Principle used

The strongest oxidizing agent is the species that gets reduced most easily.

Hence, among the given couples, the species with the highest standard reduction potential $E^\Theta$ is the strongest oxidizing agent.

Given standard reduction potentials

The oxidizing species in each reduction half-cell is:

• $O_2$ in $O_2/H_2O$ with $E^\Theta = 1.23\,\text{V}$
• $S_2O_8^{2-}$ in $S_2O_8^{2-}/SO_4^{2-}$ with $E^\Theta = 2.05\,\text{V}$
• $Br_2$ in $Br_2/Br^-$ with $E^\Theta = 1.09\,\text{V}$
• $Au^{3+}$ in $Au^{3+}/Au$ with $E^\Theta = 1.40\,\text{V}$

Compare the values

$2.05 > 1.40 > 1.23 > 1.09$

Therefore, the species with the greatest tendency to be reduced is:

$S_2O_8^{2-}$

Conclusion

So, the strongest oxidizing agent is:

$\boxed{S_2O_8^{2-}}$

Option check

• A: $O_2$ — incorrect
• B: $Au^{3+}$ — incorrect
• C: $Br_2$ — incorrect
• D: $S_2O_8^{2-}$ — correct