Electrochemistry 1 Question 4
4. Given, that $E_{\mathrm{O}{2} / \mathrm{H}{2} \mathrm{O}}^{\ominus}=+1.23 \mathrm{~V}$;
$$ \begin{aligned} E_{\mathrm{S}{2} \mathrm{O}{8}^{2-} / \mathrm{SO}{4}^{2-}}^{\ominus} & =2.05 \mathrm{~V} \ E{\mathrm{Br}{2} / \mathrm{Br}^{\ominus}}^{\ominus} & =+1.09 \mathrm{~V} \ E{\mathrm{Au}^{3+} / \mathrm{Au}}^{\ominus} & =+1.4 \mathrm{~V} \end{aligned} $$
The strongest oxidising agent is
(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 oxidising agent among the given option is as follows:
$$ \mathrm{S}{2} \mathrm{O}{8}^{2-}>\mathrm{Au}^{3+}>\mathrm{O}{2}>\mathrm{Br}{2} $$
