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Electrochemistry - 1 - JEE MCQ


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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Electrochemistry - 1

Electrochemistry - 1 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Electrochemistry - 1 questions and answers have been prepared according to the JEE exam syllabus.The Electrochemistry - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Electrochemistry - 1 below.
Solutions of Electrochemistry - 1 questions in English are available as part of our Chemistry for JEE Main & Advanced for JEE & Electrochemistry - 1 solutions in Hindi for Chemistry for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Electrochemistry - 1 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Electrochemistry - 1 - Question 1

Which of the following reactions is possible at anode ?           

[AIEEE-2002]

Detailed Solution for Electrochemistry - 1 - Question 1

The correct answer is Option C.

The anode is the electrode where oxidation (loss of electrons) takes place; in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode. 
 
The cathode is the electrode where reduction (gain of electrons) takes place; in a galvanic cell, it is the positive electrode, as less oxidation occurs, fewer ions go into solution, and fewer electrons are left on the electrode.
So, in the reaction 
2Cr3+ +7H2O→Cr2O72−+14H+
 
Cr is getting oxidised as the oxidation state is changing from +3 to +6 and hence this reaction is possible at anode.
 

*Multiple options can be correct
Electrochemistry - 1 - Question 2

For a cell given below Ag | Ag+ || Cu2+ | Cu 

—              + 

Ag+ + e- → Ag,      Eº = x 

Cu2+ +2e- → Cu,    Eº = y 

Eº cell is –                   

 [AIEEE-2002]

Detailed Solution for Electrochemistry - 1 - Question 2

The correct answer is Option C.

The left portion of the salt bridge(| |)  is anode and the right portion of the salt bridge(| |) is cathode.

Since, Ag | Ag+ is half-cell oxidation and Cu2+ | Cu is half-cell reduction. Thus,
Eo cell = E cathode - E anode
           = y - x

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Electrochemistry - 1 - Question 3

The standard e.m.f. of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is (F = 96500 C mol-1 ; R = 8.314 JK-1mol-1)      

[AIEEE-2004]

Detailed Solution for Electrochemistry - 1 - Question 3

The correct answer is option C
For a cell reaction in equilibrium at 298 K,
Eocell =0.0591/nlogKc
(Ke = equilibrium constant)
Give,     Eocell = 0.591V
Now, log Kc = (Eocell x n )/ 0.0591
                   =(0.0591 x n )/ 0.0591
log Kc = 10
Kc = antilog 10

  1. Kc = 1 x 1010
Electrochemistry - 1 - Question 4

Aluminium oxide may be electrolysed at 1000ºC to furnish aluminium metal (At. Mass=27 amu ; 1 Faraday = 96,500 Coulombs). The cathode reaction is

Al3+ + 3e-→ Alº 

To prepare 5.12 kg of aluminium metal by this method would require -    

[AIEEE-2005]

Detailed Solution for Electrochemistry - 1 - Question 4

The correct answer is option B
Al3+ + 3e− → Al
w = zQ
where w = amount of metal
=5.12kg = 5.12 × 103g
z = electrochemical equivalent

Electrochemistry - 1 - Question 5

The cell, Zn | Zn2+ (1M) | | Cu2+(1M)|Cu (Eºcell)=1.10 V), was allowed to be completely discharged at 298 K.

the relative concentration of zn2+ to 

Detailed Solution for Electrochemistry - 1 - Question 5

The correct answer is option C
If it is allowed to completely discharge, Ecell​=0

Electrochemistry - 1 - Question 6

Given EºCr3+/Cr = – 0.72 V, EºFe2+/Fe= – 0.42 V. The potential for the cell Cr |Cr3+ (0.1 M)| |Fe2+ (0.01 M) | Fe is -

[AIEEE 2008]

Detailed Solution for Electrochemistry - 1 - Question 6

The correct answer is option D
Cr∣Cr3+(0.1 M)∣∣Fe2+(0.01 M)∣Fe
Oxidation half-cell
Cr→Cr3++3e]×2
Reduction half-cell
Fe2++2e−→Fe]×3
Net cell reaction
2Cr+3Fe2+→2Cr3++3Fe, n=6
E∘​cell​= Eoxi​+E​red
=0.72−0.42 = 0.30 V

 

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