Question

Using the data given below, the strongest reducing agent is: $E_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}}^\circ = 33\text{ V}$, $E_{\text{Cl}_2/\text{Cl}^-}^\circ = 36\text{ V}$, $E_{\text{MnO}_4^-/\text{Mn}^{2+}}^\circ = 51\text{ V}$, $E_{\text{Cr}^{3+}/\text{Cr}}^\circ = -74\text{ V}$}

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

Hint:

To find the strongest reducing agent quickly, look for the most negative or lowest value in the standard reduction potential data list. The conjugate neutral metal atom or lower-valence ion matching that lowest value ($\text{Cr}$ at $-0.74\text{ V}$) is your answer.

Answer 1

The Correct Option is A

Concept: The standard reduction potential ($E^\circ$) measures the tendency of a chemical species to gain electrons and be reduced.
• A high, positive $E^\circ$ value indicates a strong tendency to gain electrons, making the species a powerful oxidizing agent.
• A low, negative $E^\circ$ value indicates a weak tendency to be reduced. Consequently, its conjugate oxidized species will readily lose electrons, functioning as a powerful reducing agent.

Step 1: Identify the reducing species and compare reduction potentials.
A reducing agent undergoes oxidation by donating electrons. Let's look at the standard reduction potential values provided:
• $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad E^\circ = +1.51\text{ V}$
• $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \quad E^\circ = +1.36\text{ V}$
• $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \quad E^\circ = +1.33\text{ V}$
• $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ = -0.74\text{ V}$

Step 2: Select the strongest reducing agent based on the lowest potential.
The reduction half-reaction for chromium exhibits the lowest potential value ($E^\circ = -0.74\text{ V}$). This large negative value shows that $\text{Cr}^{3+}$ has little tendency to gain electrons, meaning elemental metallic chromium ($\text{Cr}$) readily loses electrons to undergo oxidation ($\text{Cr} \rightarrow \text{Cr}^{3+} + 3e^-$). Therefore, elemental $\text{Cr}$ is the strongest reducing agent in this group.