Question

Given $E _{ Cl _{2} / Cl ^{-}}^{\circ}=1 . 36\, V,\, E _{ Cr ^{3+} / Cr }^{\circ}=-7 \cdot 74\, V$ $E _{ Cr _{2} O _{7}^{2-} / Cr ^{3+}}^{\circ}=1 .33\, V,\, E _{ MnO _{4}^{-} / Mn ^{2}+}^{\circ}=1. 51\, V$ Among the following, the strongest reducing agent is:

• A. $Cr^{3+}$
• B. $Cl^{-}$
• C. $Cr$
• D. $Mn^{2+}$

Answer 1

The Correct Option is C

To determine the strongest reducing agent among the given options, we need to consider the standard electrode potentials (E^{\circ}) for each species. The reducing agent is the one that gets oxidized and donates electrons easily. Generally, the species with the lowest (or most negative) standard electrode potential is the strongest reducing agent.

1. Identify the given standard electrode potentials:
   • E_{Cl_2/Cl^{-}}^{\circ} = 1.36 \, \text{V}
   • E_{Cr^{3+}/Cr}^{\circ} = -0.74 \, \text{V}
   • E_{Cr_2O_7^{2-}/Cr^{3+}}^{\circ} = 1.33 \, \text{V}
   • E_{MnO_4^{-}/Mn^{2+}}^{\circ} = 1.51 \, \text{V}
2. Standard electrode potential explanation:

   The species with the most negative E^{\circ} value functions as a stronger reducing agent. This is because a more negative potential indicates a greater tendency to lose electrons.

3. Analyzing the given options:
   • Cl^{-}: Not listed specifically but is derived from the reduction of Cl_2, so E_{Cl_2/Cl^{-}}^{\circ} = 1.36 \, \text{V}.
   • Cr^{3+}: Belongs to the E_{Cr_2O_7^{2-}/Cr^{3+}}^{\circ} = 1.33 \, \text{V} and E_{Cr^{3+}/Cr}^{\circ} = -0.74 \, \text{V} redox systems.
   • Cr: From E_{Cr^{3+}/Cr}^{\circ} = -0.74 \, \text{V}.
   • Mn^{2+}: Related to E_{MnO_4^{-}/Mn^{2+}}^{\circ} = 1.51 \, \text{V}.
4. Determining the strongest reducing agent:

   From the electrode potentials, the most negative value is E_{Cr^{3+}/Cr}^{\circ} = -0.74 \, \text{V}. This indicates that chromium metal (Cr) is the strongest reducing agent as it is willing to donate electrons readily compared to the other species.

Thus, the correct answer is: Cr.