Given below are the half-cell reactions Mn2+ + 2e- - Mn, E° = -1.18 V; 2(Mn3+ + e- - Mn2+), E° = +

Q: Why can't I just subtract the $E^\circ$ values directly like $E^\circ = E_1^\circ - E_2^\circ$?

Standard electrode potential ($E^\circ$) is an intensive property, meaning it doesn't depend on the amount of substance. When you combine half-reactions where the number of electrons doesn't perfectly cancel out to form a simple galvanic cell, you must use an extensive property like Gibbs free energy ($\Delta G^\circ$) to ensure energy conservation.

Q: How do I know the number of electrons ($n$) for the target reaction?

Look at the balanced target reaction: $3Mn^{2+} \longrightarrow Mn + 2Mn^{3+}$. You can split this back into oxidation and reduction. The reduction is $Mn^{2+} + 2e^- \longrightarrow Mn$ (2 electrons). The oxidation is $2Mn^{2+} \longrightarrow 2Mn^{3+} + 2e^-$ (2 electrons). Thus, the net electron transfer $n$ is 2.

Q: What does a negative $E^\circ$ value physically mean for this reaction?

A negative standard cell potential indicates that the standard Gibbs free energy change ($\Delta G^\circ$) is positive. This means the reaction is non-spontaneous under standard conditions. The reactants are more stable than the products.

Q: Why did we multiply the second reaction's $E^\circ$ by 2 when calculating $\Delta G^\circ$?

We didn't multiply $E^\circ$ by 2. The formula is $\Delta G^\circ = -nFE^\circ$. For the second reaction, $2Mn^{3+} + 2e^- \longrightarrow 2Mn^{2+}$, the number of electrons transferred ($n$) is 2. So we multiply the $E^\circ$ by $n=2$ and Faraday's constant $F$ to get the total energy change.

Q: Is $Mn^{2+}$ always this stable?

Yes, in aqueous solutions, $Mn^{2+}$ is exceptionally stable. This is largely due to its $d^5$ electron configuration, which is exactly half-filled. Half-filled subshells offer extra exchange energy and symmetrical charge distribution, making it energetically unfavorable to either add or remove an electron.

Question

Given below are the half-cell reactions
$Mn^{2+} + 2e^- \longrightarrow Mn, E^\circ = -1.18 	ext{ V}$
$2(Mn^{3+} + e^- \longrightarrow Mn^{2+}), E^\circ = +1.51 	ext{ V}$
The $E^\circ$ for $3Mn^{2+} \longrightarrow Mn + 2Mn^{3+}$ will be

Options:

(A) $-2.69 	ext{ V}$, the reaction will not occur

(B) $-2.69 	ext{ V}$ the reaction will occur

(C) $-0.33 	ext{ V}$, the reaction will not occur

(D) $-0.33 	ext{ V}$, the reaction will occur

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a chemical architect, and your job is to predict whether a specific reaction will happen on its own. In this problem, we are given two foundational building blocks—two half-cell reactions involving different oxidation states of Manganese.

Our first building block is the reduction of $Mn^{2+}$ to solid Manganese:
$Mn^{2+} + 2e^- \longrightarrow Mn \quad (E_1^\circ = -1.18 	ext{ V})$

Our second building block is the reduction of $Mn^{3+}$ to $Mn^{2+}$:
$2Mn^{3+} + 2e^- \longrightarrow 2Mn^{2+} \quad (E_2^\circ = +1.51 	ext{ V})$

Our ultimate goal is to find the standard potential ($E^\circ$) for a completely new reaction—the disproportionation of $Mn^{2+}$:
$3Mn^{2+} \longrightarrow Mn + 2Mn^{3+}$

### The Master Equation: Why We Can't Just Add Voltages

You might be tempted to just add or subtract the voltages directly. **Stop right there!** Standard electrode potential ($E^\circ$) is an *intensive property*. It's like temperature; you can't just add the temperatures of two cups of water to get the final temperature.

To combine these reactions safely, we must translate them into an *extensive property*—energy. Specifically, we use the standard Gibbs free energy change ($\Delta G^\circ$). The bridge between voltage and energy is given by the master equation:

$\Delta G^\circ = -nFE^\circ$

Here, $n$ is the number of electrons transferred, and $F$ is Faraday's constant. Energy is conserved, so we can add and subtract $\Delta G^\circ$ values all day long!

### Calculating the Energies

Let's calculate the energy for our first reaction. Two electrons are involved ($n=2$), so:

$\Delta G_1^\circ = -2F(-1.18) = +2.36F$

Now for the second reaction. Again, two electrons are involved ($n=2$):

$\Delta G_2^\circ = -2F(+1.51) = -3.02F$

### Constructing the Target Reaction

How do we build our target reaction from the two blocks? If we take the first reaction and subtract the second reaction from it, the electrons cancel out perfectly:

$(Mn^{2+} + 2e^- \longrightarrow Mn) - (2Mn^{3+} + 2e^- \longrightarrow 2Mn^{2+})$

Rearranging the terms by moving the negative components to the other side, we get exactly what we want:

$3Mn^{2+} \longrightarrow Mn + 2Mn^{3+}$

Since we subtracted the chemical equations, we must do the exact same mathematical operation to their energies:

$\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ$

$\Delta G_3^\circ = 2.36F - (-3.02F) = +5.38F$

### Final Calculation: Back to Voltage

We have the energy of our target reaction. Now, let's translate it back to voltage. In our target reaction, the net electron transfer is still 2 ($n_3 = 2$).

