For the cell reaction \(\text{Zn(s)} + \text{Cl}_2\text{(g, 1 atm)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)}\), E°cell = 2.12 V. The emf of the cell can be increased by:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Decreasing the concentration of Zn2+ ions
The Nernst equation gives the emf of a cell at any concentration. For this cell (n = 2), with the reaction quotient written from the products over the reactants:
\( \displaystyle E_{cell}=E^\circ_{cell}-\frac{RT}{2F}\ln\frac{[\ce{Zn^2+}][\ce{Cl^-}]^2}{p_{\ce{Cl2}}} \)
The emf rises when the fraction after “ln” becomes smaller. Lowering [\( \ce{Zn^2+} \)], a product, makes the fraction smaller, so Ecell increases. Zinc is the anode here as in the Daniell cell, where a lower [\( \ce{Zn^2+} \)] also raises the emf.
Why the other options are incorrect
Option (1) Increasing the concentration of Zn2+ ions — Incorrect. A larger [\( \ce{Zn^2+} \)] makes the fraction larger, so the emf decreases.
Option (3) Increasing the concentration of Cl− ions — Incorrect. \( \ce{Cl^-} \) is also a product and appears as [\( \ce{Cl^-} \)]2, so raising it lowers the emf.
Option (4) Decreasing the partial pressure of Cl2 — Incorrect. \( \ce{Cl2} \) is a reactant in the denominator; a lower pressure makes the fraction larger and lowers the emf.
NCERT (2026-27 edition):
2.3 Nernst Equation – “… It can be seen that E(cell) depends on the concentration of both Cu2+ and Zn2+ ions. It increases with increase in the concentration of Cu2+ ions and decrease in the concentration of Zn2+ ions.”