Question

Given : $E^{\circ}_{Au^{3+}/Au} = 1.4 V , E^{\circ}_{O_2/H_2O} = 1.23 V, E^{\circ}_{Br_2/Br^{-}} = 1.09 V , E^{\circ}_{S_2O_8^{2-} /SO_{4}^{-2} } = 2.08 V $
Which will be best oxidising agent ?

• A. $\ce{Au^{3+}}$
• B. $O_2$
• C. $\ce{Br_{2}}$
• D. $\ce{S_{2} O_{8}^{-2}}$

Answer 1

Answer: (D)

The species which has highest value of reduction potential will be best oxidising agent. Hence $\ce{S_{2} O_{8}^{-2}}$