Consider the following reaction, MnO4- + 8H+ + 5e- - Mn2+ + 4H2O, E° = 1.51 V. The quantity of elec

Q: Why do we ignore the $E^\circ = 1.51\text{ V}$ given in the question?

The standard electrode potential ($E^\circ$) tells us about the thermodynamic feasibility and the voltage of the half-cell, but it is not needed to calculate the quantity of charge (stoichiometry). It's extra information here.

Q: What exactly is one Faraday?

One Faraday is the magnitude of electrical charge carried by exactly one mole of electrons. It is approximately equal to $96485\text{ Coulombs}$.

Q: How do we know the reaction is in an acidic medium?

The presence of $H^+$ ions in the reactants side of the balanced half-reaction indicates that the reduction is taking place in an acidic medium.

Q: What if the question asked for the charge in Coulombs?

You would simply multiply the number of Faradays by Faraday's constant ($96500\text{ C/mol}$). So, the answer would be $25 \times 96500\text{ C}$.

Q: Does the number of water molecules produced affect the charge required?

No, the charge required depends strictly on the number of electrons transferred in the balanced half-reaction, not on the other byproducts like water.

Question

Consider the following reaction,
$MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O, E^\circ = 1.51	ext{ V}$.
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^-$ is ………… .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are running an electrochemical cell, and you need to reduce a specific amount of a chemical substance. How do you know exactly how much electricity to pump into the system? This is where the beautiful marriage of stoichiometry and Faraday's laws of electrolysis comes into play.

In our problem, we are given the reduction half-reaction for the permanganate ion ($MnO_4^-$) in an acidic medium:

$$MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$$

This equation is our master recipe. It tells us exactly what ingredients are needed and in what proportions. Notice the standard electrode potential ($E^\circ = 1.51	ext{ V}$) provided in the question. While this value is crucial for determining if the reaction is thermodynamically spontaneous or for calculating the cell voltage, it is actually a distractor here! To find the *quantity* of electricity, we only care about the stoichiometry—specifically, the electrons.

### The Master Equation

Let's look closely at the coefficients in our balanced equation. For every **$1$ mole** of $MnO_4^-$ that gets reduced to $Mn^{2+}$, the reaction consumes exactly **$5$ moles of electrons ($e^-$)**.

This is a direct 1-to-5 ratio. The question asks us to find the charge required to reduce **$5$ moles** of $MnO_4^-$. Using a simple unitary method, we can scale our recipe up:

$$	ext{Moles of } e^- = 5 	ext{ moles of } MnO_4^- 	imes \left( \frac{5 	ext{ moles of } e^-}{1 	ext{ mole of } MnO_4^-} ight)$$

$$	ext{Moles of } e^- = 25 	ext{ moles}$$

### Final Calculation

Now, how do we translate "moles of electrons" into a measurable electrical quantity? Enter Michael Faraday. Faraday established that the total electrical charge carried by exactly one mole of electrons is defined as **$1$ Faraday ($1	ext{ F}$)**.

Since our scaled-up reaction requires $25$ moles of electrons, the total quantity of electricity required is simply:

$$	ext{Total Charge} = 25 	ext{ Faraday}$$

And there we have it! By simply reading the balanced chemical equation and applying Faraday's principle, we arrive at our final integer answer of **$25$**.

Animated Solution for Chemistry - Electrochemistry: Consider the following reaction, $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O, E^{\circ} = 1.51 V$ . The quantity of electricity required in Faraday to reduce five moles of $M n O_{4}^{-}$ is ………… .

Enter Numerical Value:

Visualized Solution

The Half-Cell Reaction

Moles of Electrons per Mole of Reactant

Scaling to Five Moles

Faraday's Constant

Final Conclusion

The Sigma Insight: Electrolytic Conduction

Analyzing the Setup

The Master Equation

Final Calculation

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