Chemistry · 2019
JEE Main · 8 April 2019, Shift 1 · Q58
Given that E_O_2 / H_2 O^ =+1.23 V; E_S_2 O_8^2- / SO_4^2-^ =2.05 V; E_Br_2 / Br^-^ =+1.09 V; E_Au^3+ / Au^ =+1.4 V The strongest oxidizing agent is:
Given that $\displaystyle \mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\ominus}=+1.23 \mathrm{~V}$;
$$\begin{aligned}
& \mathrm{E}_{\mathrm{S}_2 \mathrm{O}_8^{2-} / \mathrm{SO}_4^{2-}}^{\ominus}=2.05 \mathrm{~V} \\
& \mathrm{E}_{\mathrm{Br}_2 / \mathrm{Br}^{-}}^{\ominus}=+1.09 \mathrm{~V} ; \\
& \mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{\ominus}=+1.4 \mathrm{~V}
\end{aligned}
$$
The strongest oxidizing agent is :
Official answer
(1)
$\displaystyle \mathrm{S}_2 \mathrm{O}_8^{2-}$
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JEE Main 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.