Chemistry

Chemistry

Problem

Given ECr3+/Cr∘=−0.74 VE_{\text{Cr}^{3+}/\text{Cr}}^{\circ} = -0.74 \text{ V} ; EMnO4−/Mn2+∘=1.51 VE_{\text{MnO}_4^-/\text{Mn}^{2+}}^{\circ} = 1.51 \text{ V} ECr2O72−/Cr3+∘=1.33 VE_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}}^{\circ} = 1.33 \text{ V} ; ECl2/Cl−∘=1.36 VE_{\text{Cl}_2/\text{Cl}^-}^{\circ} = 1.36 \text{ V} Based on the data given above, the strongest oxidising agent will be : a. Mn2+\text{Mn}^{2+} b. MnO4−\text{MnO}_4^- c. Cl−\text{Cl}^- d. Cr3+\text{Cr}^{3+}
Intuitive explanation

The strongest oxidizing agent is the species most eager to gain electrons—the one with the highest (most positive) reduction potential. Compare all the reduction half-reactions: MnO4−\text{MnO}_4^- wants electrons at +1.51+1.51 V, Cr2O72−\text{Cr}_2\text{O}_7^{2-} at +1.33+1.33 V, Cl2\text{Cl}_2 at +1.36+1.36 V, and Cr3+\text{Cr}^{3+} at −0.74-0.74 V. The higher the E°E° value, the stronger the pull for electrons—so identify which species has the largest positive potential.

Intuitive explanation

The strongest oxidizing agent is the species most eager to gain electrons—the one with the highest (most positive) reduction potential. Compare all the reduction half-reactions: MnO4−\text{MnO}_4^- wants electrons at +1.51+1.51 V, Cr2O72−\text{Cr}_2\text{O}_7^{2-} at +1.33+1.33 V, Cl2\text{Cl}_2 at +1.36+1.36 V, and Cr3+\text{Cr}^{3+} at −0.74-0.74 V. The higher the E°E° value, the stronger the pull for electrons—so identify which species has the largest positive potential.

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