Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper

JEE : Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

The document Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev is a part of the JEE Course Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper.
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Direction : question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t.

Q.

Match the reactions in Columns I with nature of the reactions/type of the products in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

Column IColumn II
Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev(p) redox reaction
 (q) one of the products has trigonal planar structure
 (r) dimeric bridged tetrahedral metal ion
 (s) disproportionation

 

Ans:  A - p, s); (B - r); (C - p, q); (D - p). 

Solution :  A→ p, s; The reaction is redox reaction because the O.N. of O in  Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev is – 0.5 and that in O2 is zero. In Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev is –1.0. It
involves reduction oxidation reaction. Since here a part of molecule is oxidised and a part is reduced so it is disproportionation.
B → r; The structure of  Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev is given below  
 

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

[NOTE : In any solution dichromate ions and chromate ions exist in equilibrium. In alkali solution, dichromate ions are converted into chromate ions and on acidification chromate ions are converted back into dichromate ion.]
C → p, q; The reaction is  Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev In involves change in O.N of Mn   Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev to
+ 2(in Mn2+), So Mn is reduced and NO2–  is oxidised to Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev  it is a redox reaction. The structure of Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev it is a redox reaction.
– (one of the products is trigonal planar)
D → p, It is a redox reaction

 

 

Direction : questions have matching lists. The codes for the lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q. 

An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation in conductivity of these reactions is given in List II. Match list I with List II and select the correct answer using the code given below the lists :                         (JEE Adv. 2013)

List IList II
Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev1. Conductivity decreases and then increases
2. Conductivity decreases and then does not change much
3. Conductivity increases and then does not change much
4. Conductivity does not change much and then increases

 

Codes: 

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Ans: a

Solution :  (a) 
 (p)

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev
Initially conductivity increases because on neutralisation ions are created. After that it becomes practically constant because X alone can
not form ions.

Q.

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Number of ions in the solution remains constant as only AgNO3 precipitated as AgI. Thereafter conductance increases due to increase in number
of ions.

(R) Initially conductance decreases due to the decrease in the number of  Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev  ions as OH is getting replaced by CH3COO which has poorer conductivity
thereafter it slowly increases due to the increase in number of H+ ions.

(S) Initially it decreases due to decrease in H+ ions and then increases due to the increase in OH ions.

 

Q.

The standard reduction potential data at 25°C is given below : (JEE Adv. 2013)
 E°(Fe3+, Fe2+) = + 0.77 V; E°(Fe2+, Fe) = – 0.44 V; E°(Cu2+, Cu) = + 0.34 V; E°(Cu+, Cu) = + 0.52 V
 E°[O2(g) + 4H+ + 4e → 2H2O] = + 1.23 V; E°[O2(g) + 2H2O + 4e → 4OH] = + 0.40 V
 E°(Cr3+, Cr) = – 0.74 V; E°(Cr2+, Cr) = – 0.91 V

 

Match E° of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists :

List IList II
Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev1. – 0.18 V
2. – 0.4 V
3. – 0.04 V
4. – 0.83 V

Codes : 

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Ans: d

Solution : 

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

 

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

 

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Matrix-Match Type Questions: Electrochemistry | JEE Advanced Notes | EduRev
– 0.74V, n = 3
x × 1 + 2 × (– 0.91) = 3 × (– 0.74)
x – 1.82 = – 2.22 ⇒ x = – 0.4V

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