The increasing order of reduction of alkyl halides – Class 11 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.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Formula of rust is :$
(a) FeO.xH2O
(b) Fe3O4.xH2O
(c) Fe2O3.xH2O
(d) None of the above
›Reveal solutionSolution

Rust is hydrated iron(III) oxide, Fe₂O₃·xH₂O, formed by the electrochemical corrosion of iron.

Corrosion of iron is an electrochemical process. Moist air acts as an electrolytic medium; at one spot on the iron surface (anodic region) iron is oxidised:

$Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-$

At another spot (cathodic region), dissolved oxygen is reduced in the presence of H⁺ (from H₂CO₃ formed by dissolved CO₂):

$O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$

The Fe²⁺ formed is further oxidised by atmospheric oxygen to Fe³⁺, which combines with water to give hydrated iron(III) oxide — rust — with a variable amount of water of hydration, written as Fe₂O₃·xH₂O.

✓Final answer
$(c) Fe₂O₃\cdotxH₂O$
CBSE 2025Set D1 markMCQ
$Q. Rust is$
(a) Powdered iron
(b) Ferrous oxide
(c) Ferric oxide
(d) Hydrated ferric oxide
›Reveal solutionSolution

Rust is hydrated ferric oxide, Fe2O3·xH2O, formed by electrochemical corrosion of iron in moist air.

Corrosion of iron in the presence of moisture and oxygen is an electrochemical process. Iron is oxidised (anode): Fe → Fe2+ + 2e-; oxygen is reduced at cathodic sites. The Fe2+ ions are further oxidised by atmospheric oxygen to Fe3+, which combines with water to form the brown, flaky deposit of hydrated ferric oxide, Fe2O3·xH2O — this is rust.

Hence rust is not simple ferrous or ferric oxide but the hydrated ferric oxide.

✓Final answer
$(D) Rust is hydrated ferric oxide, Fe2O3\cdotxH2O.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Galvanisation is applying a coating of:$
(a) Cr
(b) Cu
(c) Zn
(d) Pb
›Reveal solutionSolution

Galvanisation is the coating of iron/steel with a protective layer of zinc.

Galvanisation is an anti-corrosion technique in which a thin coating of zinc (Zn) metal is applied to iron or steel articles (e.g. by hot-dip galvanising) to protect them from rusting. Zinc, being more reactive (more easily oxidised) than iron, acts as a sacrificial anode: it corrodes preferentially and also forms a protective oxide layer, thereby saving the underlying iron from corrosion, even where the coating is locally scratched.

✓Final answer
$(c) Zn$
CBSE 2024Set ANNUAL1 mark
$Q. Write down the electrochemical reactions that takes place in the corrosion of iron.$
›Reveal solutionSolution

Rusting is an electrochemical process: iron is oxidised at anodic spots, oxygen is reduced at cathodic spots, and the resulting Fe²⁺ is later air-oxidised to hydrated Fe₂O₃.

A droplet of water on an iron surface behaves like a tiny galvanic cell, with different regions of the same iron object acting as anode and cathode (driven by impurities/stress/oxygen-concentration differences).

At the anode (oxidation):

$Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-$

At the cathode (reduction), in the presence of dissolved $H^+$ / $O_2$ :

$O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$

The electrons released at the anode flow through the metal to the cathodic region, where they reduce oxygen. The $Fe^{2+}$ formed then diffuses and is further oxidised by atmospheric oxygen to $Fe^{3+}$ , which precipitates as hydrated iron(III) oxide, $Fe_2O_3\cdot xH_2O$ — rust:

$4Fe^{2+} + O_2 + 4H_2O \rightarrow 2Fe_2O_3 + 8H^+$

✓Final answer
$Fe \rightarrow Fe^{2+}+2e^-$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Rust is a mixture of -$
(a) FeO and Fe(OH)3
(b) FeO and Fe(OH)2
(c) Fe2O3 and Fe(OH)3
(d) Fe3O4 and Fe(OH)3
›Reveal solutionSolution

Rusting is an electrochemical (corrosion) process that oxidises iron to the +3 state, giving a hydrated mixture of ferric oxide (Fe $_2$ O $_3\cdot x$ H $_2$ O) and ferric hydroxide, Fe(OH) $_3$ — option (c).

Corrosion of iron (rusting) is an electrochemical process. At an anodic site iron is oxidised:

$Fe \rightarrow Fe^{2+} + 2e^-$

and atmospheric oxygen dissolved in the surface moisture is reduced at a cathodic site:

$O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$

The Fe $^{2+}$ formed is further oxidised by dissolved oxygen to Fe $^{3+}$ . This Fe $^{3+}$ then combines with oxygen and water and deposits as hydrated ferric oxide, commonly written Fe $_2$ O $_3\cdot x$ H $_2$ O, together with ferric hydroxide, Fe(OH) $_3$ . Rust is therefore a mixture in which the iron is entirely in the +3 (ferric) state, not a single pure compound. Among the given options, the pair in which both species are ferric (Fe $^{3+}$ ) is Fe $_2$ O $_3$ and Fe(OH) $_3$ .

[!ANSWER]

(c) Fe $_2$ O $_3$ and Fe(OH) $_3$ .
CBSE 2022Set ANNUAL1 markMCQ
$Q. Rust is a mixture of$
(a) FeO and Fe(OH)3
(b) FeO and Fe(OH)2
(c) Fe2O3 and Fe(OH)3
(d) Fe3O4 and Fe(OH)2
›Reveal solutionSolution

Rusting of iron is an electrochemical corrosion process in which iron is oxidized by atmospheric O2 and moisture to a hydrated Fe2O3/Fe(OH)3 mixture.

In the presence of moisture and dissolved O₂/CO₂, iron acts as the anode of a tiny electrochemical cell: Fe → Fe²⁺ + 2e⁻ at anodic spots; O₂ + 4H⁺ + 4e⁻ → 2H₂O (or O₂ + 2H₂O + 4e⁻ → 4OH⁻) at cathodic spots. The Fe²⁺ formed is further oxidized by dissolved oxygen to Fe³⁺, which precipitates as a hydrated mixture of Fe₂O₃ and Fe(OH)₃ — this reddish-brown flaky deposit is rust, roughly represented as Fe₂O₃·xH₂O.

✓Final answer
$(c) Fe₂O₃ and Fe(OH)₃.$
CBSE 2021Set A1 markMCQ
$Q. The best way to protect iron from rusting is$
(a) making iron cathode
(b) putting it in saline water
(c) both (A) and (B)
(d) none of these
›Reveal solutionSolution
Rusting is oxidation of iron; making iron the cathode (cathodic protection) stops it being oxidised, so it is the best protection.

Rusting is an electrochemical process in which iron is oxidised (Fe → Fe2+ + 2e-) at anodic sites. To stop it, we must prevent iron from acting as the anode.

(A) Making iron the CATHODE — cathodic protection (e.g. connecting a more active sacrificial metal like Zn or Mg, or an impressed current) — forces iron to be reduced/protected rather than oxidised. This is the most effective method.

(B) Putting iron in saline water actually SPEEDS UP rusting, because dissolved salts increase the conductivity of the electrolyte.

Since (B) is harmful, (C) 'both A and B' and (D) 'none' are wrong.

✓Final answer
$(A) making iron cathode.$

CBSE 2020Set 56/1/11 mark

Q. Out of zinc and tin, whose coating is better to protect iron objects?

›Reveal solution

The key idea is Galvanic Corrosion — when two metals are in contact in an electrolyte, the more active (anodic) metal corrodes preferentially, protecting the less active (cathodic) metal. For iron, zinc coating is better because zinc is more reactive than iron and acts as a sacrificial anode, whereas tin is less reactive and only protects as long as the coating is intact.

Why This Question Matters

This isn't just a memorisation problem — it's about understanding how corrosion protection actually works in the real world. Iron objects rust because iron readily gives up electrons to oxygen and moisture. When you coat iron with another metal, you're changing the electrochemical behaviour of the surface. The question asks: which coating — zinc or tin — gives better long-term protection?

The answer depends on the electrochemical series and what happens when the coating gets scratched.

Step-by-Step Reasoning

1. Recall the reactivity order of the three metals

In the electrochemical series (from most reactive to least):

\text{Zn} > \text{Fe} > \text{Sn}

Zinc is more reactive (more anodic) than iron.
Tin is less reactive (more cathodic) than iron.

\begin{aligned} \text{Zn}^{2+} + 2e^- &\rightarrow \text{Zn} \quad E^\circ = -0.76\ \text{V} \\ \text{Fe}^{2+} + 2e^- &\rightarrow \text{Fe} \quad E^\circ = -0.44\ \text{V} \\ \text{Sn}^{2+} + 2e^- &\rightarrow \text{Sn} \quad E^\circ = -0.14\ \text{V} \end{aligned}

2. Understand what happens when the coating is intact

When the coating is perfect (no scratches), both zinc and tin protect iron by simply acting as a physical barrier — they keep oxygen and moisture away from the iron surface. In this ideal scenario, both work equally well.

But coatings never stay perfect forever. Scratches happen.

3. Analyse the scratched coating scenario — this is the crux

When a scratch exposes the underlying iron to the environment, you now have two different metals in electrical contact through the electrolyte (moist air, water). This creates a galvanic cell.

Note

In a galvanic cell, the more reactive metal becomes the anode and corrodes (loses electrons). The less reactive metal becomes the cathode and is protected (gains electrons).

