Electrochemistry 1 Question 4
4. Given, that $E_{\mathrm{O} _{2} / \mathrm{H} _{2} \mathrm{O}}^{\ominus}=+1.23 \mathrm{V}$;
$ \begin{alignedat} 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) $\mathrm{Au}^{3+}$
(b) $\mathrm{O} _{2}$
(c) $\mathrm{S} _{2} \mathrm{O} _{8}^{2-}$
(d) $\mathrm{Br} _{2}$
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Answer:
Correct Answer: 4. (b)
Solution:
- Higher the standard reduction potential $\left(E_{M^{n+} / M}^{\mathrm{o}}\right)$, better is oxidising agent. Among the given, $E_{\mathrm{S} _{2} \mathrm{O} _{8}^{2-} / \mathrm{SO} _{4}^{2-}}^{\circ}$ is highest, hence $\mathrm{S} _{2} \mathrm{O} _{8}^{2-}$ is the strongest oxidising agent.
The decreasing order of oxidizing agent among the given options is as follows:
$ S${2}$O${8}^{2-}$>Br${2}$>O${2}$>Au$^{3+}$ $
BETA
