Given the half-reactions and $ \mathrm{E}_{\mathrm{red}}^{\circ} $values in the table below, determine which species is the best oxidizing agent.
$\begin{array}{ll}
\mathrm{MnO}_{4}-(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) & \mathrm{E}_{\mathrm{red}=+1.512 \mathrm{~V}}^{\circ} \
4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}-(a q)+3 e^{-} \rightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \mathrm{E}_{\mathrm{red}=+0.964 \mathrm{~V}}^{\circ} \
\mathrm{Fe}^{3+}(a q)+e-\rightarrow \mathrm{Fe}^{2+}(s) & \mathrm{E}_{\mathrm{red}}^{\circ}=+0.771 \mathrm{~V} \
2 \mathrm{H}^{+}(a q)+2 e^{-} \rightarrow \mathrm{H}_{2}(g) & \mathrm{E}_{\text {red }}^{\circ}=0.000 \mathrm{~V}
\end{array}$

Question

Quizplus · Accepted Answer

The species with the highest standard reduction potential is the best oxidizing agent. In this case, the $ \mathrm{MnO}_4^- $ half-reaction has the highest $ \mathrm{E}_{\mathrm{red}}^{\circ} $ value (+1.512 V), indicating that $ \mathrm{MnO}_4^- $ is the strongest oxidizing agent. Therefore, the correct answer is E, which corresponds to $ \mathrm{MnO}_4^- $.

Given the Half-Reactions And Ered∘ \mathrm{E}_{\mathrm{red}}^{\circ} Ered∘​ Values in the Table Below, Determine Which Species Is the Table

Given the Half-Reactions And $\mathrm{E}_{\mathrm{red}}^{\circ}$ Values in the Table Below, Determine Which Species Is the Table