In neutral or faintly alkaline solution, 8 moles permanganate anion quantitatively oxidize thiosulph

Q: Why does permanganate go to $MnO_2$ and not $Mn^{2+}$?

The reduction product of the permanganate ion ($MnO_4^-$) depends strictly on the pH of the medium. In a strongly acidic medium, it reduces to $Mn^{2+}$. However, the question explicitly states the medium is 'neutral or faintly alkaline', where it only reduces to $MnO_2$.

Q: Why is the n-factor of $SO_4^{2-}$ taken as 4 and not 8?

The n-factor is defined as the total change in oxidation state *per mole of the substance you are considering*. Sulphur changes from +2 to +6, a change of 4. Since one molecule of $SO_4^{2-}$ contains only one Sulphur atom, its n-factor is $4 \times 1 = 4$. If you were finding the n-factor of the reactant $S_2O_3^{2-}$, it would be $4 \times 2 = 8$ because it has two Sulphur atoms.

Q: Can we solve this by balancing the chemical equation?

Yes, absolutely. The balanced equation is $8 MnO_4^- + 3 S_2O_3^{2-} + H_2O \rightarrow 8 MnO_2 + 6 SO_4^{2-} + 2 OH^-$. From the stoichiometry, 8 moles of $MnO_4^-$ produce exactly 6 moles of $SO_4^{2-}$. However, using the equivalent concept is much faster and less prone to balancing errors.

Q: What is the oxidation state of S in thiosulphate?

Structurally, the two sulphur atoms in thiosulphate are in different environments (one is central with a +6 state, the other is terminal with a -2 state). However, for the purpose of redox stoichiometric calculations, using the average oxidation state of +2 works perfectly and yields the correct n-factor.

Q: Why do we equate equivalents of a reactant to a product?

The Law of Equivalence states that in any chemical reaction, all reactants are consumed and all products are formed in equal numbers of equivalents. Therefore, you can equate the equivalents of any reactant to any other reactant, or to any product.

Question

In neutral or faintly alkaline solution, 8 moles permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. the magnitude of X is

Sigma Percentile · Accepted Answer

The beauty of redox chemistry lies in its predictable patterns, provided you know the rules of the game. When dealing with stoichiometric calculations in redox reactions, balancing the entire chemical equation can often feel like navigating a maze blindfolded. It is time-consuming, prone to algebraic errors, and frankly, unnecessary if you wield the right conceptual tools.

In this problem, we are tasked with finding the amount of a sulphur-containing product formed when permanganate reacts with thiosulphate. Let's embark on this journey by breaking down the chemistry and the mathematics behind it.

### Analyzing the Setup: The Role of the Medium

The very first phrase of the question is a massive hint: **"In neutral or faintly alkaline solution"**. In redox chemistry, the medium is the director of the play; it dictates the final state of the actors.

The permanganate ion ($MnO_4^-$) is a versatile and powerful oxidizing agent. Its behavior changes drastically depending on the pH of the solution:
- In a **strongly acidic** medium, it reduces to $Mn^{2+}$ (a color change from purple to colorless).
- In a **strongly alkaline** medium, it reduces to the manganate ion, $MnO_4^{2-}$ (green).
- In a **neutral or faintly alkaline** medium, it reduces to manganese dioxide, $MnO_2$ (a brown precipitate).

Simultaneously, the thiosulphate ion ($S_2O_3^{2-}$) is oxidized. In a neutral or faintly alkaline medium, permanganate is strong enough to oxidize thiosulphate all the way up to the sulphate ion ($SO_4^{2-}$).

So, our skeletal reaction looks like this:
$$MnO_4^- + S_2O_3^{2-} \xrightarrow{	ext{faintly alkaline}} MnO_2 + SO_4^{2-}$$

### The Master Equation: The Law of Equivalence

We could try to balance this equation by splitting it into half-reactions, balancing oxygen with water, hydrogen with $H^+$ or $OH^-$, and equating electrons. But why take the long road when a shortcut exists?

Enter the **Law of Equivalence**. This fundamental principle states that in any chemical reaction, substances react and are produced in the exact same ratio of their equivalents.

Mathematically, for our reaction:
$$	ext{Equivalents of } MnO_4^- 	ext{ consumed} = 	ext{Equivalents of } SO_4^{2-} 	ext{ produced}$$

To use this, we need to calculate the number of equivalents, which is given by the product of the number of moles and the n-factor (valency factor).
$$	ext{Equivalents} = 	ext{Moles} 	imes n	ext{-factor}$$

### Calculating the n-factors

The n-factor for a substance in a redox reaction is the net change in oxidation state per molecule or ion of that substance.

**1. For Permanganate ($MnO_4^-$):**
Let's find the oxidation state of Manganese (Mn). Oxygen is typically $-2$. 
$$x + 4(-2) = -1 \implies x = +7$$
In the product, $MnO_2$, the oxidation state of Mn is:
$$y + 2(-2) = 0 \implies y = +4$$
The change in oxidation state is $|+7 - 4| = 3$. Since there is only one Mn atom in $MnO_4^-$, the n-factor is **$3$**.

**2. For Sulphate ($SO_4^{2-}$):**
We need the n-factor of the *product* because the question asks for the moles of the product formed. 
First, let's find the average oxidation state of Sulphur (S) in the reactant, thiosulphate ($S_2O_3^{2-}$):
$$2s + 3(-2) = -2 \implies 2s = +4 \implies s = +2$$
Now, let's find the oxidation state of S in the product, sulphate ($SO_4^{2-}$):
$$z + 4(-2) = -2 \implies z = +6$$
The change in oxidation state per Sulphur atom is $|+6 - 2| = 4$. 
Since we are calculating the n-factor for $SO_4^{2-}$, and there is exactly **one** Sulphur atom in one molecule of $SO_4^{2-}$, the n-factor is simply $4 	imes 1 = \mathbf{4}$.

*(Note: If we were calculating the n-factor for the reactant $S_2O_3^{2-}$, it would be $4 	imes 2 = 8$, because there are two Sulphur atoms in one molecule of thiosulphate. But the Law of Equivalence allows us to directly equate the equivalents of the reactant to the equivalents of the product!)*

### Final Calculation

Now, we bring it all together. We are given $8$ moles of $MnO_4^-$, and we need to find $X$ moles of $SO_4^{2-}$.

$$	ext{Moles of } MnO_4^- 	imes n	ext{-factor of } MnO_4^- = 	ext{Moles of } SO_4^{2-} 	imes n	ext{-factor of } SO_4^{2-}$$

Substituting our calculated values:
$$8 	imes 3 = X 	imes 4$$

$$24 = 4X$$

$$X = \frac{24}{4} = 6$$

**The magnitude of X is 6.**

By trusting the Law of Equivalence and carefully calculating the n-factors based on the specific medium, we bypassed a tedious balancing process and arrived at the answer with elegant simplicity. Always remember: in redox chemistry, the medium is the message, and the n-factor is the key.

Animated Solution for Chemistry - Redox Reactions: In neutral or faintly alkaline solution, 8 moles permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. the magnitude of X is

Enter Numerical Value:

Visualized Solution

$R e a c t i o n in F ain tl y A l k a l in e M e d i u m$

$O x i d a t i o n S t a t es & n - f a c t or$

$L a w o f E q u i v a l e n ce$

$S u b s t i t u t in g V a l u es$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Oxidation and Reduction

Analyzing the Setup

The Master Equation

Calculating the n-factors

Final Calculation

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