Given that $\mathrm{E}_{\mathrm{O}_{2}/\mathrm{H}_{2}\mathrm{O}}^{0}=+1.23 \mathrm{~V}$; $\mathrm{E}_{\mathrm{S}_{2}\mathrm{O}_{8}^{2-}/\mathrm{SO}_{4}^{2-}}^{0}=2.05 \mathrm{~V}$; $\mathrm{E}_{\mathrm{Br}_{2}/\mathrm{Br}^{-}}^{0}=+1.09 \mathrm{~V}$; $\mathrm{E}_{\mathrm{Au}^{3+}/\mathrm{Au}}^{0}=1.4 \mathrm{~V}$. The strongest oxidizing agent is:

(B) $\mathrm{S}_{2}\mathrm{O}_{8}^{2-}$

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Redox Reactions and Electrochemistry

Given that

E_{O_{2} / H_{2} O}^{0} = + 1.23 V

;

E_{S_{2} O_{8}^{2 -} / SO_{4}^{2 -}}^{0} = 2.05 V

;

E_{Br_{2} / Br^{-}}^{0} = + 1.09 V

;

E_{Au^{3 +} / Au}^{0} = 1.4 V

. The strongest oxidizing agent is:

(A) $O_{2}$
(B) $S_{2} O_{8}^{2 -}$ ✓ Correct
(C) $Au^{3 +}$
(D) $Br_{2}$

Answer: (B)

Solution

The stronger an oxidising agent, the more positive its standard reduction potential, since a higher potential means a greater pull for electrons. Comparing the four couples,

O_{2} / H_{2} O

+ 1.23

B r_{2} /B r^{-}

+ 1.09

V and

A u^{3 +} /Au

+ 1.4

V all lie below the peroxydisulphate couple

S_{2} O_{8}^{2 -} /S O_{4}^{2 -}

+ 2.05

V. With the highest reduction potential,

S_{2} O_{8}^{2 -}

is the strongest oxidising agent.

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Redox Reactions and Electrochemistry

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