How do I identify if a reaction is a disproportionation reaction or a comproportionation reaction?

In a disproportionation reaction, a single reactant element in one oxidation state is simultaneously oxidized to a higher oxidation state and reduced to a lower oxidation state (e.g., $2H_2O_2 \rightarrow 2H_2O + O_2$). In a comproportionation reaction, two reactants containing the same element in different oxidation states react to form a single product with an intermediate oxidation state (e.g., $2H_2S + SO_2 \rightarrow 3S + 2H_2O$).

Why does peroxodisulfuric acid ($H_2S_2O_8$) have an oxidation state of $+6$ for sulfur instead of $+7$?

Using the naive algebraic formula $2(+1) + 2(x) + 8(-2) = 0$ yields $x = +7$, which exceeds sulfur's maximum valence electron count ($6$). $H_2S_2O_8$ contains one peroxide linkage ($-O-O-$). Structurally, two oxygen atoms have an oxidation state of $-1$ and six have $-2$. Thus, $2(+1) + 2(x) + 6(-2) + 2(-1) = 0$, giving $x = +6$.

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Solved High-Yield Sample Questions

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Question #1NEET 2024

What is the individual oxidation state of sulfur atoms in $Na_2S_4O_6$ (sodium tetrathionate) calculated structurally?

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Step-by-Step Solution & Concept

Step 1

Draw the structure of tetrathionate ion $S_4O_6^{2-}$ as $O_3S-S-S-SO_3^{2-}$ . Note the central $S-S$ covalent linkage.

Step 2

Assign oxidation states based on electronegativity differences. The two central sulfur atoms are bonded only to sulfur, contributing an oxidation state of $0$ each.

Step 3

The two terminal sulfur atoms are each bonded to three oxygen atoms and one sulfur atom, giving $x + 3(-2) = -1 \implies x = +5$ . Hence, two sulfur atoms have $0$ and two have $+5$ (giving an average of $+2.5$ ).

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Question #2JEE Main 2023

In the disproportionation reaction $3Br_2 + 6OH^- \rightarrow 5Br^- + BrO_3^- + 3H_2O$ , what is the equivalent weight of $Br_2$ in terms of its molecular weight $M$ ?

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Step-by-Step Solution & Concept

Step 1

Calculate $n_1$ for reduction of $Br_2$ ( $0$ ) to $2Br^-$ ( $-1$ ). Total electron gain per mole of $Br_2$ is $n_1 = 2 \times 1 = 2$ .

Step 2

Calculate $n_2$ for oxidation of $Br_2$ ( $0$ ) to $2BrO_3^-$ ( $+5$ ). Total electron loss per mole of $Br_2$ is $n_2 = 2 \times 5 = 10$ .

Step 3

For disproportionation, total $n\text{-factor} = \frac{n_1 \times n_2}{n_1 + n_2} = \frac{2 \times 10}{2 + 10} = \frac{20}{12} = \frac{5}{3}$ . Equivalent weight $E = \frac{M}{n\text{-factor}} = \frac{5M}{3}$ .

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Mastering Redox Reactions: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Oxidation Number Rule for Fractional Oxidation States

Ion-Electron Method (Charge Balance)

Equivalent Weight in Redox Reactions

n-Factor of Oxidizing or Reducing Agent

n-Factor in Disproportionation Reactions

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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