The reaction Cl2(g) + 2OH^-(aq) → ClO^-(aq) + Cl^-(aq) + H2O(l) represents the process of bleaching. Identify

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Q. The reaction Cl2(g) + 2OH^-(aq) \to ClO^-(aq) + Cl^-(aq) + H2O(l) represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.

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Concept understanding — Redox Reaction Stoichiometry

Redox Reaction Stoichiometry – From Intuition to Precision

Imagine you're balancing a seesaw. On one side, electrons are being lost; on the other, they're being gained. The seesaw stays level only when the number of electrons lost equals the number gained. That's the core idea behind redox stoichiometry.

The Intuition: Electrons Are the Currency

In any redox reaction, two things happen simultaneously:

Oxidation: a substance loses electrons (its oxidation state increases).
Reduction: a substance gains electrons (its oxidation state decreases).

Think of electrons as money. If one person gives away ₹10, another must receive exactly ₹10. You can't have ₹5 floating in the air. Similarly, the total number of electrons lost in oxidation must equal the total number of electrons gained in reduction.

This simple equality is what makes redox stoichiometry work. It's not about balancing atoms first — it's about balancing electrons first.

The Precise Statement

\text{Total electrons lost (by reducing agent)} = \text{Total electrons gained (by oxidising agent)}

This equality is the foundation of the half-reaction method (also called the ion-electron method) for balancing redox equations.

How It Works in Practice

Let's walk through a classic example: the reaction between permanganate ions ( $\text{MnO}_4^-$ ) and iron(II) ions ( $\text{Fe}^{2+}$ ) in acidic medium.

Step 1: Write the two half-reactions (unbalanced).

Oxidation half: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Reduction half: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Notice: iron loses 1 electron per atom, while permanganate gains 5 electrons per ion.

Step 2: Balance electrons between the halves.

To make electrons lost = electrons gained, multiply the oxidation half by 5:

$5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$

Now both halves involve 5 electrons.

Step 3: Add the halves and cancel common terms.

$5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}$

The electrons cancel because they appear on opposite sides. The equation is now balanced in both atoms and charge.

Tip

Always check that the net charge on both sides is equal after balancing. In the example above: left side charge = $5(+2) + (-1) + 8(+1) = +17$ ; right side = $5(+3) + (+2) + 0 = +17$ . Matches perfectly.

Why This Matters for Exams

In Indian board exams (CBSE, ICSE, state boards), redox stoichiometry appears in two main forms:

Balancing equations using the half-reaction method (acidic or basic medium).
Titration calculations where you use the electron equality to find unknown concentrations.

For titrations, the key formula is:

\frac{n_1}{n_2} = \frac{M_1 V_1}{M_2 V_2}

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The bleaching action in this reaction is due to the hypochlorite ion, $ClO^-$ , which acts as a strong oxidising agent.

Bleaching is a chemical process that removes colour from substances, typically by oxidising the coloured compounds into colourless ones. In the given reaction, we need to identify which product is responsible for this oxidising action. To do this, we will analyse the changes in oxidation states of the elements involved.

The reaction provided is:

$Cl_2(g) + 2OH^-(aq) \rightarrow ClO^-(aq) + Cl^-(aq) + H_2O(l)$

This is a disproportionation reaction, where a single element (chlorine in this case) is simultaneously oxidised and reduced. The species that causes oxidation in other substances is called an oxidising agent.

Assign oxidation states to all elements:
- In $Cl_2$ : Chlorine is in its elemental form, so its oxidation state is $0$ .
- In $OH^-$ : Oxygen is $-2$ , Hydrogen is $+1$ .
- In $ClO^-$ : Let the oxidation state of Cl be $x$ . Oxygen is $-2$ . The overall charge is $-1$ . So, $x + (-2) = -1$ , which means $x = +1$ .
- In $Cl^-$ : The oxidation state of Cl is $-1$ .
- In $H_2O$ : Oxygen is $-2$ , Hydrogen is $+1$ .
Identify changes in oxidation states for chlorine:
- Chlorine in $Cl_2$ starts at an oxidation state of $0$ .
- In $ClO^-$ , chlorine has an oxidation state of $+1$ . This is an increase in oxidation state, meaning $Cl_2$ has been oxidised to $ClO^-$ .
- In $Cl^-$ , chlorine has an oxidation state of $-1$ . This is a decrease in oxidation state, meaning $Cl_2$ has been reduced to $Cl^-$ .
Determine the role of the products:
- The question asks for the species that bleaches due to its oxidising action. This means we are looking for a product that can act as an oxidising agent.
- An oxidising agent is a species that itself gets reduced while causing another species to be oxidised. …

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Q.The reaction Cl2(g) + 2OH^-(aq) → ClO^-(aq) + Cl^-(aq) + H2O(l) represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.

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Q.The reaction
Cl2(g) + 2OH^-(aq) → ClO^-(aq) + Cl^-(aq) + H2O(l)
represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.

Unlock everything free for 14 days