Case 1: Zinc coating (scratched)

Zinc is more reactive than iron → Zinc becomes the anode.
Iron becomes the cathode.
Result: Zinc corrodes instead of iron. Even at the scratch site, the iron is protected because electrons flow from zinc to iron. This is called sacrificial protection or cathodic protection.

Case 2: Tin coating (scratched)

Tin is less reactive than iron → Iron becomes the anode.
Tin becomes the cathode.
Result: Iron corrodes at the scratch site, and the corrosion actually accelerates because the tin coating acts as a large cathode, driving the anodic reaction on the exposed iron. This is worse than having no coating at all!

Watch out

A common mistake is to think "any coating is better than none." For tin on iron, a scratched coating actually promotes rusting because it sets up a galvanic cell where iron is the sacrificial metal. Tin cans are made of steel coated with tin, but they rely on the tin being intact — once scratched, the can rusts quickly.

4. Consider practical examples

Galvanised iron (zinc-coated iron) is used for roofing sheets, buckets, and water pipes. Even when scratched, the zinc continues to protect the iron.
Tin cans (tin-coated steel) are used for food containers because tin is non-toxic and resists food acids. But they must be kept dry and intact — a dented tin can rusts rapidly.

5. Summarise the comparison

Property	Zinc coating	Tin coating
Reactivity relative to iron	More reactive (anodic)	Less reactive (cathodic)
Protection when intact	Physical barrier	Physical barrier
Protection when scratched	Sacrificial protection — zinc corrodes, iron safe	Accelerated corrosion — iron corrodes at scratch
Long-term reliability	Excellent — self-healing in a sense	Poor — requires perfect coating

Final Answer

✓Final answer

Zinc coating is better to protect iron objects because zinc is more reactive than iron and provides sacrificial protection even when the coating is scratched, whereas tin accelerates iron corrosion once the coating is damaged.

CBSE 2019Set ANNUAL1 mark
$Q. What is corrosion?$
›Reveal solutionSolution

Corrosion is a naturally occurring, spontaneous electrochemical process by which a metal is gradually converted into its oxide/other compound, deteriorating the metal — the most familiar example being iron rusting.

Corrosion occurs when a metal surface is attacked by substances in its environment — mainly oxygen and moisture, often assisted by dissolved CO $_2$ , SO $_2$ , or other acidic pollutants — converting the free metal into an oxide, sulphide, carbonate, or other compound. It is fundamentally an electrochemical process: small regions of the metal surface act as tiny anodes (where the metal is oxidised, M → M $^{n+}$ +ne $^-$ ) and other regions act as cathodes (where atmospheric O $_2$ , dissolved in a film of moisture, is reduced), completing a tiny galvanic cell on the metal's surface. The classic example is the rusting of iron, where iron is oxidised and the Fe $^{2+}$ /Fe $^{3+}$ ions combine with water and oxygen to form hydrated iron(III) oxide, Fe $_2$ O $_3\cdot xH_2O$ (rust), which flakes off, continually exposing fresh metal to further attack.

✓Final answer
$_2$
CBSE 2018Set ANNUAL1 markMCQ
$Q. Match the following. Group A item: Corrosion. Choose its correct match from Group B: (A) Lanthanides and Actinides (B) Used in treatment of cancer (C) Purple of Cassius (D) Barbituric acid (E) Hydrated ferric oxide$
›Reveal solutionSolution

Corrosion of iron (rusting) produces hydrated ferric oxide.

Corrosion is an electrochemical process in which a metal is oxidised by atmospheric oxygen and moisture, forming a coating of its oxide/hydroxide on the surface. In the specific case of iron, this process (rusting) forms hydrated ferric oxide, Fe2O3.xH2O.

✓Final answer
$Corrosion matches with E — Hydrated ferric oxide.$
CBSE 2017Set ANNUAL1 mark
$Q. Zinc protects iron from corrosion better than tin. Give reason.$
›Reveal solutionSolution

Zinc coats iron sacrificially (self-corrodes to protect the exposed metal), whereas a scratched tin coating instead accelerates iron's corrosion by making iron the anode.

Zinc lies above iron in the reactivity/electrochemical series (more negative reduction potential), so in the electrochemical corrosion cell formed at a scratch, zinc acts as the anode and is preferentially oxidised, while iron is protected as the cathode — this continues to protect iron even after the coating is locally broken (galvanisation, "sacrificial protection").

Tin, on the other hand, is less reactive than iron (less electropositive). As long as the tin coating is intact it protects iron by simple physical barrier action, but once scratched, iron (being more reactive) becomes the anode of the resulting cell and tin the cathode — this actually accelerates the corrosion (rusting) of the exposed iron, making tin a worse protective coating than zinc once damaged.

✓Final answer
$Zinc, being more reactive than iron, corrodes preferentially (sacrificial anode) even when scratched; tin, being less reactive, instead makes exposed iron corrode faster once scratched.$

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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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