Assertion (A): The equilibrium constant of the reaction \( \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \) is very large, about 2 × 1037 at 298 K.
Reason (R): The standard hydrogen electrode is assigned a zero potential at all temperatures.
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Both (A) and (R) are true but (R) is not the correct explanation of (A).
Assertion (A): The equilibrium constant of the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) is very large, about 2 × 1037 at 298 K. — True. For the Daniell cell E°cell = 1.1 V and n = 2:
\( \displaystyle \log K_C = \frac{2 \times 1.1\ \mathrm{V}}{0.059\ \mathrm{V}} = 37.288, \qquad K_C = 2 \times 10^{37} \)
Reason (R): The standard hydrogen electrode is assigned a zero potential at all temperatures. — True. The potential of a single half-cell cannot be measured, so by convention the standard hydrogen electrode is taken as the reference with zero potential, and the potentials of other electrodes are found with respect to it.
Why R does not explain A?
R only fixes the zero of the scale of electrode potentials; E°cell is a difference of two electrode potentials, so it does not depend on where the zero is placed and R cannot tell why this K is so large.
The correct reason is the large positive E°cell of the Daniell cell (1.1 V), because E°cell = (2.303RT/nF) log KC relates the equilibrium constant to the standard potential of the cell.
NCERT (2026-27 edition):
2.3.1 Equilibrium Constant from Nernst Equation – “…KC = 2 × 1037 at 298K.”
\( \displaystyle E^\circ_{\text{cell}} = \frac{2.303RT}{nF}\log K_C \qquad (2.14) \)
“Thus, Eq. (2.14) gives a relationship between equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place.”
2.2.1 Measurement of Electrode Potential – “The potential of individual half-cell cannot be measured. We can measure only the difference between the two half-cell potentials that gives the emf of the cell. If we arbitrarily choose the potential of one electrode (half-cell) then that of the other can be determined with respect to this. According to convention, a half-cell called standard hydrogen electrode … represented by Pt(s)|H2(g)|H+(aq), is assigned a zero potential at all temperatures corresponding to the reaction H+(aq) + e– → ½ H2(g)”