$\Delta G_3^\circ = -n_3 F E_3^\circ$

$+5.38F = -2 F E_3^\circ$

The Faraday constants cancel out, leaving us with:

$E_3^\circ = \frac{5.38}{-2} = -2.69 	ext{ V}$

### The Ninja Shortcut

Did you notice a beautiful mathematical coincidence? The number of electrons in the first reaction ($n_1=2$), the second reaction ($n_2=2$), and the final target reaction ($n_3=2$) are all identical!

**Here is a powerful shortcut:** Whenever the number of electrons transferred is identical across all steps, the $nF$ terms factor out completely. In these rare cases, you *can* directly add or subtract the standard potentials!

$E_3^\circ = E_1^\circ - E_2^\circ = -1.18 - 1.51 = -2.69 	ext{ V}$

This gets you the answer in seconds, but always verify that $n$ is constant before using it!

### Conclusion on Spontaneity

We found our standard potential to be $-2.69 	ext{ V}$. What does a negative voltage physically mean? It means the standard Gibbs free energy change is positive ($\Delta G^\circ > 0$).

Nature loves to minimize energy. A positive $\Delta G^\circ$ means the reaction requires an input of energy to proceed. Therefore, under standard conditions, this disproportionation reaction is **non-spontaneous**. It will not occur.

Chemically, this tells us a profound truth: $Mn^{2+}$ is incredibly stable in aqueous solutions. Thanks to its perfectly half-filled $d^5$ electron configuration, it fiercely resists both oxidation and reduction.

Animated Solution for Chemistry - Electrochemistry: Given below are the half-cell reactions $M n^{2 +} + 2 e^{-} ⟶ M n, E^{\circ} = - 1.18 V$ $2 (M n^{3 +} + e^{-} ⟶ M n^{2 +}), E^{\circ} = + 1.51 V$ The $E^{\circ}$ for $3 M n^{2 +} ⟶ M n + 2 M n^{3 +}$ will be

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Visualized Solution

$Given Half-Cell Reactions$

$The Target Reaction$

$Δ G^{\circ} for Reaction (i)$

$Δ G^{\circ} for Reaction (ii)$

$Constructing the Target Reaction$

$Calculating E_{3}^{\circ} (The Foolproof Way)$

$The Ninja Shortcut$

$Conclusion on Spontaneity$

$The Way Forward$

The Sigma Insight: Electrochemical Cells

Analyzing the Setup

The Master Equation

Calculating the Energies

Constructing the Target Reaction

Final Calculation

The Ninja Shortcut

Conclusion on Spontaneity

Similar Questions

Consider the following $E^{\circ}$ values :\n $E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V$ \n $E_{S n^{2 +} / S n}^{\circ} = - 0.14 V$ \nUnder standard conditions, the potential for the reaction\n $S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)$ is

Given, $E_{C r^{3 +} / C r}^{\circ} = - 0.72 V$ , $E_{F e^{2 +} / F e}^{\circ} = - 0.42 V$ . The potential for the cell $C r ∣ C r^{3 +} (0.1 M) ∣∣ F e^{2 +} (0.01 M) ∣ F e$ is

For the redox reaction $Z n (s) + C u^{2 +} (0.1 M) ⟶ Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{ce l l}^{\circ}$ is $1.10 V$ . $E_{ce l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

For the galvanic cell, $Z n (s) + C u^{2 +} (0.02 M) ⟶ Z n^{2 +} (0.04 M) + C u (s)$ , $E_{C e l l} = \dots \times 1 0^{- 2} V$ . (Nearest integer) [Use $E_{C u / C u^{2 +}}^{\circ} = - 0.34 V$ , $E_{Z n / Z n^{2 +}}^{\circ} = + 0.76 V$ , $\frac{2.303 R T}{F} = 0.059 V$ ]

The reduction potential ( $E^{0}$ , in V) of $MnO_{4}^{-} (aq) / Mn (s)$ is _______. [Given : $E_{(MnO_{4}^{-} (aq) / MnO_{2} (s))}^{0} = 1.68 V$ ; $E_{(MnO_{2} (s) / Mn^{2 +} (aq))}^{0} = 1.21 V$ ; $E_{(Mn^{2 +} (aq) / Mn (s))}^{0} = - 1.03 V$ ]

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

The standard reduction potentials for $Zn^{2 +} / Zn$ , $Ni^{2 +} / Ni$ and $Fe^{2 +} / Fe$ are $- 0.76$ , $- 0.23$ and $- 0.44 V$ , respectively. The reaction $X + Y^{2 +} ⟶ X^{2 +} + Y$ will be spontaneous when

Given the electrochemical cell: $E_{Cu^{2 +} ∣ Cu}^{\circ} = + 0.34 V$ $E_{Zn^{2 +} ∣ Zn}^{\circ} = - 0.76 V$ Identify the incorrect statement from the options below for the above cell.

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Half-Cell Reactions

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔG∘ for Reaction (i)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔG∘ for Reaction (ii)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constructing the Target Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating E3∘​ (The Foolproof Way)

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

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

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