The total EMF of a voltaic cell is greater than zero. Which of the following statements is true?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\Delta G < 0\) and \(E_{cell}\) is intensive property
The Gibbs energy change of a cell reaction is related to its emf by
\( \Delta_r G = -nFE_{cell} \)
n and F are positive, so if \( E_{cell} \gt 0 \), then \( \Delta_r G \) is negative and the cell reaction is spontaneous. \( E_{cell} \) is an intensive property: it does not depend on how much of the reaction is written or carried out. \( \Delta_r G \) is extensive: for \( \ce{Zn + Cu^2+ -> Zn^2+ + Cu} \) it is \( -2FE_{cell} \), but for the reaction written twice it is \( -4FE_{cell} \).
Why the other options are incorrect
Option (1) \(\Delta G = 0\) and \(E_{cell}\) is extensive property — Incorrect. \( \Delta G = 0 \) only when \( E_{cell} = 0 \) (the cell is at equilibrium), and \( E_{cell} \) is not extensive.
Option (2) \(\Delta G > 0\) and \(E_{cell}\) is intensive property — Incorrect. \( \Delta G \gt 0 \) would need \( E_{cell} \lt 0 \), a non-spontaneous cell reaction.
Option (4) \(\Delta G < 0\) and \(E_{cell}\) is extensive property — Incorrect. \( \Delta G \lt 0 \) is right, but \( E_{cell} \) is intensive, not extensive.
NCERT (2026-27 edition):
Electrochemical Cell and Gibbs Energy of the Reaction – “Electrical work done in one second is equal to electrical potential multiplied by total charge passed. If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf of the cell is E and nF is the amount of charge passed and ΔrG is the Gibbs energy of the reaction, then
ΔrG = – nFE(cell) (2.15)
It may be remembered that E(cell) is an intensive parameter but ΔrG is an extensive thermodynamic property and the value depends on n.”