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JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced PDF Download

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
JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced(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  JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced is – 0.5 and that in O2 is zero. In JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced 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  JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced is given below  
 

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

[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  JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced In involves change in O.N of Mn   JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced to
+ 2(in Mn2+), So Mn is reduced and NO2–  is oxidised to JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced  it is a redox reaction. The structure of JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced 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
JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced1. 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: 

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

Ans: a

Solution :  (a) 
 (p)

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced
Initially conductivity increases because on neutralisation ions are created. After that it becomes practically constant because X alone can
not form ions.

Q.

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

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  JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced  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
JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced1. – 0.18 V
2. – 0.4 V
3. – 0.04 V
4. – 0.83 V

Codes : 

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

Ans: d

Solution : 

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

 

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

 

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electrochemistry | Chapter-wise Tests for JEE Main & Advanced
– 0.74V, n = 3
x × 1 + 2 × (– 0.91) = 3 × (– 0.74)
x – 1.82 = – 2.22 ⇒ x = – 0.4V

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