Read the passage carefully and answer.
When a metal is placed in a solution of its ions, the metal can lose electron and can go in the solution as ion or the metal ion in the solution can take electron from the electrode and get deposited as metal. Thus, the electrode acquires either positive or negative charge with respect to the solution leading to the development of a potential difference between the metal electrode and the solution. This potential difference is called the electrode potential of the metal. The electrode potential of a metal is a measure of relative tendency to undergo oxidation (loss of electron) or reduction (gain of electron). When the concentration of the ions in the solution is unity, the electrode potential is called standard electrode potential. The magnitude of potential depends upon the nature of electrode, concentration of ions in solution and the temperature. In a galvanic cell, the electrode at which oxidation occurs is known as anodic half-cell and the electrode at which reduction occurs is known as cathodic half cell. The emf of the cell in terms of standard reduction potential is given by \( E°_{cell} = E°_{cathode} - E°_{anode} \). The standard free energy change of the redox reaction taking place in the cell is related to \( E°_{cell} \) by \( ΔG°= -nFE° \). A redox reaction would occur spontaneously if the free energy change is negative. The equilibrium constant of the reaction is related to the standard free energy change by \( ΔG°= -RT \ln K \).
The equilibrium constant of a feasible cell reaction is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → greater than one
The standard Gibbs energy change of a cell reaction is linked both to the cell potential and to the equilibrium constant:
\( \displaystyle \Delta G^\circ = -nFE^\circ_\mathrm{cell} \)
\( \displaystyle \Delta G^\circ = -RT\ln K \)
A feasible (spontaneous) cell reaction has a positive \( E^\circ_\mathrm{cell} \) and therefore a negative \( \Delta G^\circ \). Equating the two expressions:
\( \displaystyle -RT\ln K = -nFE^\circ_\mathrm{cell} \)
\( \displaystyle \ln K = \frac{nFE^\circ_\mathrm{cell}}{RT} \)
Since \( n \), \( F \), \( R \) and \( T \) are all positive, a positive \( E^\circ_\mathrm{cell} \) makes \( \ln K \) positive, and so \( K \gt 1 \). So the equilibrium constant of a feasible cell reaction is greater than one: at equilibrium the products are favoured.
Why the other options are incorrect
Option (1) less than one — Incorrect. \( K \lt 1 \) gives a negative \( \ln K \), hence a positive \( \Delta G^\circ \) and a negative \( E^\circ_\mathrm{cell} \): the reaction is not feasible.
Option (3) equal to one — Incorrect. \( K = 1 \) gives \( \Delta G^\circ = 0 \) and \( E^\circ_\mathrm{cell} = 0 \); such a cell has no driving force under standard conditions.
Option (4) zero — Incorrect. \( K \) is a ratio of concentrations, which are never zero at equilibrium, so it cannot be zero.
NCERT (2026-27 edition):
Electrochemical Cell and Gibbs Energy of the Reaction – “… If the concentration of all the reacting species is unity, then \( E_{(cell)} = E^{o}_{(cell)} \) and we have
\( \Delta_r G^{o} = -nFE^{o}_{(cell)} \)
Thus, from the measurement of \( E^{o}_{(cell)} \) we can obtain an important thermodynamic quantity, \( \Delta_r G^{o} \), standard Gibbs energy of the reaction. From the latter we can calculate equilibrium constant by the equation: \( \Delta_r G^{o} = -RT \ln K \).”