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JEE MAIN 2019
08-04-2019 S1
Question
Given that $\mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2 \mathrm{O}}^{\ominus}=+1.23 \mathrm{~V}$; $$ \begin{aligned} & \mathrm{E}_{\mathrm{S}_2 \mathrm{O}_8^{2-} / \mathrm{SO}_4^{2-}}^{\odot}=2.05 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Br}_2 / \mathrm{Br}^{-}}^{\odot}=+1.09 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{\odot}=+1.4 \mathrm{~V} \end{aligned} $$
The strongest oxidizing agent is
Select the correct option:
A
$\mathrm{Br}_2$
B
$\mathrm{Au}^{3+}$
C
$\mathrm{S}_2 \mathrm{O}_8^{2-}$
D
$\mathrm{O}_2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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Solution
More positive is the reduction potential stronger is the oxidising agent. Reduction potential is maximum for $\mathrm{S}_2 \mathrm{O}_8{ }^{2-}$
Question Tags
JEE Main
Chemistry
Easy
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