Q.Balance the following redox reaction by ion-electron method: MnO4-(aq) + I-(aq) → MnO2(s) + I2(s) (In basic medium). OR Consider the elements: Cs, Ne, I and F. Identify the element that:

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

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.

Why This Matters for Exams

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

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

For titrations, the key formula is: