### Core Logic Let's find the moles of I^-$I^{-}$ in 200text mL$200\text{ mL}$ of 0.1text M$0.1\text{ M}$ KI: textMoles = 0.1 times 0.2 = 0.02text mol$\text{Moles} = 0.1 \times 0.2 = 0.02\text{ mol}$ According to the stoichiometry: * 5text moles I^- ightarrow 1text mole IO_3^-$5\text{ moles } I^{-} ightarrow 1\text{ mole } IO_{3}^{-}$ Therefore, 0.02text mol I^-$0.02\text{ mol } I^{-}$ reacts with frac0.025 = 0.004text mol$\frac{0.02}{5} = 0.004\text{ mol}$ of KIO_3$KIO_{3}$. Statement (A) is correct. * For Statement (D), Iodine goes from +5$+5$ state in IO_3^-$IO_{3}^{-}$ to 0$0$ state in I_2$I_{2}$. The change in oxidation state per iodine atom is 5$5$. textEquivalent weight = fractextMolecular weight5$\text{Equivalent weight} = \frac{\text{Molecular weight}}{5}$ Statement (D) is correct. ### Step 1: Check Statement B and C For 200text mL$200\text{ mL}$ KI (0.02text mol$0.02\text{ mol}$): 5text moles I^- text requires 3text moles H_2SO_4 implies 0.02 times frac35 = 0.012text mol$5\text{ moles } I^{-} \text{ requires } 3\text{ moles } H_2SO_4 implies 0.02 \times \frac{3}{5} = 0.012\text{ mol}$ Hence, Statement (B) is false. For 0.5text L$0.5\text{ L}$ KI (0.05text mol$0.05\text{ mol}$): 0.05 times frac35 = 0.03text mol I_2$0.05 \times \frac{3}{5} = 0.03\text{ mol } I_2$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions

### Related Formula Oxalic acid titration equation: 2mathrmMnO_4^- + 5(mathrmCOO)_2^2- + 16mathrmH^+ rightarrow 10mathrmCO_2 + 2mathrmMn^2+ + 8mathrmH_2mathrmO$2\mathrm{MnO}_4^- + 5(\mathrm{COO})_2^{2-} + 16\mathrm{H}^+ \rightarrow 10\mathrm{CO}_2 + 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$ ### Core Logic Statement I is true because the reaction between oxalic acid and mathrmKMnO_4$\mathrm{KMnO}_4$ is slow at room temperature and requires initial heating to around 60^circmathrmC$60^{\circ}\mathrm{C}$. No heating is required for FAS titration. Statement II is true because mathrmMn^2+$\mathrm{Mn}^{2+}$ acts as an autocatalyst. Heating FAS would cause atmospheric oxygen to prematurely oxidize mathrmFe^2+$\mathrm{Fe}^{2+}$ to mathrmFe^3+$\mathrm{Fe}^{3+}$, leading to experimental errors. ### Pattern Recognition Sees: Oxalic acid vs FAS titration with permanganate. Shortcut: Oxalic acid requires heat + autocatalysis. FAS titration must be kept cold to avoid aerial oxidation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions

### Core Logic Let us evaluate each redox pair: - **A:** mathrmCH_4 + 2mathrmO_2 rightarrow mathrmCO_2 + 2mathrmH_2mathrmO$\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$ involves elements combining with oxygen (combustion/combination), corresponding to **(II)**. - **B:** 2mathrmNaH rightarrow 2mathrmNa + mathrmH_2$2\mathrm{NaH} \rightarrow 2\mathrm{Na} + \mathrm{H}_2$ shows a single compound breaking down into its elements, matching **(III)** Decomposition. - **C:** mathrmV_2mathrmO_5 + 5mathrmCa rightarrow 2mathrmV + 5mathrmCaO$\mathrm{V}_2\mathrm{O}_5 + 5\mathrm{Ca} \rightarrow 2\mathrm{V} + 5\mathrm{CaO}$ represents calcium displacing vanadium from its oxide, matching **(IV)** Displacement. - **D:** 2mathrmH_2mathrmO_2 rightarrow 2mathrmH_2mathrmO + mathrmO_2$2\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{O}_2$ represents oxygen in -1$-1$ state going simultaneously to -2$-2$ and 0$0$ states, which is **(I)** Disproportionation. ### Pattern Recognition Sees: Classic classification table of redox classes. Shortcut: Identify mathrmH_2mathrmO_2$\mathrm{H}_2\mathrm{O}_2$ decomposition as the benchmark example of disproportionation (D-I) to immediately eliminate wrong choices. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions