Question

Electrode potential data are given below: \[ {Fe}^{3+}({aq}) + e^- \rightarrow {Fe}^{2+}({aq}); E^\circ = +0.77 \, {V} \] \[ {Al}^{3+}({aq}) + 3e^- \rightarrow {Al}({s}); E^\circ = -1.66 \, {V} \] \[ {Br}_2({aq}) + 2e^- \rightarrow 2{Br}^-({aq}); E^\circ = +1.08 \, {V} \] Based on the data, the reducing power of \( {Fe}^{2+} \), Al, and \( {Br}^- \) will increase in the order:

• A. \( {Br}^-<{Fe}^{2+}<{Al} \)
• B. \( {Fe}^{2+}<{Al}<{Br}^- \)
• C. \( {Al}<{Br}^-<{Fe}^{2+} \)
• D. \( {Al}<{Fe}^{2+}<{Br}^- \)

Hint:

A more negative electrode potential corresponds to a stronger reducing agent, as it indicates a greater tendency to donate electrons.

Answer 1

The Correct Option is A

Step 1: The reducing power of a species is related to the magnitude of its reduction potential. A more negative reduction potential indicates a stronger tendency to lose electrons (thus, a stronger reducing agent). 
Step 2: The species with the lowest reduction potential will be the best reducing agent: 
- \( {Fe}^{3+} + e^- \rightarrow {Fe}^{2+} \), with \( E^\circ = +0.77 \, {V} \), indicates that \( {Fe}^{2+} \) can be reduced to \( {Fe}^{3+} \), so \( {Fe}^{2+} \) is a relatively weak reducing agent. 
- \( {Al}^{3+} + 3e^- \rightarrow {Al} \), with \( E^\circ = -1.66 \, {V} \), indicates that Al is the strongest reducing agent due to the very negative reduction potential. 
- \( {Br}_2 + 2e^- \rightarrow 2{Br}^- \), with \( E^\circ = +1.08 \, {V} \), shows that \( {Br}^- \) has the weakest reducing power among the three. Thus, the correct order of reducing power is: \[ {Br}^-<{Fe}^{2+}<{Al} \]