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Redox And Equivalent Concept Quiz-2

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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(N3-) 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 O72-+2H2 O⟶2SO42-+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 O42- ⟶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 
 ∴wCaO=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.


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