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 standard reduction potential of Zinc is represented by the reaction
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(Zn^{2+}(aq) + 2e^- → Zn (s)\)
A standard reduction potential (standard electrode potential) refers to the reduction half-reaction of an electrode in contact with its ions in aqueous solution at unit concentration (1 M) at 298 K. The oxidised form is written on the left with the electrons gained.
Zinc forms the \( \ce{Zn^2+} \) ion, so two electrons are gained, the ion is in solution and the metal is solid:
\( \displaystyle \ce{Zn^2+(aq) + 2e- -> Zn(s)} \qquad E^\circ = -0.76\ \mathrm{V} \)
So Option (1) represents the standard reduction potential of zinc.
Why the other options are incorrect
Option (2) \(Zn(s) → Zn^+ (aq) + e^-\) — Incorrect. This is an oxidation, and zinc does not form a \( \ce{Zn+} \) ion; its stable ion is \( \ce{Zn^2+} \).
Option (3) \(Zn^{2+} (g) + 2e^-→Zn(s)\) — Incorrect. Electrode potentials refer to ions in solution, not to gaseous ions.
Option (4) \(Zn^{2+} (s) + 2e^-→Zn(s)\) — Incorrect. The ion must be in aqueous solution; a solid ion cannot form an electrode–solution half-cell.
NCERT (2026-27 edition):
Measurement of Electrode Potential – “… If the concentrations of the oxidised and the reduced forms of the species in the right hand half-cell are unity, then the cell potential is equal to standard electrode potential, \( E^{o}_{R} \) of the given half-cell. … Similarly, the measured emf of the cell: Pt(s) | \( \ce{H2} \)(g, 1 bar) | \( \ce{H+} \)(aq, 1 M) || \( \ce{Zn^2+} \)(aq, 1 M) | Zn is −0.76 V corresponding to the standard electrode potential of the half-cell reaction: Zn2+ (aq, 1 M) + 2e– → Zn(s)”