Given :
\( E^{\circ}_{\mathrm{K^+/K}}=-2.93\ \mathrm{V} \),
\( E^{\circ}_{\mathrm{Ag^+/Ag}}=0.80\ \mathrm{V} \),
\( E^{\circ}_{\mathrm{Hg^{2+}/Hg}}=0.79\ \mathrm{V} \),
\( E^{\circ}_{\mathrm{Mg^{2+}/Mg}}=-2.37\ \mathrm{V} \),
\( E^{\circ}_{\mathrm{Cr^{3+}/Cr}}=-0.74\ \mathrm{V} \),
Choose the correct order :
(A) Ag < Hg < Cr < Mg < K (increasing reducing power)
(B) Ag < Hg < Cr < Mg < K (decreasing oxidising power)
(C) Ag > Hg > Cr > Mg > K (decreasing oxidising power)
(D) Hg > Cr > Ag > K > Mg (increasing reactivity)
(E) K > Mg > Cr > Hg > Ag (decreasing reactivity order)
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (E) only
Standard reduction potentials compare the tendency of couples to be reduced. The lower (more negative) the E°, the more easily the metal is oxidised: the metal is a stronger reducing agent and more reactive. The higher (more positive) the E°, the stronger the ion as an oxidising agent. Arranging the given values:
E°/V: \( \ce{Ag^{+}} \)/Ag (+0.80) > \( \ce{Hg^{2+}} \)/Hg (+0.79) > \( \ce{Cr^{3+}} \)/Cr (−0.74) > \( \ce{Mg^{2+}} \)/Mg (−2.37) > \( \ce{K^{+}} \)/K (−2.93)
(A) Ag < Hg < Cr < Mg < K (increasing reducing power) — Correct. Reducing power rises as E° falls, so it increases from Ag (+0.80 V) to K (−2.93 V).
(B) Ag < Hg < Cr < Mg < K (decreasing oxidising power) — Incorrect. The < signs say the oxidising power rises from Ag to K, which is the reverse; \( \ce{Ag^{+}} \) (highest E°) is the strongest oxidising agent of the set.
(C) Ag > Hg > Cr > Mg > K (decreasing oxidising power) — Correct. Oxidising power follows E°: highest for \( \ce{Ag^{+}} \)/Ag, lowest for \( \ce{K^{+}} \)/K.
(D) Hg > Cr > Ag > K > Mg (increasing reactivity) — Incorrect. This sequence does not follow the E° values at all; reactivity increases Ag < Hg < Cr < Mg < K.
(E) K > Mg > Cr > Hg > Ag (decreasing reactivity order) — Correct. K has the most negative E° and is the most reactive; Ag the least.
NCERT (2026-27 edition):
Measurement of Electrode Potential – “The standard electrode potentials are very important and we can extract a lot of useful information from them. The values of standard electrode potentials for some selected half-cell reduction reactions are given in Table 2.1. … It can be seen that the standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. 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.”