The E°_{M^{3+}/M^{2+}} values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 V, respectivel

Q: Why are the given potentials considered reduction potentials?

By IUPAC convention, standard electrode potentials are always expressed as reduction potentials unless specified otherwise. The notation $M^{3+}/M^{2+}$ implies the reduction half-reaction: $M^{3+} + e^{-} \rightarrow M^{2+}$.

Q: What does a negative standard reduction potential physically mean?

A negative reduction potential means that the species is less likely to gain electrons (get reduced) compared to the standard hydrogen electrode. Instead, its reduced form is highly likely to lose electrons and get oxidized.

Q: Can I solve this by converting the values to oxidation potentials?

Absolutely! The oxidation potential is simply the negative of the reduction potential ($E^{\circ}_{ox} = -E^{\circ}_{red}$). The metal with the highest positive oxidation potential will be the easiest to oxidize. For Chromium, $E^{\circ}_{ox} = +0.41\text{ V}$, which is the highest among the four.

Q: Why is Manganese's reduction potential so high ($+1.57\text{ V}$)?

Manganese in the $+2$ oxidation state has a stable half-filled $3d^5$ electronic configuration. It strongly resists losing another electron to become $Mn^{3+}$ ($3d^4$). Therefore, $Mn^{3+}$ readily accepts an electron to achieve the stable $Mn^{2+}$ state, resulting in a high positive reduction potential.

Q: Does the number of electrons transferred affect the $E^{\circ}$ value?

No, the standard electrode potential ($E^{\circ}$) is an intensive property. It depends on the nature of the substance and its concentration, but not on the stoichiometric coefficients or the number of electrons in the balanced half-reaction.

Question

The $E^{\circ}_{M^{3+}/M^{2+}}$ values for Cr, Mn, Fe and Co are $-0.41, +1.57, +0.77$ and $+1.97	ext{ V}$, respectively. For which one of these metals, the change in oxidation state from $+2$ to $+3$ is easiest ?

Options:

(A) Cr

(B) Mn

(C) Fe

(D) Co

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a battlefield where transition metals are fighting over electrons. Some metals are desperate to grab electrons, while others are more than happy to give them away. In electrochemistry, we measure this "desperation" using the **Standard Reduction Potential** ($E^{\circ}$).

In our problem, we are given the $E^{\circ}_{M^{3+}/M^{2+}}$ values for four metals: Chromium (Cr), Manganese (Mn), Iron (Fe), and Cobalt (Co). By IUPAC convention, these values represent the tendency of the $M^{3+}$ ion to gain an electron and reduce to the $M^{2+}$ state.

$$M^{3+} + e^{-} ightarrow M^{2+}$$

### The Master Equation of Tendency

Here is the golden rule you must always remember: **The more positive the reduction potential, the stronger the tendency to get reduced.**

If a metal has a high positive $E^{\circ}$ value, it acts as a strong oxidizing agent because it loves to pull electrons towards itself. Conversely, if a metal has a low or negative $E^{\circ}$ value, it hates getting reduced. Instead, it prefers the exact opposite—it wants to lose electrons and undergo **oxidation**.

### Final Calculation

The question asks us to find the metal for which the change in oxidation state from $+2$ to $+3$ is the easiest. Let's write down this process:

$$M^{2+} ightarrow M^{3+} + e^{-}$$

This is clearly an oxidation reaction! Therefore, we are looking for the metal that is most easily oxidized. According to our golden rule, the easiest metal to oxidize will be the one with the **lowest (most negative) standard reduction potential**.

Let's compare the given values:
*   **Cr:** $-0.41	ext{ V}$
*   **Mn:** $+1.57	ext{ V}$
*   **Fe:** $+0.77	ext{ V}$
*   **Co:** $+1.97	ext{ V}$

Chromium is the only metal with a negative reduction potential ($-0.41	ext{ V}$), making it the lowest value among the four. This means that $Cr^{2+}$ has the strongest tendency to lose an electron and oxidize to $Cr^{3+}$.

Thus, the oxidation from $+2$ to $+3$ is easiest for **Chromium (Cr)**.

Animated Solution for Chemistry - Electrochemistry: The $E_{M^{3 +} / M^{2 +}}^{\circ}$ values for Cr, Mn, Fe and Co are $- 0.41, + 1.57, + 0.77$ and $+ 1.97 V$ , respectively. For which one of these metals, the change in oxidation state from $+ 2$ to $+ 3$ is easiest ?

Select Answer:

Visualized Solution

$E_{M^{3 +} / M^{2 +}}^{\circ}$

$M^{3 +} + e^{-} \to M^{2 +}$

$E_{r e d}^{\circ} vs Oxidation$

$M^{2 +} \to M^{3 +} + e^{-}$

$Cr^{2 +} \to Cr^{3 +}$

The Sigma Insight: Electrochemical Series, Electrode Potential and EMF

Analyzing the Setup

The Master Equation of Tendency

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EM3+/M2+∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M3++e−→M2+

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2+→M3++e−

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cr2+→Cr3+