redox-and-equivalent-concept-quiz-10

Redox And Equivalent Concept Quiz-10

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 not a redox reaction?

•  H₂+Br₂ ⟶2HBr
•  NH₄ Cl ⟶NH₃+HCl
•  NH₃ NO₃ ⟶N₂ O+2H₂ O
•  Fe+S ⟶FeS

Solution
(b) No change in ox.no. of any species.  

Q2.The halogen that shows same oxidation state in all its compounds with other elements is:

•  I₂
•  F₂
•  Cl₂
•  Br₂

Solution

(b) F₂ shows only −1 ox.no.

Q3. Oxidation states of X,Y,Z are +2, +5 and -2 respectively. Formula of the compound formed by these wii be  

•  X₂ YZ₆
•  XY₂ Z₆
•  XY₅
•  X₃ YZ₄

Solution
(b) The sum of the oxidation states is always zero in neutral compound. 

 The oxidation state of X,Y, and Z are +2, +5 and -2 respectively.

 In X₂ YZ₆ 2×2+5+6(-2)≠0 In XY₂ Z₆ 2+5×2+6(-2)=0 

 In XY₅ 2+5×5≠0 In X₃ YZ₄ 3×2+5+4(-2)≠0 

 Hence, the formula of the compound is XY₂ Z₆.

Q4. In acid medium Zn reduces nitrate ion to NH₄⁺ ion according to the reaction Zn+NO₃ Zn²⁺+NH₄⁺+H₂ O (unbalanced) How many moles of HCl are required to teduce half a mole of NaNO₃ completely? Assume the availability of sufficient Zn.

•  5
•  4
•  3
•  2

Solution
(a) ∴4Zn+NO₃^-+10H⁺⟶4Zn²⁺+NH₄⁺+3H₂ O(Net equation)

 4Zn+NO₃^-+10HCl⟶4Zn²⁺+NH₄⁺+5Cl₂+3H₂ O 

 ∵1 mole of NO₃^- (0r NaNO₃) is reduced by =10 moles of HCl 

 ∴1/2 mole of No₃^- will be reduced by = 10×1/2 moles of HCl = 5 moles of HCl

Q5.Which one of the compound does not decolourised an acidified solution of KMnO₄?

•  SO₂
•  FeCl₃
•  H₂ O₂
•  FeSO₄

Solution
  (b) FeCl₃ cannot be oxidised because Fe has highest oxidation state.  

Q6. In the aluminothermic process, aluminium acts as :

•  An oxidising agent
•  A flux
•  A reducing agent
•  A solder

Solution
(c) Cr₂ O₃+2Al ⟶Al₂ O₃+2Cr.  

Q7.An element A in a compound ABD has oxidation number A^n-. It is oxidized by Cr₂ O₇^2- in acidic medium. In the experiment 1.68 × 10^-3 mole of K₂ Cr₂ O₇ were used for 3.26 × 10^-3 mole of ABD. The new oxidation number of A after oxidation is :

•  3
•  3-n
•  n-3
•  +n

Solution

(b) A^n-⟶A^a++(a+n)e Cr₂⁶⁺+6e ⟶2Cr⁺³ 

 Also, Meq of A=Meq.of K₂ Cr₂ O₇ 3.26 × 10^-3 (a+n)=1.68 × 10^-3 × 6 Or a+n=3 

 ∴ a=3-n

Q8.CrO₅ reacts with H₂ SO₄ to give Cr₂ (SO₄ )₃,H₂ O and O₂. Moles of O₂ liberated by 1 mole of CrO₅ in this reaction are :

•  2.5
•  1.25
•  4.5
•  1.75

Solution
(d) 4CrO₅+6H₂ SO₄ ⟶2Cr₂ (SO₄ )₃+6H₂ O+7O₂  

Q9.Reaction of Br₂ with Na₂ CO₃ in aqueous solution gives sodium bromide and sodium bromate with evolution of CO₂ gas. The number of sodium bromide molecules involved in the balanced chemical equation is

•  1
•  3
•  5
•  7

Solution

(c) Br₂ is disproportionated in basic medium as 3Br₂+3Na₂ CO₃⟶5NaBr+NaBrO₃+3CO₂

Q10. MnO₄^- is a good oxidising agent in different medium changing to MnO₄^- ⟶Mn²⁺ ⟶ MnO₄^2- ⟶ MnO₂ ⟶Mn₂ O₃ Changes in oxidation number respectively are

•  1,3,4,5
•  5,4,3,2
•  5,1,3,4
•  2,6,4,3

Solution

(c) MnO₄^-=Mn=+7 

 MnO₄^2-=Mn=+6 

 MnO₂=Mn=+4 

 Mn₂ O₃=Mn=+3 

 Hence, changes in oxidation number are 5,1,3,4.