ChemistryMedium86×since 2002Q989Given that EΘO2/H2O=1.23 V{E^\Theta }_{{O_2}/{H_2}O} = 1.23\,VEΘO2/H2O=1.23V ; EΘS2O82−/SO42−=2.05 V{E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,VEΘS2O82−/SO42−=2.05V EΘBr2/Br−=1.09 V{E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,VEΘBr2/Br−=1.09V EΘAu3+/Au=1.4 V{E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,VEΘAu3+/Au=1.4V The strongest oxidizing agent is :AO₂BAu³⁺CBr₂DS2O82−{S_2}O_8^{2 - }S2O82−Check answerSkip