Given that
\(E^{\ominus}_{\mathrm{O_2/H_2O}} = 1.23\ \mathrm{V}\);
\(E^{\ominus}_{\mathrm{S_2O_8^{2-}/SO_4^{2-}}} = 2.05\ \mathrm{V}\)
\(E^{\ominus}_{\mathrm{Br_2/Br^-}} = 1.09\ \mathrm{V}\)
\(E^{\ominus}_{\mathrm{Au^{3+}/Au}} = 1.4\ \mathrm{V}\)
The strongest oxidizing agent is :
\(E^{\ominus}_{\mathrm{O_2/H_2O}} = 1.23\ \mathrm{V}\);
\(E^{\ominus}_{\mathrm{S_2O_8^{2-}/SO_4^{2-}}} = 2.05\ \mathrm{V}\)
\(E^{\ominus}_{\mathrm{Br_2/Br^-}} = 1.09\ \mathrm{V}\)
\(E^{\ominus}_{\mathrm{Au^{3+}/Au}} = 1.4\ \mathrm{V}\)
The strongest oxidizing agent is :
Show Hint
- \(\mathrm{O_2}\)
- \(\mathrm{Au^{3+}}\)
- \(\mathrm{Br_2}\)
- \(\mathrm{S_2O_8^{2-}}\)
The Correct Option is D
Solution and Explanation
Compare: \(\mathrm{O_2}\) (1.23 V), \(\mathrm{S_2O_8^{2-}}\) (2.05 V), \(\mathrm{Br_2}\) (1.09 V), \(\mathrm{Au^{3+}}\) (1.4 V).
Order of oxidizing power: \(\mathrm{S_2O_8^{2-}} \gt \mathrm{Au^{3+}} \gt \mathrm{O_2} \gt \mathrm{Br_2}\).
The highest value is 2.05 V, belonging to \(\mathrm{S_2O_8^{2-}}\).
Answer: (4) \(\mathrm{S_2O_8^{2-}}\)
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