Consider the following $$\begin{aligned} & \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(\mathrm{~s}) ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V} \\ & \mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}(\mathrm{~s}) ; \mathrm{E}^{\circ}=-2.87 \mathrm{~V} \\ & \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}(\mathrm{~s}) ; \mathrm{E}^{\circ}=-2.36 \mathrm{~V} \\ & \mathrm{Ni}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{~s}) ; \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \end{aligned}$$ The reducing power of the metals increases in the order

(B) $\mathrm{Ni}<\mathrm{Zn}<\mathrm{Mg}<\mathrm{Ca}$

(A) $\mathrm{Ca}<\mathrm{Zn}<\mathrm{Mg}<\mathrm{Ni}$

(C) $\mathrm{Zn}<\mathrm{Mg}<\mathrm{Ni}<\mathrm{Ca}$

(D) $\mathrm{Ca}<\mathrm{Mg}<\mathrm{Zn}<\mathrm{Ni}$

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Redox Reactions and Electrochemistry

Consider the following

Zn^{2 +} + 2 e^{-} ⟶ Zn (s); E^{\circ} = - 0.76 V Ca^{2 +} + 2 e^{-} ⟶ Ca (s); E^{\circ} = - 2.87 V Mg^{2 +} + 2 e^{-} ⟶ Mg (s); E^{\circ} = - 2.36 V Ni^{2 +} + 2 e^{-} ⟶ Ni (s); E^{\circ} = - 0.25 V

The reducing power of the metals increases in the order

(A) $Ca < Zn < Mg < Ni$
(B) $Ni < Zn < Mg < Ca$ ✓ Correct
(C) $Zn < Mg < Ni < Ca$
(D) $Ca < Mg < Zn < Ni$

Answer: (B)

Solution

Reducing power reflects how readily a metal gives up electrons, which is greatest when its standard reduction potential is most negative. Lining up the figures, calcium at

- 2.87

V is the most negative and so the strongest reducer, then magnesium at

- 2.36

V, then zinc at

- 0.76

V, with nickel at

- 0.25

V the weakest. Reading from weakest to strongest, the reducing power rises as

Ni < Zn < Mg < Ca

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Redox Reactions and Electrochemistry

Solution

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