Match the items of Column I and Column II. Column I: (i) Lead storage battery (ii) Mercury cell (iii) Fuel cel

NCERT Exemplar · Q55

Q. Match the items of Column I and Column II. Column I:

(i) Lead storage battery

(ii) Mercury cell

(iii) Fuel cell

(iv) Rusting
Column II:

(a) maximum efficiency

(b) prevented by galvanisation

(d)

Pb

is anode,

PbO_2

is cathode

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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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This is a matching question linking electrochemical devices/processes (Column I) with their defining characteristics (Column II). The key is to recall the specific features of each: lead storage battery uses Pb anode and PbO₂ cathode; mercury cell gives a steady potential; fuel cells have maximum efficiency; rusting is prevented by galvanisation.

Let’s understand the why behind each match before we pair them. Every electrochemical cell or process has a unique signature — its electrode materials, electrolyte, and purpose determine its properties.

Standard Electrode Potentials tell us which metal will oxidise (act as anode) and which will reduce (act as cathode). But here, we’re not calculating cell potentials — we’re matching real-world devices to their known characteristics. This is pure recall from electrochemistry, but with a logical thread.

Lead storage battery → (d) Pb is anode, PbO₂ is cathode

In a lead-acid battery (used in cars), the anode is lead (Pb) which gets oxidised to PbSO₄, and the cathode is lead dioxide (PbO₂) which gets reduced to PbSO₄. This is the defining electrode pair. No other option fits here.
Mercury cell → (c) gives steady potential

A mercury cell (Zn-HgO with KOH electrolyte) delivers a remarkably constant voltage (~1.35 V) throughout its life because the cell reaction doesn’t change the electrolyte concentration. This steady potential is its hallmark — unlike a Leclanché cell which droops.
Fuel cell → (a) maximum efficiency

Fuel cells (e.g., H₂-O₂ cell) convert chemical energy directly to electrical energy without the intermediate heat step of a heat engine. This bypasses Carnot limitations, giving theoretical efficiencies up to 70-80% — the highest among electrochemical devices.
Rusting → (b) prevented by galvanisation …

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Method: Electrochemical Cell Identification & Matching

This method uses the defining characteristics of each electrochemical device or process to match it with its correct property.

Steps:

Step 1: Identify the key feature of each item in Column I.

(i) Lead storage battery — A rechargeable (secondary) cell where:
- Anode: $Pb$ (oxidation)
- Cathode: $PbO_2$ (reduction)
- Electrolyte: $H_2SO_4$
(ii) Mercury cell — A primary (non-rechargeable) cell known for constant voltage throughout its life.
(iii) Fuel cell — Converts chemical energy directly to electrical energy; not limited by Carnot efficiency, hence has the highest theoretical efficiency.
(iv) Rusting — Corrosion of iron; prevented by galvanisation (coating with zinc, which acts as a sacrificial anode).

Step 2: Match each item with the correct property from Column II.

Column I	Key property	Column II match
(i) Lead storage battery	$Pb$ is anode, $PbO_2$ is cathode	(d)
(ii) Mercury cell	Gives steady (constant) potential	(c)

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Here are the common mistakes students make when matching these items, along with the concept-first reasoning to avoid each.

Mistake 1: Confusing the anode in a lead storage battery

The error:

Students often think $Pb$ is the cathode because it "looks like" a metal that should be reduced. They match (i) with something other than (d).

Why it happens:

In a lead storage battery (during discharge):

Anode (oxidation): $Pb(s) + SO_4^{2-} \rightarrow PbSO_4(s) + 2e^-$
Cathode (reduction): $PbO_2(s) + 4H^+ + SO_4^{2-} + 2e^- \rightarrow PbSO_4(s) + 2H_2O$

So $Pb$ is the anode and $PbO_2$ is the cathode.

How to avoid:

Always write the half-reactions. Remember: the more negative electrode is the anode (oxidation). In this cell, $Pb$ has a lower reduction potential than $PbO_2$ .

✓ Correct match: (i) → (d)

Mistake 2: Thinking a mercury cell gives "maximum efficiency"

The error:

Students match (ii) with (a) because they confuse "steady potential" with "high efficiency."

Why it happens:

A mercury cell is known for its constant voltage (steady potential) over its lifetime — not for maximum efficiency. Maximum efficiency is a hallmark of fuel cells (they convert chemical energy directly to electrical energy with minimal losses).

How to avoid:

Mercury cell → steady potential (due to constant electrolyte concentration)
Fuel cell → maximum efficiency (no Carnot cycle limitation)

✓ Correct match: (ii) → (c), (iii) → (a)

Mistake 3: Associating rusting with galvanisation

The error:

Students match (iv) with (b) because they know galvanisation prevents rusting — but the column says "prevented by galvanisation," which describes rusting itself.

Why it happens:

The phrasing is tricky. Column II says:

(b) prevented by galvanisation

This is a property of rusting — rusting is the process that galvanisation prevents.

How to avoid:

Read the description as: "Which item in Column I is described by this phrase?"

Rusting is prevented by galvanisation.
Galvanisation is not a property of the battery or cell.

✓ Correct match: (iv) → (b)

Mistake 4: Mismatching fuel cell with steady potential

The error:

Students think fuel cells give steady potential because they run continuously.

Why it happens: …

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