redox-and-equivalent-concept-quiz-2

Redox And Equivalent Concept Quiz-2

Dear Readers,

The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry..

Q1. Which is the best description of behaviour of bromine in the reaction given below?

 H₂ O+Br₂⟶HBr+HOBr

•  Proton accepted only
•  Both oxidised and reduced
•  Oxidised only
•  Reduced only

Solution


 

Q2.The equivalent weight of SnCl₂ in the reaction, 

SnCl₂+Cl₂ ⟶SnCl₄ is :

•  49
•  95
•  45
•  59

Solution

(b) Sn²⁺ ⟶Sn⁴⁺+2e ∴E=M/2=(119+71)/2=95

Q3. The number of Fe²⁺ ion oxidised by one mole of MnO₄^- ions is : 

•  1/5
•  2/3
•  5
•  3/2

Solution
(c) Mn⁷⁺+5e ⟶Mn²⁺; Fe²⁺ ⟶ Fe³⁺+ e

Q4. How many litre a 0.5 N solution of an oxidising agent are reduced by 2 litre of 2.0 N solution of a reducing agent?

•   8 litre
•   4 litre
•   6 litre
•   7 litre

Solution
(a) Meq.of oxidant=Meq.of reductant 

 0.5 × V=2 × 2000 

 ∴ V=8 litre

Q5.For the reaction : N₂+3H₂ ⟶2NH₃ ; if E₁ and E₂ are equivalent masses of NH₃ and N₂ respectively, then E₁-E₂ is :

•  1
•  2
•  3
•  4

Solution
  (a) (N⁰ )₂+6e ⟶2(N^3-) 3(H⁰ )₂ ⟶2(H¹⁺ )₃+6e 

E_{(N2 )} =28/6;E_{(NH3 )}=17/3  

Q6. 25 mL of 0.50 M H₂ O₂ solution is added to 50 mL of 0.20 M KMnO₄ in acidic solution. Which of the following statements is true?

•  0.010 mole of oxygen is liberated
•  0.005 mole of KMnO₄ are left
•  0.030 g atom of oxygen gas is evolved
•  0.0025 mole H₂ O₂ does not react with KMnO₄

Solution
(b) Meq.of H₂ O₂=25 × 0.5 × 2=25; 

 Meq.of KMnO₄=50 × 0.2 × 5=50; 

 ∴25 Meq.or 5 milli mole of KMnO₄ are left.  

Q7.Of the following reactions, only one is a redox reaction. Identify this reaction.

•  Ca(OH)₂+2HCl⟶CaCl₂+2H₂ O
•  2S₂ O₇^2-+2H₂ O⟶2SO₄^2-+4H⁺
•  BaCl₂+MgSO₄⟶BaSO₄+MgCl₂
•  Cu₂ S+2FeO⟶2Cu+2Fe+SO₂

Solution

Q8.How many g of KMnO₄ are needed to prepare 3.75 litre of 0.850 N solution if KMnO₄ is reduced as, 

MnO₄^-+8H⁺ +5e ⟶Mn²⁺+4H₂ O ?

•  101 g
•  202 g
•  50.5 g
•  303.0 g KMnO₄

Solution
(a) Meq.of KMnO₄=3750 × 0.85 

 ∴ w/31.6 × 1000=3750 × 0.85 

 ∴ w=100.7 g  

Q9.A 0.518 g sample of lime stone is dissolved in HCl and then the calcium is precipitated as CaC₂ O₄. After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO₄, solution acidified with H₂ SO₄ to titrate is as, MnO₄^-+H⁺ +C₂ O₄^2- ⟶Mn²⁺+CO₂+2H₂ O. The percentage of CaO in the sample is :

•  54.0 %
•  27.1 %
•  42%
•  84%

Solution

(a) Meq.of lime stone=Meq.of CaC₂ O₄=Meq.of KMnO₄ = Meq. Of CaO 

 ∴40 × 0.250=w/(56/2) × 1000 

 ∴w_CaO=0.28 

 ∴per cent of CaO=(0.28 × 100)/0.518 = 54%

Q10. A compound contains atoms X,Y,Z. The oxidation number of X is +2, Y is +5 and Z is −2. The possible formula of the compound is :

•  XY₁ Z₂
•  Y₂ (XZ₃ )₂
•  X₃ (YZ₄ )₂
•  X₃ (Y₄ Z)₂

Solution

(c) Sum of oxidation no. of atoms in it is zero.