In basic medium CrO4(2-) oxidises S2O3(2-) to form SO4(2-) and itself changes into Cr(OH)4(-). The v

Q: Why do we multiply by 2 for the n-factor of thiosulfate?

Because one molecule of $\text{S}_2\text{O}_3^{2-}$ contains two sulfur atoms. The change in oxidation state per sulfur atom is 4, so for two atoms, the total electron transfer is $2 \times 4 = 8$.

Q: Can we balance the full chemical equation instead of using n-factors?

Yes, you can balance the redox reaction using the ion-electron method and then use mole ratios. However, the equivalent concept ($N_1 V_1 = N_2 V_2$) is much faster and less prone to balancing errors.

Q: How did you find the oxidation state of Cr in $\text{Cr(OH)}_4^-$?

The overall charge of the complex is -1. Hydroxide ($\text{OH}^-$) has a charge of -1. So, $x + 4(-1) = -1$, which gives $x = +3$.

Q: What if the medium was acidic instead of basic?

The products might be different. For example, in an acidic medium, $\text{CrO}_4^{2-}$ would convert to $\text{Cr}_2\text{O}_7^{2-}$ and then reduce to $\text{Cr}^{3+}$. The n-factor would still be 3 per Cr atom, but the reaction pathway changes.

Q: Why do we use molarity $\times$ n-factor instead of just molarity?

Because substances react in the ratio of their equivalents, not necessarily their moles. The n-factor accounts for the number of electrons transferred per molecule, converting moles to equivalents.

Question

In basic medium $	ext{CrO}_4^{2-}$ oxidises $	ext{S}_2	ext{O}_3^{2-}$ to form $	ext{SO}_4^{2-}$ and itself changes into $	ext{Cr(OH)}_4^-$. The volume of $0.154 	ext{ M } 	ext{CrO}_4^{2-}$ required to react with $40 	ext{ mL}$ of $0.25 	ext{ M } 	ext{S}_2	ext{O}_3^{2-}$ is ............ mL (Rounded off to the nearest integer).

Sigma Percentile · Accepted Answer

### The Core Principle: Law of Chemical Equivalence

When dealing with redox titrations, balancing the entire chemical equation can be a tedious and error-prone process. Instead, we rely on a much more elegant and powerful tool: the **Law of Chemical Equivalence**.

This law states that in any chemical reaction, the number of gram equivalents of the oxidizing agent must perfectly equal the number of gram equivalents of the reducing agent.

Mathematically, this is expressed as $N_1 V_1 = N_2 V_2$, where $N$ represents normality and $V$ represents volume. Since normality is simply the product of molarity and the n-factor ($N = M 	imes n$), our master equation becomes $M_1 n_1 V_1 = M_2 n_2 V_2$.

### Decoding the Oxidizing Agent

Our oxidizing agent here is the chromate ion, $	ext{CrO}_4^{2-}$. To use our master equation, we first need to determine its n-factor, which is the total change in oxidation state per molecule.

Let's calculate the oxidation state of chromium in $	ext{CrO}_4^{2-}$. Oxygen is typically $-2$, so we have $x + 4(-2) = -2$, which gives us $x = +6$.

During the reaction in a basic medium, chromate is reduced to $	ext{Cr(OH)}_4^-$. Here, the hydroxide ion has a charge of $-1$, so $x + 4(-1) = -1$, yielding $x = +3$.

The change in oxidation state for the single chromium atom is $|6 - 3| = 3$. Therefore, the n-factor for chromate ($n_1$) is **$3$**.

### Decoding the Reducing Agent

Now, let's turn our attention to the reducing agent: the thiosulfate ion, $	ext{S}_2	ext{O}_3^{2-}$. This is where many students fall into a classic trap!

First, we find the average oxidation state of sulfur in $	ext{S}_2	ext{O}_3^{2-}$. Setting up the equation $2x + 3(-2) = -2$, we get $2x = 4$, so $x = +2$.

Thiosulfate is oxidized to the sulfate ion, $	ext{SO}_4^{2-}$. In sulfate, the oxidation state of sulfur is $x + 4(-2) = -2$, which means $x = +6$.

The change in oxidation state per sulfur atom is $|6 - 2| = 4$. However, **here is the catch**: there are *two* sulfur atoms in every thiosulfate molecule! We must account for both.

The total change in oxidation state, and thus the n-factor ($n_2$), is $2 	imes 4 = \mathbf{8}$.

### The Master Equation

With both n-factors in hand, we are ready to deploy our master equation: $M_1 n_1 V_1 = M_2 n_2 V_2$.

Let's list out our known variables:
- For Chromate: $M_1 = 0.154 	ext{ M}$, $n_1 = 3$, $V_1 = ?$
- For Thiosulfate: $M_2 = 0.25 	ext{ M}$, $n_2 = 8$, $V_2 = 40 	ext{ mL}$

### Final Calculation

Substituting these values into the equation, we get:
$$0.154 	imes 3 	imes V_1 = 0.25 	imes 8 	imes 40$$

Simplifying the right side:
$$0.25 	imes 8 = 2$$
$$2 	imes 40 = 80$$

Now, simplifying the left side:
$$0.154 	imes 3 = 0.462$$

So, our equation becomes:
$$0.462 	imes V_1 = 80$$

Solving for $V_1$:
$$V_1 = \frac{80}{0.462} \approx 173.16 	ext{ mL}$$

The question asks us to round off to the nearest integer. Since $173.16$ is closest to $173$, our final required volume is **$173 	ext{ mL}$**.

Enter Numerical Value:

Visualized Solution

$CrO_{4}^{2 -} + S_{2} O_{3}^{2 -} \to SO_{4}^{2 -} + Cr(OH)_{4}^{-}$

$n_{1} = 3$

$n_{2} = 8$

$M_{1} n_{1} V_{1} = M_{2} n_{2} V_{2}$

$V_{1} = \frac{80}{0.462}$

$V_{1} \approx 173$

The Sigma Insight: Oxidation and Reduction

The Core Principle

Decoding the Oxidizing Agent

Decoding the Reducing Agent

The Master Equation

Final Calculation

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