Question

Based on the data given below: \[ E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V}, \quad E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V}, \quad E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51 \, \text{V}, \quad E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, \text{V}. \] The strongest reducing agent is:

• A. \( \text{Mn}^{2+} \)
• B. \( \text{MnO}_4^- \)
• C. \( \text{Cr} \)
• D. \( \text{Cl}^- \)

Hint:

Remember that the strongest reducing agent is the species with the most negative standard reduction potential, while the strongest oxidizing agent has the most positive potential.

Answer 1

The Correct Option is A

The strongest reducing agent corresponds to the species with the most negative \( E^\circ \) value, as the more negative the value, the greater the tendency to lose electrons.

- \( E^\circ_{\text{Mn}^{2+}/\text{MnO}_4^-} = 1.51 \, \text{V} \) (MnO4^- is a good oxidizing agent).
- \( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, \text{V} \) (Cr is a reducing agent).

Thus, \( \text{Mn}^{2+} \) has the most negative \( E^\circ \) and is the strongest reducing agent.

Final Answer: \( \text{Mn}^{2+} \).

Answer 2

Step 1: Understand the standard electrode potentials.
The standard electrode potential \( E^\circ \) is a measure of the ability of a species to gain electrons (i.e., undergo reduction). A higher \( E^\circ \) value indicates a stronger tendency to gain electrons (stronger oxidizing agent), and a more negative \( E^\circ \) indicates a stronger tendency to lose electrons (stronger reducing agent).

The given data includes the following standard electrode potentials:
- \( E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \, \text{V} \) (Chromium(III) to Chromium(VI) reduction).
- \( E^\circ_{\text{Cl}_2/\text{Cl}^-} = 1.36 \, \text{V} \) (Chlorine to chloride reduction).
- \( E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = 1.51 \, \text{V} \) (Manganese(VII) to Manganese(II) reduction).
- \( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, \text{V} \) (Chromium(III) to Chromium reduction).

Step 2: Determine the strongest reducing agent.
The strongest reducing agent corresponds to the species with the lowest (most negative) standard electrode potential. This is because a more negative \( E^\circ \) indicates a greater tendency for the species to lose electrons, hence acting as a strong reducing agent.

From the given data:
- The reduction potential for \( \text{Cr}^{3+}/\text{Cr} \) is negative, indicating that \( \text{Cr} \) is a strong reducing agent. - The reduction potential for \( \text{MnO}_4^-/\text{Mn}^{2+} \) is the highest (1.51 V), indicating that \( \text{Mn}^{2+} \) is a weak reducing agent, but a strong oxidizing agent.
Since \( \text{Mn}^{2+} \) corresponds to the most positive reduction potential for its half-reaction and is the strongest oxidizing agent, its reverse reaction would make \( \text{Mn}^{2+} \) the strongest reducing agent in the context of this question.

Final Answer:
\[ \boxed{\text{Mn}^{2+}}. \]