Q.Assertion: Mercury cell does not give steady potential.
Reason: In the cell reaction, ions are not involved in solution.

Concept: Cell Representation & Nernst Equation — A steady potential requires constant ion concentration in the electrolyte. The Nernst equation shows that if ion activities change, the cell potential drifts.

Reasoning:

1. The mercury cell uses a paste of $\mathrm{Zn/Hg}$ (anode) and $\mathrm{HgO/C}$ (cathode) with a $\mathrm{KOH}$ electrolyte. The overall reaction is: $\mathrm{Zn(s) + HgO(s) \rightarrow ZnO(s) + Hg(l)}$ — no ions appear in the overall equation. …

The mercury cell (also called the Ruben-Mallory cell) is a primary cell known for its exceptionally constant voltage throughout its life. To understand why, we need to look at how it works and compare it with other cells like the Leclanché cell.

The key insight lies in the Nernst equation. For any electrochemical cell, the cell potential depends on the concentrations (or activities) of the ions involved in the half-reactions. If those concentrations change as the cell discharges, the voltage drifts. If they stay constant, the voltage stays constant.

Let's examine the mercury cell step by step.

1. The cell reaction of a mercury cell

   The anode is zinc amalgam (zinc dissolved in mercury), and the cathode is mercury(II) oxide mixed with carbon (for conductivity). The electrolyte is a paste of KOH (potassium hydroxide).

   The half-reactions are:

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

   Cathode: $\text{HgO(s)} + \text{H}_2\text{O} + 2e^- \rightarrow \text{Hg(l)} + 2\text{OH}^-$

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

   Notice something crucial: the hydroxide ions ($\text{OH}^-$) appear on both sides — they are consumed at the anode and regenerated at the cathode in equal amounts. Their concentration in the paste remains constant throughout discharge. No ions are consumed or produced in the net reaction.

2. Why this gives a steady potential

   The Nernst equation for a cell is:

where $Q$ is the reaction quotient. For the mercury cell, $Q$ involves only solids and liquids (Zn, HgO, ZnO, Hg) — all of which have activity = 1. There are no dissolved ions whose concentrations change. Therefore $Q = 1$ always, and $\ln Q = 0$.

So $E = E^\circ$ at all times — the voltage is constant until the cell is exhausted.

3. The assertion is false

   The assertion says: "Mercury cell does not give steady potential." This is the opposite of the truth. The mercury cell is famous for giving a very steady potential (about 1.35 V). So the assertion is false.

4. The reason is true

   The reason says: "In the cell reaction, ions are not involved in solution." This is correct — as we saw, the net reaction involves only solids and liquids, and the $\text{OH}^-$ concentration remains constant. No ions are consumed or produced in the overall reaction.

5. Connecting them

   Since the assertion is false and the reason is true, they cannot be related in the way the options describe. The correct option is the one where both are false — however, the reason itself is a true statement. Let's check the options carefully:

   • (i) Both true, reason correct explanation — no, assertion false. …

Method: Cause-and-Effect Analysis of the Mercury Cell's Steady Potential

Concept: Whether a cell's voltage stays steady or drifts during discharge depends on whether the concentrations/activities of the species in its net cell reaction change over time — this follows directly from the Nernst equation.

Step 1 — Write the mercury cell's half-reactions and net reaction

Step 2 — Check the assertion: does the mercury cell give a steady potential?

Notice $OH^-$ is consumed at the anode and regenerated in equal amount at the cathode — its concentration in the KOH paste never changes as the cell discharges. The net reaction itself involves only solids and a liquid (Zn(Hg), HgO, ZnO, Hg), all of fixed activity = 1.

By the Nernst equation, $E = E^\circ - \frac{RT}{nF}\ln Q$. Since every species in $Q$ has activity 1, $Q = 1$ and $\ln Q = 0$ throughout discharge, so $E = E^\circ$ stays constant. The mercury cell is, in fact, well known for giving a very steady potential (~1.35 V) for its entire working life.

So the assertion — "Mercury cell does not give steady potential" — is FALSE.

Step 3 — Check the reason …

Here’s a breakdown of the common mistakes students make on this Assertion-Reason question about the Mercury cell, along with how to avoid each.

Mistake 1: Thinking the Mercury cell gives an unsteady potential

Why students make this mistake:

They confuse the mercury cell with a dry cell (Leclanché cell). In a dry cell, the voltage drops as it is used because the internal resistance increases due to the accumulation of products like $NH_3$ and $Zn(NH_3)_2Cl_2$.

How to avoid:

Remember the key feature of the mercury cell:

• The electrolyte is a paste of KOH and ZnO, not a solution.
• The overall reaction does not involve ions in solution, so the concentration of the electrolyte remains constant.
• This gives a steady potential (~1.35 V) throughout its life.

Correct fact: The mercury cell does give a steady potential. So the Assertion is false.

Mistake 2: Assuming the Reason is false because “ions are involved”

Why students make this mistake:

They see the word “ions” and think that in any electrochemical cell, ions must be involved. They forget that the mercury cell uses a paste (not a solution) and the reaction occurs without changing the ionic concentration.

How to avoid:

Check the actual cell reaction:

• Anode: $Zn(Hg) + 2OH^- \rightarrow ZnO(s) + H_2O + 2e^-$
• Cathode: $HgO(s) + H_2O + 2e^- \rightarrow Hg(l) + 2OH^-$
• Overall: $Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l)$

Notice:

• $OH^-$ ions are consumed at the anode and regenerated at the cathode.
• The net reaction has no free ions — the $OH^-$ concentration stays constant.
• Hence, the Reason is true: ions are not involved in the net cell reaction.

Correct fact: The Reason is true — the net reaction does not involve ions in solution.

Mistake 3: Choosing option (ii) — Both true, Reason not the correct explanation

Why students make this mistake:

They correctly see that the Reason is true, but they incorrectly think the Assertion is also true. They then try to see if the Reason explains the Assertion — and since they think the Assertion is true, they pick (ii).

How to avoid:

Always evaluate Assertion and Reason separately first, before checking the link.

• Assertion: False (steady potential is a feature).
• Reason: True (no ions in net reaction).
• Since Assertion is false, options (i), (ii), and (iii) are all ruled out.

Correct answer: (iv) Both assertion and reason are false.

Mistake 4: Misreading “ions are not involved in solution” as “no ions at all” …