Assertion (A): When all the coefficients in the balanced equation of a cell reaction are doubled, the value of ΔrG for the reaction is doubled, but Ecell remains the same.
Reason (R): Ecell is an intensive parameter, whereas ΔrG is an extensive thermodynamic property.
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both (A) and (R) are true and (R) is the correct explanation of (A).
The Gibbs energy of a cell reaction is linked to the cell potential by \( \Delta_r G = -nFE_{cell} \), where n is the number of moles of electrons in the reaction as it is written.
Assertion (A): When all the coefficients in the balanced equation of a cell reaction are doubled, the value of ΔrG for the reaction is doubled, but Ecell remains the same. — True. For the Daniell cell, \( \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \) has n = 2, so ΔrG = −2FEcell. Written as \( \ce{2Zn(s) + 2Cu^2+(aq) -> 2Zn^2+(aq) + 2Cu(s)} \), n = 4 and ΔrG = −4FEcell, twice as much, while the same cell still shows the same Ecell.
Reason (R): Ecell is an intensive parameter, whereas ΔrG is an extensive thermodynamic property. — True. The value of ΔrG depends on n, that is, on the amount of reaction written; Ecell does not.
Why R explains A?
Doubling the equation doubles the amount of reaction and the charge nF passed.
An extensive property such as ΔrG grows with it, but an intensive parameter such as Ecell does not change with the amount, so R gives the cause of A.
NCERT (2026-27 edition):
2.3.2 Electrochemical Cell and Gibbs Energy of the Reaction – “…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. Thus, if we write the reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (2.1)
ΔrG = – 2FE(cell)
but when we write the reaction
2 Zn (s) + 2 Cu2+(aq) → 2 Zn2+(aq) + 2Cu(s)
ΔrG = – 4FE(cell)”