Given, E°(Cl2/Cl-) = 1.36 V, E°(Cr3+/Cr) = -0.74 V, E°(Cr2O72-/Cr3+) = 1.33 V, E°(MnO4-/Mn2+) = 1.51

Q: Why do we look for the lowest reduction potential for the strongest reducing agent?

A reducing agent gets oxidized. The tendency to get oxidized is given by the oxidation potential. Since $E^\circ_{ox} = -E^\circ_{red}$, a lower (more negative) reduction potential means a higher (more positive) oxidation potential, indicating a stronger tendency to act as a reducing agent.

Q: How do I know which species in the couple is the reducing agent?

In a standard reduction couple written as $Ox/Red$ (e.g., $Cr^{3+}/Cr$), the species on the right ($Red$) is the reduced form and acts as the reducing agent, while the species on the left ($Ox$) is the oxidized form and acts as the oxidizing agent.

Q: Is $Cr^{3+}$ a reducing agent?

$Cr^{3+}$ can act as a reducing agent if it gets further oxidized (e.g., to $Cr_2O_7^{2-}$). In the given data, the couple $Cr_2O_7^{2-}/Cr^{3+}$ has $E^\circ = 1.33\text{ V}$, so the oxidation of $Cr^{3+}$ to $Cr_2O_7^{2-}$ has an oxidation potential of $-1.33\text{ V}$, which is much lower than the oxidation potential of $Cr$ to $Cr^{3+}$ ($+0.74\text{ V}$).

Q: What if the question asked for the strongest oxidizing agent?

You would look for the highest standard reduction potential. In this case, $MnO_4^-$ has the highest $E^\circ_{red}$ ($1.51\text{ V}$), making it the strongest oxidizing agent among the given species.

Q: Does the number of electrons transferred affect the reducing power?

No, the standard electrode potential $E^\circ$ is an intensive property and does not depend on the stoichiometry or the number of electrons transferred in the balanced half-reaction. It solely reflects the inherent thermodynamic tendency of the process.

Question

Given, $E^\circ_{Cl_2/Cl^-} = 1.36	ext{ V}$, $E^\circ_{Cr^{3+}/Cr} = -0.74	ext{ V}$, $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33	ext{ V}$, $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51	ext{ V}$. Among the following, the strongest reducing agent is

Options:

(A) Cr

(B) $Mn^{2+}$

(C) $Cr^{3+}$

(D) $Cl^-$

Sigma Percentile · Accepted Answer

### The Core Concept: Redox and Agents

To master questions involving the electrochemical series, we must first build a rock-solid foundation on what oxidizing and reducing agents actually do. A **reducing agent** is a chemical species that brings about reduction in another substance. But how does it do that? It does so by donating its own electrons. In the process of giving away electrons, the reducing agent itself undergoes **oxidation**.

Therefore, the strength of a reducing agent is directly proportional to its tendency to get oxidized. The more eagerly a substance wants to lose electrons, the stronger a reducing agent it is.

### Decoding the Electrochemical Series

The electrochemical series is essentially a leaderboard of standard reduction potentials ($E^\circ_{red}$). These values tell us how much a species "wants" to gain electrons and get reduced.

If a species has a very high, positive $E^\circ_{red}$, it is desperate to gain electrons. This makes it a fantastic oxidizing agent. Conversely, if a species has a very low, or highly negative $E^\circ_{red}$, it absolutely hates gaining electrons. Instead, it prefers the reverse process: losing electrons (oxidation).

Mathematically, the standard oxidation potential is simply the negative of the standard reduction potential:

$$E^\circ_{ox} = -E^\circ_{red}$$

Thus, to find the **strongest reducing agent**, we must look for the species with the **highest oxidation potential**, which corresponds to the **lowest standard reduction potential**.

### Analyzing the Given Data

Let's line up the standard reduction potentials provided in the problem:

1. $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51	ext{ V}$
2. $E^\circ_{Cl_2/Cl^-} = 1.36	ext{ V}$
3. $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33	ext{ V}$
4. $E^\circ_{Cr^{3+}/Cr} = -0.74	ext{ V}$

When we compare these values, it is glaringly obvious that the $Cr^{3+}/Cr$ couple has the lowest reduction potential at $-0.74	ext{ V}$.

### The Final Verdict

Because the $Cr^{3+}/Cr$ couple has the lowest reduction potential, the reverse reaction (oxidation of $Cr$ to $Cr^{3+}$) has the highest oxidation potential ($+0.74	ext{ V}$).

$$Cr ightarrow Cr^{3+} + 3e^- \quad (E^\circ_{ox} = +0.74	ext{ V})$$

This means that among all the species listed, solid Chromium ($Cr$) has the greatest thermodynamic drive to lose electrons. Consequently, **Chromium ($Cr$) is the strongest reducing agent**.

As a thought experiment, if the question had asked for the strongest oxidizing agent, we would look at the top of our leaderboard. The $MnO_4^-/Mn^{2+}$ couple has the highest reduction potential ($1.51	ext{ V}$), meaning the Permanganate ion ($MnO_4^-$) has the strongest tendency to gain electrons, making it the most powerful oxidizing agent in this set.

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Visualized Solution

$E_{r e d}^{\circ} Values$

$Reducing Power Logic$

$Comparing Potentials$

$Identifying the Lowest E_{r e d}^{\circ}$

$Final Answer$

$The Way Forward$

The Sigma Insight: Electrochemical Cells

The Core Concept

Decoding the Electrochemical Series

Analyzing the Given Data

The Final Verdict

Similar Questions

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Given that the standard potentials ( $E^{\circ}$ ) of $C u^{2 +} / C u$ and $C u^{+} / C u$ are $0.34 V$ and $0.522 V$ respectively, the $E^{\circ}$ of $C u^{2 +} / C u^{+}$ is

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ered∘​ Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reducing Power Logic

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Potentials

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying the Lowest Ered∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward