Electro Chemistry Part 3

Concept:A species acts as a reducing agent by getting oxidised (losing electrons).For a half‑cell, a more negative standard reduction potential ($E^\circ_{\text{red}}$) means its reduced form is a stronger reducing agent.Equivalently, the higher the standard oxidation potential ($E^\circ_{\text{ox}} = -E^\circ_{\text{red}}$), the stronger the reducing power.Explanation:Write the given reduction half‑reactions with their $E^\circ_{\text{red}}$ values.1. $\mathrm{Al}^{3+} + 3e^- \rightarrow \mathrm{Al}(s)$, $E^\circ_{\text{red}} = -1.66\ \text{V}$Oxidation: $\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+} + 3e^-$, $E^\circ_{\text{ox}} = +1.66\ \text{V}$2. $\mathrm{Cr}^{3+} + 3e^- \rightarrow \mathrm{Cr}(s)$, $E^\circ_{\text{red}} = -0.74\ \text{V}$Oxidation: $\mathrm{Cr}(s) \rightarrow \mathrm{Cr}^{3+} + 3e^-$, $E^\circ_{\text{ox}} = +0.74\ \text{V}$3. $\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}$, $E^\circ_{\text{red}} = +0.77\ \text{V}$Oxidation: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-$, $E^\circ_{\text{ox}} = -0.77\ \text{V}$4. $\mathrm{Co}^{3+} + e^- \rightarrow \mathrm{Co}^{2+}$, $E^\circ_{\text{red}} = +1.81\ \text{V}$Oxidation: $\mathrm{Co}^{2+} \rightarrow \mathrm{Co}^{3+} + e^-$, $E^\circ_{\text{ox}} = -1.81\ \text{V}$Arrange the reducing agents ($\mathrm{Al}$, $\mathrm{Cr}$, $\mathrm{Fe}^{2+}$, $\mathrm{Co}^{2+}$) in decreasing order of $E^\circ_{\text{ox}}$:$+1.66\ \text{V} > +0.74\ \text{V} > -0.77\ \text{V} > -1.81\ \text{V}$Thus, the tendency to act as a reducing agent decreases as: $\mathrm{Al} > \mathrm{Cr} > \mathrm{Fe}^{2+} > \mathrm{Co}^{2+}$.Answer:Option D: $\mathrm{Al} > \mathrm{Cr} > \mathrm{Fe}^{2+} > \mathrm{Co}^{2+}$