Assertion: Cu is less reactive than hydrogen. Reason: E^°Cu²⁺/Cu is negative. (i) Both assertion and reason ar

NCERT Exemplar · Q59

Cu

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.

(iii) Assertion is true but the reason is false.

(iv) Both assertion and reason are false.

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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. …

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 assertion that Cu is less reactive than hydrogen is true, but the reason given ( $E^\circ_{Cu^{2+}/Cu}$ is negative) is false. The correct option is (iii).

This is a classic assertion-reason question from electrochemistry. Let’s unpack it properly.

The core idea: Reactivity and electrode potentials

The reactivity of a metal with hydrogen (or acids) is directly linked to its standard reduction potential ( $E^\circ$ ). A more negative $E^\circ$ means the metal is more easily oxidised — i.e., it is more reactive. Hydrogen has an $E^\circ$ of $0.00\ \text{V}$ by definition (for the reaction $2H^+ + 2e^- \rightarrow H_2$ ).

If a metal has $E^\circ < 0$ , it can displace hydrogen from acids (it is more reactive than hydrogen). If $E^\circ > 0$ , it cannot — it is less reactive.

E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}

This is a standard value you must memorise for exams.

Step-by-step reasoning

1. Check the assertion: Is Cu less reactive than hydrogen?

Yes. Copper does not react with dilute acids like HCl or H₂SO₄ to produce H₂ gas. This is because its reduction potential is positive (+0.34 V), meaning the reverse reaction (oxidation of Cu to Cu²⁺) is less favourable than the oxidation of H₂ to H⁺. So the assertion is true.

2. Check the reason: Is $E^\circ_{Cu^{2+}/Cu}$ negative?

No. The standard reduction potential for copper is +0.34 V, not negative. The reason given is factually incorrect. So the reason is false.

Watch out

A common mistake is to confuse the sign convention. Remember: $E^\circ$ for Cu²⁺/Cu is positive. Only metals like Zn ( $-0.76\ \text{V}$ ), Fe ( $-0.44\ \text{V}$ ), and Al ( $-1.66\ \text{V}$ ) have negative $E^\circ$ values and are more reactive than hydrogen.

3. Relate assertion and reason …

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Here’s a breakdown of the common mistakes students make on this assertion-reason question and how to avoid each.

Mistake 1: Thinking $E^\circ_{Cu^{2+}/Cu}$ is negative

Why students do this:

They confuse the sign convention for standard reduction potentials. Many remember that zinc has a negative $E^\circ$ and is reactive, so they assume copper, being less reactive, must also have a negative value.

Correct fact:

$E^\circ_{Cu^{2+}/Cu} = +0.34\ \text{V}$ (positive).

A positive reduction potential means $Cu^{2+}$ is easily reduced — copper is a less reactive metal (it does not readily lose electrons).

How to avoid:

Memorise the standard reduction potential series for common metals:
- $Zn^{2+}/Zn$ : $-0.76\ \text{V}$ (reactive)
- $Fe^{2+}/Fe$ : $-0.44\ \text{V}$
- $H^+/H_2$ : $0.00\ \text{V}$ (reference)
- $Cu^{2+}/Cu$ : $+0.34\ \text{V}$ (noble)
Remember: more negative $E^\circ$ = more reactive metal (easier to oxidise).
For copper, the positive value tells you it is below hydrogen in the reactivity series.

Mistake 2: Assuming the reason correctly explains the assertion

Why students do this:

They see “Assertion true” and “Reason true” and tick option (i) without checking if the reason actually causes the assertion.

Correct logic:

Assertion: “Cu is less reactive than hydrogen” → True (copper does not displace $H^+$ from acids).
Reason: “ $E^\circ_{Cu^{2+}/Cu}$ is negative” → False (it is positive).
So the correct answer is (iii): Assertion true, reason false.

How to avoid:

Always check the truth value of each statement separately first.
Then, only if both are true, ask: “Does the reason directly cause the assertion?”
In this case, the reason is factually wrong, so no further analysis is needed.

Mistake 3: Misinterpreting “less reactive” as “more easily reduced”

Why students do this:

“Less reactive” means the metal does not lose electrons easily. But students sometimes think “less reactive” means it gains electrons easily — which is actually the same as “more noble.” They get the direction wrong.

Correct understanding:

Reactivity refers to the tendency to lose electrons (be oxidised).
Copper has a high reduction potential (+0.34 V), meaning $Cu^{2+}$ is easily reduced — so copper metal is hard to oxidise, i.e., less reactive.
Hydrogen has $E^\circ = 0.00\ \text{V}$ , so it is more easily oxidised than copper.

How to avoid:

Draw a simple reactivity series: K > Ca > Na > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Ag > Au
The more negative the $E^\circ$ , the higher the metal is in this series (more reactive).
Copper is below hydrogen → less reactive.

--- …

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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. …
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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

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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). …
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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). …
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←Previous questionQ58 — Match the items of Column I and Column II on the basis of data given b…→Next questionQ60 — Assertion: ECell should have a positive value for the cell to function…

Q.Assertion: $Cu$ is less reactive than hydrogen.
Reason: $E^\circ_{Cu^{2+}/Cu}$ is negative.

Concept understanding — Galvanic Corrosion

Galvanic Corrosion: From Intuition to Precision

The Intuition: A "Battery" That Eats Metal

The Precise Statement

The Galvanic Series: The "Who Eats Whom" Chart

The Three Conditions for Galvanic Corrosion

Real-World Examples

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: Reactivity and electrode potentials

Step-by-step reasoning

Method: Standard Electrode Potential Comparison

Steps:

Mistake 1: Thinking $E^\circ_{Cu^{2+}/Cu}$ is negative

Mistake 2: Assuming the reason correctly explains the assertion

Mistake 3: Misinterpreting “less reactive” as “more easily reduced”

Q.Assertion: CuCuCu is less reactive than hydrogen. Reason: ECu2+/Cu∘E^\circ_{Cu^{2+}/Cu}ECu2+/Cu∘​ is negative.

The core idea: Reactivity and electrode potentials

Step-by-step reasoning

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Q.Assertion: $Cu$ is less reactive than hydrogen.
Reason: $E^\circ_{Cu^{2+}/Cu}$ is negative.

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