Given below are two statements : Statement I: In the oxalic acid vs mathrmKMnO_4$\mathrm{KMnO}_4$ (in the presence of dil mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ ) titration the solution needs to be heated initially to 60^circmathrmC$60^{\circ}\mathrm{C}$ , but no heating is required in Ferrous ammonium sulphate (FAS) vs mathrmKMnO_4$\mathrm{KMnO}_4$ titration (in the presence of dil mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ ) Statement II : In oxalic acid vs mathrmKMnO_4$\mathrm{KMnO}_4$ titration, the initial formation of mathrmMnSO_4$\mathrm{MnSO}_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs mathrmKMnO_4$\mathrm{KMnO}_4$ , heating oxidizes mathrmFe^2+$\mathrm{Fe}^{2+}$ into mathrmFe^3+$\mathrm{Fe}^{3+}$ by oxygen of air and error may be introduced in the experiment. In the light of the above statements, choose the correct answer from the options given below:

### 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

### 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

### Related Formula textReducing Power propto frac1textStandard Reduction Potential (E^circ_textred)$\text{Reducing Power} \propto \frac{1}{\text{Standard Reduction Potential } (E^{\circ}_{\text{red}})}$ ### Core Logic A stronger reducing agent undergoes oxidation more easily, which corresponds to the lowest standard reduction potential value among the given options. Comparing the given values: * E^circ_mathrmMnO_4^-/mathrmMn^2+ = +1.51mathrm V$E^{\circ}_{\mathrm{MnO}_{4}^{-}/\mathrm{Mn}^{2+}} = +1.51\mathrm{ V}$ * E^circ_mathrmCl_2/mathrmCl^- = +1.36mathrm V$E^{\circ}_{\mathrm{Cl}_{2}/\mathrm{Cl}^{-}} = +1.36\mathrm{ V}$ * E^circ_mathrmCr_2mathrmO_7^2-/mathrmCr^3+ = +1.33mathrm V$E^{\circ}_{\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}/\mathrm{Cr}^{3+}} = +1.33\mathrm{ V}$ * E^circ_mathrmCr^3+/mathrmCr = -0.74mathrm V$E^{\circ}_{\mathrm{Cr}^{3+}/\mathrm{Cr}} = -0.74\mathrm{ V}$ Since mathrmCr^3+/mathrmCr$\mathrm{Cr}^{3+}/\mathrm{Cr}$ has the lowest standard reduction potential (-0.74mathrm V$-0.74\mathrm{ V}$), elemental metallic mathrmCr$\mathrm{Cr}$ is the most easily oxidized and is therefore the strongest reducing agent. ### Pattern Recognition To find the strongest reducing agent, simply look for the lowest or most negative reduction potential. Ensure you pick the species on the right side of the reduction half-reaction (the reduced form, which will act as the reducer). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 12 Chemistry: Electrochemistry

### Core Logic When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, the species is said to undergo a disproportionation reaction. ### Step 1: Check Reaction (A) Reaction (A): Cu^+ rightarrow Cu^2+ + Cu$Cu^+ \rightarrow Cu^{2+} + Cu$ Oxidation state of Cu$Cu$ goes from +1 to +2 (oxidation) and +1 to 0 (reduction). This is disproportionation. ### Step 2: Check Reaction (B) Reaction (B): 3MnO_4^2- + 4H^+ rightarrow 2MnO_4^- + MnO_2 + 2H_2O$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ Oxidation state of Mn$Mn$ in MnO_4^2-$MnO_4^{2-}$ is +6. In MnO_4^-$MnO_4^-$ it is +7, and in MnO_2$MnO_2$ it is +4. The +6 state oxidizes to +7 and reduces to +4. This is disproportionation. ### Step 3: Check Reactions (C) and (D) Reaction (C): 2KMnO_4 rightarrow K_2MnO_4 + MnO_2 + O_2$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$ Mn$Mn$ in KMnO_4$KMnO_4$ is +7, forming +6 (K_2MnO_4$K_2MnO_4$) and +4 (MnO_2$MnO_2$). Here, Oxygen is oxidized from -2 to 0. This is an intramolecular redox reaction, NOT disproportionation. Reaction (D): 2MnO_4^- + 3Mn^2+ + 2H_2O rightarrow 5MnO_2 + 4H^+$2MnO_4^- + 3Mn^{2+} + 2H_2O \rightarrow 5MnO_2 + 4H^+$ Mn$Mn$ +7 and Mn$Mn$ +2 both convert to Mn$Mn$ +4. This is a comproportionation reaction, the reverse of disproportionation. ### Pattern Recognition Disproportionation requires the SAME element in a SINGLE oxidation state in the reactants to split into a HIGHER and LOWER oxidation state in the products. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions