200 text cc$200 \text{ cc}$ of x times 10^-3 text M$x \times 10^{-3} \text{ M}$ potassium dichromate is required to oxidise 750 text cc$750 \text{ cc}$ of 0.6 text M$0.6 \text{ M}$ Mohr's salt solution in acidic medium. Here x =

### Related Formula textMoles = fractextGiven MasstextMolar Mass$$\text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}}$$ ### Core Logic Balanced reaction: textMnO_2 + 4textHCl rightarrow textMnCl_2 + textCl_2 + 2textH_2textO$$\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}$$ Molar mass of textMnO_2 = 55 + 32 = 87 text g/mol$\text{MnO}_2 = 55 + 32 = 87 \text{ g/mol}$. Moles of textMnO_2 = frac8.787 = 0.1 text mol$\text{MnO}_2 = \frac{8.7}{87} = 0.1 \text{ mol}$. From stoichiometry, 1 mole of textMnO_2$\text{MnO}_2$ produces 1 mole of textCl_2$\text{Cl}_2$. Therefore, moles of textCl_2$\text{Cl}_2$ produced = 0.1 text mol$0.1 \text{ mol}$. ### Step 1: Final Calculation Weight of textCl_2 = 0.1 times 71 = 7.1 text g$\text{Cl}_2 = 0.1 \times 71 = 7.1 \text{ g}$. ### Pattern Recognition Sees: stoichiometry calculation with manganese dioxide and hydrochloric acid. Trap: Forgetting stoichiometric coefficients for chlorine gas. ### Chapter Mix Class 11 Chemistry: Redox Reactions

### Core Logic Dichromate ion (Cr_2O_7^2-$Cr_2O_7^{2-}$) in acidic medium acts as a strong oxidizing agent, getting reduced to Cr^3+$Cr^{3+}$. This reduction process consumes 6$6$ electrons per dichromate ion: Cr_2O_7^2- + 14H^+ + 6e^- rightarrow 2Cr^3+ + 7H_2O$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ Reaction 1: Oxidation of Iodide 6I^- rightarrow 3I_2 + 6e^-$6I^- \rightarrow 3I_2 + 6e^-$ Overall balanced reaction: Cr_2O_7^2- + 14H^+ + 6I^- rightarrow 2Cr^3+ + 3I_2 + 7H_2O$Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$ Number of electrons involved (X) = 6. Reaction 2: Oxidation of Sulfide 3S^2- rightarrow 3S + 6e^-$3S^{2-} \rightarrow 3S + 6e^-$ Overall balanced reaction: Cr_2O_7^2- + 14H^+ + 3S^2- rightarrow 2Cr^3+ + 3S + 7H_2O$Cr_2O_7^{2-} + 14H^+ + 3S^{2-} \rightarrow 2Cr^{3+} + 3S + 7H_2O$ 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 (Cr^6+$Cr^{6+}$ to Cr^3+$Cr^{3+}$, two atoms) always demands 6 electrons in a balanced equation block. The reducing agent must therefore supply exactly 6 electrons to match. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The d- and f-Block Elements

### 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