Tiêu đề Redox Reaction _ Practice Sheet _ Varun JEE Advanced 2024.pdf ✅

JEE JEE Advanced (2024) Q1 The reaction, is an example of (A) oxidation reaction (B) reduction reaction (C) disproportionation reaction (D) decomposition reaction Q2 In the standardization of Na S O using K Cr O by iodometry, the equivalent weight of K Cr O is (A) (molecular weight)/2 (B) (molecular weight)/6 (C) (molecular weight)/3 (D) same as molecular weight Q3 The half-cell reactions for rusting of iron are: Fe + 2e → Fe(s); E°=– 0.44 VΔG° (in kJ) for the reaction is (A) –76 (B) –322 (C) –122 (D) –176 Q4 For the following cell, Zn(s) | ZnSO (aq) || CuSO (aq) I Cu(s) when the concentration of Zn is 10 times the concentration of Cu , the expression for ΔG (in J mol ) is [F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1 V] (A) 2.303RT + 11 F (B) 1.1 F (C) 2.303RT – 2.2 F (D) –2.2 F Q5 An element A in a compound ABD has oxidation number A . It is oxidised by Cr O in acidic medium. In the experiment 1.68 × 10 moles of K Cr O were used for 3.26 × 10 moles of ABD. The new oxidation number of A after oxidation is: (A) 3 (B) 3 – n (C) n – 3 (D) +n Q6 If 10 g of V O is dissolved in acid and is reduced to V by zinc metal, how many mole of I could be reduced by the resulting solution if it is further oxidised to VO ions? [Assume no change in state of Zn ions] (Atomic mass: V = 51, O = 16, I = 127): (A) 0.11 mole of I (B) 0.22 mole of I (C) 0.055 mole of I (D) 0.44 mole of I Q7 A 150 mL of solution of I is divided into two unequal parts. Part-I reacts with hypo solution in acidic medium. 15 mL of 0.4 M hypo was consumed.Part-II was added with 100 mL of 0.3 M NaOH solution. Residual base required 10 mL of 0.3 M H SO solution for complete neutralization. What was the initial concentration of I ? (A) 0.08 M (B) 0.1 M (C) 0.2 M (D) None of these Q8 A mixture of H SO and H C O (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre, 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment 100 mL of the same solution in hot condition required 4 mL of 0.02M KMnO solution for complete reaction. The wt. % of H SO in the mixture was: (A) 40 (B) 50 (C) 60 (D) 80 Q9 0.80 g of sample of impure potassium dichromate was dissolved in water and made upto 500 mL solution. 25 mL of this solution treated with excess of KI in acidic medium and I liberated required 24 mL of a sodium thiosulphate solution. 30 mL of this sodium thiosulphate solution required 15 mL of N/20 solution of pure potassium dichromate. What was the percentage of K Cr O in given sample? (A) 73.5 % (B) 75.3 % (C) 36.75 % (D) None of these Directions (10) Read the following passage and answer the given questions. Equivalent weight = Chemistry Redox Reaction 3 (aq) → (aq) + 2 (aq) ClO− ClO− 3 Cl− 2 2 3 2 2 7 2 2 7 2H + 2 + → O (l); = + e− 1 2 O2 H2 E ∘ +1.23 V 2+ – 4 4 2+ 2+ –1 n– 2 7 2– –3 2 2 7 –3 2 5 2+ 2 2+ 2+ 2 2 2 2 2 2 4 2 2 4 2 2 4 4 2 4 2 2 2 7 Molecular weight n−factor
JEE n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predicts the molar ratio of the reactant species taking part in reactions. The reciprocal of n-factor's ratio of the reactants is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H /OH furnished per mole of acid/base. n-factor of a reactant is no. of moles of electrons lost or gained per mole of reactant. Example 1 : 1. In acidic medium : KMnO (n = 5) → Mn 2. In basic medium : KMnO (n = 3) → Mn In neutral medium : KMnO (n = 1) → Mn Example 2: FeC O → Fe + 2CO Total no. of moles of e lost by 1 mole of FeC O = 1 + 1 × 2 ⇒ 3 ⸫ n-factor of FeC O = 3 Q10 n-factor of Ba(MnO ) in acidic medium is : (A) 2 (B) 6 (C) 10 (D) None of these Directions (11) Read the following passage and answer the given questions. Equivalent weight = n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predicts the molar ratio of the reactant species taking part in reactions. The reciprocal of n-factor's ratio of the reactants is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H /OH furnished per mole of acid/base. n-factor of a reactant is no. of moles of electrons lost or gained per mole of reactant. Example 1 : 1. In acidic medium : KMnO (n = 5) → Mn 2. In basic medium : KMnO (n = 3) → Mn 3. In neutral medium : KMnO (n = 1) → Mn Example 2: FeC O → Fe + 2CO Total no. of moles of e lost by 1 mole of FeC O = 1 + 1 × 2 ⇒ 3 ⸫ n-factor of FeC O = 3 Q11 For the reaction, H PO + NaOH → NaH PO + H O What is the equivalent weight of H PO ? (mol. wt. is M) (A) M (B) M/2 (C) M/3 (D) None of these Directions (12) Read the following passage and answer the given questions. Equivalent weight = n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predicts the molar ratio of the reactant species taking part in reactions. The reciprocal of n-factor's ratio of the reactants is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H /OH furnished per mole of acid/base. n-factor of a reactant is no. of moles of electrons lost or gained per mole of reactant. Example 1 : 1. In acidic medium : KMnO (n = 5) → Mn 2. In basic medium : KMnO (n = 3) → Mn 3. In neutral medium : KMnO (n = 1) → Mn Example 2: FeC O → Fe + 2CO Total no. of moles of e lost by 1 mole of FeC O = 1 + 1 × 2 ⇒ 3 ⸫ n-factor of FeC O = 3 Q12 For the reaction, Fe O (molar mass : M) → Fe O . What is the eq. wt. of Fe O? (A) (B) (C) (D) None of these Q13 H PO is a tri basic acid and one of its salt is NaH PO . What volume of 1 M NaOH solution should be added to 12 g of NaH PO to convert it into Na PO ? Q14 1.64 g of mixture of CaCO and MgCO was dissolved in 50 mL of 0.8 M HCl. The excess of acid required 16 mL of 0.25 M NaOH for neutralization. Calculate the percentage of CaCO and MgCO in the sample. Q15 1.5 g of chalk were treated with 10 mL of 4N HCl. The chalk was dissolved and the solution made to 100 mL, 25 mL of this solution required 18.75 mL of 0.2 N NaOH solution for complete neutralisation. Calculate the percentage of pure CaCO in the sample of chalk? Q16 + – 4 2+ 4 2+ 4 6+ 2 4 3+ 2 – 2 4 2 4 4 2 Molecular weight n−factor + – 4 2+ 4 2+ 4 6+ 2 4 3+ 2 – 2 4 2 4 3 2 2 2 2 3 2 Molecular weight n−factor + – 4 2+ 4 2+ 4 6+ 2 4 3+ 2 – 2 4 2 4 0.95 2 3 0.95 M 0.85 M 0.95 M 0.8075 3 4 2 4 2 4 3 4 3 3 3 3 3
JEE A 100 mL solution contains Na CO and NaHCO . 20 mL of this solution required 4 mL of 1N HCl for titration with Ph indicator. The titration was repeated with the same volume of the solution but with MeOH. 10.5 mL of 1 N HCl was required this time. If x and y are the amount (in g) of Na CO & NaHCO respectively, the value of (x + y) is: Q17 Pottasium acid oxalate K C O .3H C O .4H O can be oxidized by KMnO in acid medium. Calculate the volume (in mL) of 0.1 M KMnO reacting in acid solution with one gram of the acid oxalate. Q18 A volume of 12.53 mL of 0.05093 M SeO reacted with exactly 25.52 mL of 0.1 M CrSO . In the reaction, Cr was oxidized to Cr . To what oxidation state was selenium converted by the reaction. Q19 1 g of a moist sample of a mixture of KClO and KCl was dissolved in water and made upto 250 mL. 25 mL of this solution was treated with SO to reduced chlorate into chloride and the excess SO was boiled off. When the total chloride was precipitated, 0.1435 g of AgCl was obtained. In another experiment 25 mL of the original solution was treated with 30 mL of 0.2 N solution of FeSO and unreacted FeSO required 37.5 mL of 0.08 N solution of an oxidizing agent for complete oxidation. Calculate the molar ratio of chlorate and chloride in the given mixture. Fe reacts with according to equation: Also calculate the mass percent of moisture present in the moist sample. Q20 A steel sample is to be analysed for Cr and Mn simultaneously. By suitable treatment the Cr is oxidized to and the Mn to . A 10.00 g sample of steel is used to produce 250.0 mL of a solution containing and . A 10.00 mL portion of this solution is added to a BaCl solution and by proper adjustment of the acidity, the chromium is completely precipitated as BaCrO ; 0.0549 g is obtained. A second 10.00 mL portion of this solution requires exactly 15.95 mL of 0.0750 M standard Fe solution for its titration (in acid solution). Calculate the % of Mn and % of Cr in the steel sample. (Atomic mass: Cr = 52, Mn = 55) 2 3 3 2 3 3 2 2 4 2 2 4 2 4 4 2 4 2+ 3+ 3 2 2 4 4 2+ ClO− 3 ClO + 6 + 6 → + 6 − 3 Fe2+ H + Cl− Fe3+ + 3H2O Cr2 O2− 7 MnO− 4 Cr2 O2− 7 MnO− 4 2 4 2+
JEE Answer Key Q1 (C) Q2 (B) Q3 (B) Q4 (C) Q5 (B) Q6 (A) Q7 (B) Q8 (A) Q9 (A) Q10 (C) Q11 (A) Q12 (A) Q13 (200 mL) Q14 (MgCO = 51.22%, CaCO = 48.78%) Q15 (83.33) Q16 (3.17) Q17 (31.5) Q18 (Zero) Q19 Q20 (Cr = 2.821%, Mn = 1.498%) 3 3 (ClO / = 1, 1.5 − 3 Cl− % moisture by mass)