with zinc and dilute HCl is

Why bond strength governs reduction rate

Resolving the question's wording

Matching the options

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