Statement I: Zn2+ is a stronger oxidising agent than Sn2+.
Statement II: Of two metal ions, the one whose metal has the higher (less negative) standard reduction potential is the stronger oxidising agent.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Statement I is false but Statement II is correct
An oxidising agent is itself reduced. The higher (less negative) the standard reduction potential E° of a couple, the greater the tendency of its oxidised form to take up electrons. The values at 298 K are:
\( \displaystyle E^{\circ}_{\ce{Zn^2+/Zn}}=-0.76\,\mathrm{V} \qquad E^{\circ}_{\ce{Sn^2+/Sn}}=-0.14\,\mathrm{V} \)
Statement I: Zn2+ is a stronger oxidising agent than Sn2+. — False. E° for \( \ce{Sn^2+/Sn} \) (−0.14 V) is higher than for \( \ce{Zn^2+/Zn} \) (−0.76 V), so \( \ce{Sn^2+} \) is reduced more readily. \( \ce{Sn^2+} \) is the stronger oxidising agent, not \( \ce{Zn^2+} \).
Statement II: Of two metal ions, the one whose metal has the higher (less negative) standard reduction potential is the stronger oxidising agent. — True. As E° decreases down the series, the oxidising power of the species on the left (the metal ion) decreases; a higher E° means a stronger oxidising agent.
NCERT (2026-27 edition):
Statement II – 2.2.1 Measurement of Electrode Potential – “… It may be seen that as we go from top to bottom in Table 2.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction.”
Statement I – Table 2.1: Standard Electrode Potentials at 298 K
