Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?

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Q. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?

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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 mercury cell maintains a constant voltage because its overall cell reaction does not involve a change in the concentration of the electrolyte (KOH) — the electrolyte acts only as a conducting medium, not as a reactant or product. The cell potential is therefore independent of the extent of discharge.

The key to understanding this lies in the Nernst equation and what actually changes inside a cell as it runs.

For any electrochemical cell, the cell potential $E_{\text{cell}}$ is given by:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q

where $Q$ is the reaction quotient. As the cell discharges, reactants are consumed and products are formed, so $Q$ changes — and normally, $E_{\text{cell}}$ drops.

In a dry cell (Leclanché cell), the electrolyte is a paste of $\text{NH}_4\text{Cl}$ and $\text{ZnCl}_2$ . During discharge, $\text{NH}_4^+$ ions are consumed and $\text{NH}_3$ is produced, which changes the ionic composition of the electrolyte. The concentration of the electrolyte itself changes, so $Q$ changes significantly, and the voltage falls steadily.

In a mercury cell, the situation is fundamentally different. Let's see why.

The cell construction and reactions

The mercury cell has:
- Anode: Zinc amalgam (Zn/Hg)
- Cathode: Mercury(II) oxide ( $\text{HgO}$ ) mixed with graphite
- Electrolyte: Concentrated potassium hydroxide ( $\text{KOH}$ ) paste
The half-reactions are:

Anode (oxidation):

\text{Zn(Hg)} + 2\text{OH}^- \rightarrow \text{ZnO} + \text{H}_2\text{O} + 2e^-

Cathode (reduction):

\text{HgO} + \text{H}_2\text{O} + 2e^- \rightarrow \text{Hg} + 2\text{OH}^-

Overall cell reaction:

\text{Zn(Hg)} + \text{HgO} \rightarrow \text{ZnO} + \text{Hg}

What happens to the electrolyte?

Look carefully at the overall reaction. The $\text{OH}^-$ ions appear on both sides — they are consumed at the anode and regenerated in equal amount at the cathode. The net change in $\text{OH}^-$ concentration is zero. Water also cancels out.

The electrolyte $\text{KOH}$ is not consumed or produced in the overall reaction. It simply provides a medium for ion transport. Its concentration remains constant throughout the cell's life.
What does the Nernst equation say?

For the overall reaction:

\text{Zn(Hg)} + \text{HgO} \rightarrow \text{ZnO} + \text{Hg}

The reaction quotient $Q$ is:

Q = \frac{[\text{ZnO}][\text{Hg}]}{[\text{Zn(Hg)}][\text{HgO}]}

But here's the crucial point: all four species are solids (or solid amalgams). For pure solids, activity is taken as 1. So $Q = 1$ at all times during discharge.

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log(1) = E^\circ_{\text{cell}}

The log term is zero. The cell potential equals the standard cell potential and does not change as the cell runs.

Contrast with the dry cell …

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Here are the common mistakes students make when answering “Why does a mercury cell have a constant potential throughout its life, unlike a dry cell?” — and how to avoid each.

✗ Mistake 1: Saying “the electrolyte does not get used up”

Why it’s wrong:

In both cells, the electrolyte participates in the reaction. In a dry cell, the electrolyte ( $\text{NH}_4\text{Cl}$ ) is consumed and the paste dries out, causing voltage drop. In a mercury cell, the electrolyte ( $\text{KOH}$ ) is not consumed — it only acts as a medium for ion transfer.

How to avoid:

Be precise:

In a mercury cell, the electrolyte ( $\text{KOH}$ ) remains unchanged in concentration because it does not take part in the net cell reaction. The overall reaction is:

$\text{Zn (s)} + \text{HgO (s)} \rightarrow \text{ZnO (s)} + \text{Hg (l)}$

KOH is a catalyst-like medium — it only conducts ions.

✗ Mistake 2: Confusing “constant potential” with “no internal resistance change”

Why it’s wrong:

Internal resistance does increase slightly in a mercury cell, but the electrode potentials remain constant because the activities of reactants/products stay nearly constant.

How to avoid:

Use the Nernst equation:

E = E^\circ - \frac{0.059}{n} \log Q

Since both reactants (Zn, HgO) and products (ZnO, Hg) are solids, their activities are 1. So $Q = 1$ and $\log Q = 0$ , giving $E = E^\circ$ throughout discharge.

✗ Mistake 3: Saying “the dry cell has a liquid electrolyte that evaporates”

Why it’s wrong:

The dry cell’s electrolyte is actually a paste (not a free liquid). The voltage drop is due to consumption of $\text{NH}_4^+$ ions and formation of $\text{NH}_3$ gas, which polarises the electrode — not evaporation.

How to avoid:

State the real reason:

In a dry cell, $\text{NH}_4^+$ is reduced at the cathode, producing $\text{NH}_3$ gas. This gas accumulates and increases internal resistance, lowering the potential. In a mercury cell, no gas is produced — the products are solids (ZnO) and liquid mercury.

✗ Mistake 4: Forgetting to mention the amalgam formation

Why it’s wrong: …

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Q.Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?

Concept understanding — Standard Electrode Potentials

Standard Electrode Potentials: A Number for "How Badly It Wants Electrons"

The Core Idea

Reading the Table

Predicting Whether a Reaction Goes

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

Concept: Electrode Potential and Electrolyte Concentration Stability

Method: Nernst Equation Analysis

✗ Mistake 1: Saying “the electrolyte does not get used up”

✗ Mistake 2: Confusing “constant potential” with “no internal resistance change”

✗ Mistake 3: Saying “the dry cell has a liquid electrolyte that evaporates”

✗ Mistake 4: Forgetting to mention the amalgam formation

Q.Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?

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