REDOX REACTIONS Quiz-10

Chemistry Aptitude Quiz

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IIT JEE exam which consists of JEE Main and JEE Advanced is one of the most important entrance exams for engineering aspirants. The exam is held for candidates who are aspiring to pursue a career in the field of engineering and technical studies. Chemistry is important because everything you do is chemistry! Even your body is made of chemicals. Chemical reactions occur when you breathe, eat, or just sit there reading. All matter is made of chemicals, so the importance of chemistry is that it's the study of everything..

Q1. 1-Propanol and 2-propanol can be best distinguished by:

•  Oxidation with alkaline KMnO_4followed by reaction with Fehling’s solution
•  Oxidation with acidic dichromate followed by reaction with Fehling’s solution
•  Oxidation by heating with copper followed by reaction with Fehling’s solution
•  Oxidation with concentrated H_2 SO_4followed by reaction with Fehling solution

Solution

Q2.Which one of the following will most readily by dehydrated in acidic conditions?

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Solution

Q3.  Compound (A)(→┴(2 mol of))┬(HIO_4 ) 2 mol of glyoxalic acid.The compound (A) is:

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Solution

Q4.   When KMnO_4 acts as an oxidizing agent and ultimately from MnO_4^(2-), MnO_2, Mn_2 O_3, and Mn^(2+) then the numbers of electrons transferred in each case, respectively, are

•  4, 3, 1, 5
•  1, 5, 3, 7
•  1, 3, 4, 5
•  3, 5, 7, 1

Solution
(c)
e^-+MnO_4^⊖⟶MnO_4^( 2-)
3e^-+MnO_4^⊖⟶MnO_2
4e^-+2MnO_4^⊖⟶Mn_2 O_3
5e^-+MnO_4^⊖⟶Mn^(2+)

Q5.The number of moles of K_2 Cr_2 O_7 reduced by 1 mol of Sn^(2+) is

•  1/6
•  1/3
•  2/3
•  1

Solution
  (c) Eq of Cr_2 O_7^(2-)≡Eq ofSn^(2+) (n=6)(n=2) 1/6 mol=1/2 mol 1/3 mol of Cr_2 O_7^(2-)=1 mol ofSn^(2+)

Q6. Which of the following acids possesses oxidising, reducing, and complex forming properties?

•  HNO_3
•  H_2 SO_4
• HCL
•  HNO_2

Solution
(d) In HNO_2, the oxidation state of N is +3. So it can go to a higher or lower oxidation state

Q7.

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Solution

Q8.

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Solution

Q9.The oxidation states of sulphur in the anions SO_3^(2-),S_2 O_4^(2-)andS_2 O_6^(2-) follow the order

•  S_2 O_4^(2-)< SO_3^(2-)< S_2 O_6^(2-)
•  SO_3^(2-)< S_2 O_4^(2-)< S_2 O_6^(2-)
•  S_2 O_4^(2-)< S_2 O_6^(2-)< SO_3^(2-)
•  S_2 O_6^(2-)< S_2 O_4^(2-)< SO_3^(2-)

Solution
(a)
S_2 O_6^(2-):2x-12=-2⇒x=5
SO_3^(2-):x-6=-2⇒x=4
S_2 O_4^(2-):2x-8=-2⇒x=3

Q10. In Which of the following cases is the oxidation state of N atom wrongly calculated?

Compound     Oxidation state

•  NH_4 Cl       -3
•  (N_2 H_5 )_2 SO_4   +2
•  Mg_3 N_2       -3
• NH_2 OH       -1

Solution

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