\(E^o_{K^+/K} = -2.93V, \text{ } E^o_{Ag^+/Ag} =0.8V, \text{ }E^o_{Cu^{2+}/Cu} = 0.34V\)
\(E^o_{Ni^{2+}/Ni} = -0.25V, \text{ } E^o_{Li^+/Li} =-3.05V\)
Arrange the following metals in increasing order of their reducing power.
A. K
B. Ag
C. Cu
D. Ni
E. Li
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B < C < D < A < E
The standard reduction potential E° measures how easily the ion is reduced. The lower (more negative) the E°, the harder it is to reduce the ion and the more easily the metal gives up electrons. So reducing power increases as E° becomes more negative.
Arranging the given values from the most positive to the most negative:
Ag (+0.80 V) → Cu (+0.34 V) → Ni (−0.25 V) → K (−2.93 V) → Li (−3.05 V)
Shortcut: lithium, with the most negative E°, is the strongest reducing agent, so the order must end with E. Only Options (1) and (3) do, and Ni (−0.25 V) must come before K (−2.93 V), which leaves Option (3).
NCERT (2026-27 edition):
Measurement of Electrode Potential – …
