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JEE Exam  >  JEE Tests  >  Test: Electrochemistry(14 Oct) - JEE MCQ

Test: Electrochemistry(14 Oct) - JEE MCQ


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15 Questions MCQ Test - Test: Electrochemistry(14 Oct)

Test: Electrochemistry(14 Oct) for JEE 2024 is part of JEE preparation. The Test: Electrochemistry(14 Oct) questions and answers have been prepared according to the JEE exam syllabus.The Test: Electrochemistry(14 Oct) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Electrochemistry(14 Oct) below.
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Test: Electrochemistry(14 Oct) - Question 10
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*Multiple options can be correct
Test: Electrochemistry(14 Oct) - Question 11
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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 Test: Electrochemistry(14 Oct) - Question 11

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

Test: Electrochemistry(14 Oct) - Question 12
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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 Test: Electrochemistry(14 Oct) - Question 12

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

Test: Electrochemistry(14 Oct) - Question 13
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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 Test: Electrochemistry(14 Oct) - Question 13

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​= E∘​oxi​+E∘​red
=0.72−0.42 = 0.30 V

 

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