Aqueous HCl$\text{HCl}$ reacts with MnO₂(s)$\text{MnO}_2(\text{s})$ to form MnCl₂(aq)$\text{MnCl}_2(\text{aq})$, Cl₂(g)$\text{Cl}_2(\text{g})$ and H₂O(l)$\text{H}_2\text{O}(\text{l})$. What is the weight (in g) of Cl₂$\text{Cl}_2$ liberated when 8.7 g of MnO₂(s)$\text{MnO}_2(\text{s})$ is reacted with excess aqueous HCl$\text{HCl}$ solution? (Given Molar mass in g mol⁻¹: Mn = 55, Cl = 35.5, O = 16, H = 1$\text{g mol}^{-1}: \text{Mn} = 55, \text{Cl} = 35.5, \text{O} = 16, \text{H} = 1$)

Related Formula

Core Logic

In acidic medium, potassium dichromate (K₂Cr₂O₇$K_2Cr_2O_7$) reduces from Cr⁺⁶$Cr^{+6}$ to Cr⁺³$Cr^{+3}$. Change in oxidation state = 3 per Cr atom. For K₂Cr₂O₇$K_2Cr_2O_7$, n-factor = 2 × 3 = 6$n\text{-factor} = 2 \times 3 = 6$. Normality of dichromate = 6 × (x × 10⁻³) N$6 \times (x \times 10^{-3}) \text{ N}$.

Mohr's salt is (NH₄)₂Fe(SO₄)₂ · 6H₂O$(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$. The active reducing species is Fe²⁺$Fe^{2+}$, which oxidizes to Fe³⁺$Fe^{3+}$. Change in oxidation state = 1. For Mohr's salt, n-factor = 1$n\text{-factor} = 1$. Normality of Mohr's salt = 1 × 0.6 = 0.6 N$1 \times 0.6 = 0.6 \text{ N}$.

Step 1: Applying Law of Equivalence

1.2x = 450$1.2x = 450$

Pattern Recognition

Memorize the n$n$-factors for redox titrations: K₂Cr₂O₇$K_2Cr_2O_7$ in acid is always n=6$n=6$. KMnO₄$KMnO_4$ in acid is n=5$n=5$. Mohr's salt (Fe²⁺$Fe^{2+}$) is always n=1$n=1$.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: d and f Block Elements

Core Logic

Dichromate ion (Cr₂O₇²⁻$Cr_2O_7^{2-}$) in acidic medium acts as a strong oxidizing agent, getting reduced to Cr³⁺$Cr^{3+}$. This reduction process consumes 6$6$ electrons per dichromate ion: Cr₂O₇²⁻ + 14H^+ + 6e^- arrow 2Cr³⁺ + 7H₂O$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

Reaction 1: Oxidation of Iodide 6I^- arrow 3I₂ + 6e^-$6I^- \rightarrow 3I_2 + 6e^-$ Overall balanced reaction: Cr₂O₇²⁻ + 14H^+ + 6I^- arrow 2Cr³⁺ + 3I₂ + 7H₂O$Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$ Number of electrons involved (X) = 6.

Reaction 2: Oxidation of Sulfide 3S²⁻ arrow 3S + 6e^-$3S^{2-} \rightarrow 3S + 6e^-$ Overall balanced reaction: Cr₂O₇²⁻ + 14H^+ + 3S²⁻ arrow 2Cr³⁺ + 3S + 7H₂O$Cr_2O_7^{2-} + 14H^+ + 3S^{2-} \rightarrow 2Cr^{3+} + 3S + 7H_2O$ Number of electrons involved (Y) = 6.

Step 1: Final Summation

Sum of X + Y = 6 + 6 = 12$X + Y = 6 + 6 = 12$.

Pattern Recognition

Dichromate (Cr⁶⁺$Cr^{6+}$ to Cr³⁺$Cr^{3+}$, two atoms) always demands 6 electrons in a balanced equation block. The reducing agent must therefore supply exactly 6 electrons to match.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The d- and f-Block Elements

Related Formula

Core Logic

In a basic medium, permanganate (MnO₄^-$MnO_4^-$) is reduced to MnO₂$MnO_2$, with an n-factor of 3. Iodide (I^-$I^-$) is oxidized to I₂$I_2$.\nThe liberated I₂$I_2$ is titrated with thiosulfate (S₂O₃²⁻$S_2O_3^{2-}$), which is oxidized to tetrathionate (S₄O₆²⁻$S_4O_6^{2-}$) with an n-factor of 1 per thiosulfate molecule.

Step 1: Determine the Limiting Reagent

Equivalents of KMnO₄ = M × V(L) × n-factor = 0.2 × 0.5 × 3 = 0.3~eq$KMnO_4 = M \times V(L) \times n\text{-factor} = 0.2 \times 0.5 \times 3 = 0.3\mathrm{~eq}$.\nEquivalents of KI = M × V(L) × n-factor = 1.2 × 0.5 × 1 = 0.6~eq$KI = M \times V(L) \times n\text{-factor} = 1.2 \times 0.5 \times 1 = 0.6\mathrm{~eq}$.\nKMnO₄$KMnO_4$ is the limiting reagent. It will completely react to liberate exactly 0.3$0.3$ equivalents of I₂$I_2$.

Step 2: Titration with Thiosulfate

Equivalents of Na₂S₂O₃$Na_2S_2O_3$ consumed = Equivalents of I₂$I_2$ liberated = 0.3~eq$0.3\mathrm{~eq}$.\n0.3 = M × V(L) × n-factor$$0.3 = M \times V(L) \times n\text{-factor}$$\n0.3 = 0.1 × V × 1$$0.3 = 0.1 \times V \times 1$$\nV = (0.3)/(0.1) = 3~L$$V = \frac{0.3}{0.1} = 3\mathrm{~L}$$

Pattern Recognition

In iodometric titrations, equivalents transfer sequentially 1:1. The n-factor of KMnO₄$KMnO_4$ shifts radically based on medium (5$5$ acidic, 3$3$ neutral/weakly basic, 1$1$ strongly basic). Here, basic medium means n=3$n=3$.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Core Logic

Let's find the moles of I⁻$I^{-}$ in 200 mL$200\text{ mL}$ of 0.1 M$0.1\text{ M}$ KI:

According to the stoichiometry: * 5 moles I⁻ arrow 1 mole IO₃⁻$5\text{ moles } I^{-} \rightarrow 1\text{ mole } IO_{3}^{-}$ Therefore, 0.02 mol I⁻$0.02\text{ mol } I^{-}$ reacts with (0.02)/(5) = 0.004 mol$\frac{0.02}{5} = 0.004\text{ mol}$ of KIO₃$KIO_{3}$. Statement (A) is correct.

• For Statement (D), Iodine goes from +5$+5$ state in IO₃⁻$IO_{3}^{-}$ to 0$0$ state in I₂$I_{2}$. The change in oxidation state per iodine atom is 5$5$.

Statement (D) is correct.

Step 1: Check Statement B and C

For 200 mL$200\text{ mL}$ KI (0.02 mol$0.02\text{ mol}$):

Hence, Statement (B) is false. For 0.5 L$0.5\text{ L}$ KI (0.05 mol$0.05\text{ mol}$):

Hence, Statement (C) is false.

Pattern Recognition

Always focus on checking the n-factor calculation directly from oxidation states for quick elimination of equivalence statements.

Chapter Mix

Class 11 Chemistry: Redox Reactions