Chemistry Q127 – AP-EAMCET 2026

Home Chemistry Redox ReactionsQuestion #543

Observe the following unbalanced equation

a S_{8} (s) + b O H^{-} (a q) \to c S^{2 -} (a q) + d S_{2} O_{3}^{2 -} (a q) + e H_{2} O (l)

. From the balanced equation, identify the correct relations from the following sets I.

a + b = c + d + e + 1

II.

b = 2 e + d

III.

b - a = c + e + 1

IV.

b + d = 2 c + e - 1

The correct answer is

Clue 1 of 3: Balancing Redox Reactions

This reaction is a disproportionation reaction where sulfur is simultaneously oxidized and reduced. It occurs in a basic medium, which affects how hydrogen and oxygen atoms are balanced.

🧠 Full Solution Path

Step 1: Separate into Half-Reactions and Assign Oxidation States✦ Active

The given unbalanced equation is: $a S_{8} (s) + b O H^{-} (a q) \to c S^{2 -} (a q) + d S_{2} O_{3}^{2 -} (a q) + e H_{2} O (l)$ . This is a disproportionation reaction of sulfur ( $S_{8}$ ) because sulfur is oxidized to $S_{2} O_{3}^{2 -}$ and reduced to $S^{2 -}$ . Oxidation state of S in $S_{8}$ is $0$ . Oxidation state of S in $S^{2 -}$ is $- 2$ (Reduction). Oxidation state of S in $S_{2} O_{3}^{2 -}$ ( $2 x + 3 (- 2) = - 2 \Rightarrow 2 x = 4 \Rightarrow x = + 2$ ) is $+ 2$ (Oxidation).

💡 Teacher's Secret Hint

Remember to identify if the reaction is in acidic or basic medium, as this dictates the method for balancing hydrogen and oxygen atoms.

Step 2: Balance the Half-Reactions○ Expand

1. Reduction Half-Reaction ( $S_{8} \to S^{2 -}$ ): Balance S atoms: $S_{8} \to 8 S^{2 -}$ Balance charge with electrons: $S_{8} + 16 e^{-} \to 8 S^{2 -}$ (Equation A) 2. Oxidation Half-Reaction ( $S_{8} \to S_{2} O_{3}^{2 -}$ ): Balance S atoms: $S_{8} \to 4 S_{2} O_{3}^{2 -}$ The change in oxidation state for $S_{8} \to 4 S_{2} O_{3}^{2 -}$ is from $0$ to $8 \times (+ 2) = + 16$ , so $16$ electrons are lost. Balance oxygen atoms with $H_{2} O$ : $S_{8} + 12 H_{2} O \to 4 S_{2} O_{3}^{2 -}$ (since $4 \times 3 = 12$ oxygen atoms on the right). Balance hydrogen atoms with $H^{+}$ : $S_{8} + 12 H_{2} O \to 4 S_{2} O_{3}^{2 -} + 24 H^{+}$ . Balance charge with electrons: $S_{8} + 12 H_{2} O \to 4 S_{2} O_{3}^{2 -} + 24 H^{+} + 16 e^{-}$ (Equation B)

💡 Teacher's Secret Hint

Ensure the number of atoms and charges are balanced in each half-reaction before proceeding. The number of electrons gained must equal the number of electrons lost when combining.

Step 3: Combine Half-Reactions and Balance for Basic Medium○ Expand

Add Equation A and Equation B. The electrons are already balanced (16e-): $S_{8} + 16 e^{-} + S_{8} + 12 H_{2} O \to 8 S^{2 -} + 4 S_{2} O_{3}^{2 -} + 24 H^{+} + 16 e^{-}$ $2 S_{8} + 12 H_{2} O \to 8 S^{2 -} + 4 S_{2} O_{3}^{2 -} + 24 H^{+}$ Since the reaction is in a basic medium, add $O H^{-}$ ions to neutralize $H^{+}$ ions. Add $24 O H^{-}$ to both sides: $2 S_{8} + 12 H_{2} O + 24 O H^{-} \to 8 S^{2 -} + 4 S_{2} O_{3}^{2 -} + 24 H^{+} + 24 O H^{-}$ Combine $H^{+}$ and $O H^{-}$ to form $H_{2} O$ : $2 S_{8} + 12 H_{2} O + 24 O H^{-} \to 8 S^{2 -} + 4 S_{2} O_{3}^{2 -} + 24 H_{2} O$ Simplify the $H_{2} O$ molecules: $2 S_{8} (s) + 24 O H^{-} (a q) \to 8 S^{2 -} (a q) + 4 S_{2} O_{3}^{2 -} (a q) + 12 H_{2} O (l)$ Divide all coefficients by 2 to get the simplest integer coefficients: $S_{8} (s) + 12 O H^{-} (a q) \to 4 S^{2 -} (a q) + 2 S_{2} O_{3}^{2 -} (a q) + 6 H_{2} O (l)$

Step 4: Identify Coefficients and Evaluate Relations○ Expand

Comparing the balanced equation with the general form $a S_{8} (s) + b O H^{-} (a q) \to c S^{2 -} (a q) + d S_{2} O_{3}^{2 -} (a q) + e H_{2} O (l)$ , we get the coefficients: $a = 1$ , $b = 12$ , $c = 4$ , $d = 2$ , $e = 6$ . Now, let's check the given relations: I. $a + b = c + d + e + 1$ $1 + 12 = 4 + 2 + 6 + 1$ $13 = 13$ (Correct) II. $b = 2 e + d$ $12 = 2 (6) + 2$ $12 = 12 + 2$ $12 = 14$ (Incorrect) III. $b - a = c + e + 1$ $12 - 1 = 4 + 6 + 1$ $11 = 11$ (Correct) IV. $b + d = 2 c + e - 1$ $12 + 2 = 2 (4) + 6 - 1$ $14 = 8 + 6 - 1$ $14 = 13$ (Incorrect) Therefore, only relations I and III are correct.

💡 Teacher's Secret Hint

Always perform a final check of mass and charge balance for the overall reaction. For example, count all atoms and total charge on both sides of the final balanced equation.

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Chemistry Question 127 – AP-EAMCET 2026

Clue 1 of 3: Balancing Redox Reactions

🧠 Full Solution Path