Electrochemistry - Result Question 55
4. Given, that $E _{O _2 / H _2 O}^{\ominus}=+1.23 V$;
$$ \begin{aligned} E _{S _2 O _8^{2-} / SO _4^{2-}}^{\ominus} & =2.05 V \ E _{Br _2 / Br^{\ominus}}^{\ominus} & =+1.09 V \ E _{Au^{3+} / Au}^{\ominus} & =+1.4 V \end{aligned} $$
The strongest oxidising agent is fluorine
(2019 Main, 8 April I)
(a) $Au^{3+}$
(b) $O _2$
(c) $S _2 O _8^{2-}$
(d) $Br _2$
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Answer:
Correct Answer: 4. (b)
Solution:
- Higher the standard reduction potential $\left(E _{M^{n+} / M}^{o}\right)$, better is oxidising agent. Among the given, $E _{S _2 O _8^{2-} / SO _4^{2-}}^{\circ}$ is highest, hence $S _2 O _8^{2-}$ is the strongest oxidising agent.
The decreasing order of oxidizing agent among the given options is as follows:
$$ S₂O₈²⁻>Au³⁺>O₂>Br₂ $$
BETA
