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Q. No.Electrolysis of aqueous solution of \(CuSO_4\) is carried out, where 300 mg of copper is deposited (atomic mass of Cu = 63.54). After this 600 milli amp. current is further passed for 28 minutes. Calculate total volume of \(O_2\) released (in ml), given that 1 mole of a gas occupy 22.4 litres.
Given half reactions:
\(\mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{~s})\)
\(2 \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-}\)
1. 110
2. 111.15
3. 101.5
4. 80.5
Given half reactions:
\(\mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{~s})\)
\(2 \mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-}\)
1. 110
2. 111.15
3. 101.5
4. 80.5
Subtopic: Electrode & Electrode Potential | Faraday’s Law of Electrolysis |
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If one coulomb of electric charge is passed through a solution of silver nitrate (\(\mathrm{AgNO}_3\)) during electrolysis, what is the mass of silver (in milligrams) deposited at the electrode?
(Given: Molar mass of Ag = 108 g/mol)
(Given: Molar mass of Ag = 108 g/mol)
| 1. | 1.12 mg | 2. | 3.26 mg |
| 3. | 2.54 mg | 4. | 4.02 mg |
Subtopic: Faraday’s Law of Electrolysis |
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Level 1: 80%+
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If one Faraday of electricity is used in the discharging of \(\text{Cu}^{2+},\) then find the mass (in g) of Cu deposited (In nearest integer):
1. 32
2. 19
3. 90
4. 78
1. 32
2. 19
3. 90
4. 78
Subtopic: Faraday’s Law of Electrolysis |
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Level 1: 80%+
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In the aqueous solution of \([\text{AuCl}_4]^-\), the current is passed for 10 minutes during the electrolysis, the mass Au deposited at Cathode is 1.97 gm. Find out current required (in A): (Nearest integer)
1. 6
2. 5
3. 1
4. 9
1. 6
2. 5
3. 1
4. 9
Subtopic: Faraday’s Law of Electrolysis |
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Level 2: 60%+
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What is the mass of\(\text { Zn }\)(in mg) is deposited when a current of \(0.015~\text{A}\) was passed for \(15\) minutes through a \(\mathrm{ZnSO}_4\) solution.
[In nearest integer]
[Given: Atomic weight of \(\text {Zn }=65.4\) ]
1. Seven (7)
2. Eight (8)
3. Ten (10)
4. Five (5)
[In nearest integer]
[Given: Atomic weight of \(\text {Zn }=65.4\) ]
1. Seven (7)
2. Eight (8)
3. Ten (10)
4. Five (5)
Subtopic: Faraday’s Law of Electrolysis |
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How many moles of electrons are required to reduce 1 mole of permanganate ions into manganese dioxide?
1. 3
2. 6
3. 4
4. 7
1. 3
2. 6
3. 4
4. 7
Subtopic: Faraday’s Law of Electrolysis |
80%
Level 1: 80%+
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How many charges in Faraday is required in the conversion of 1 mole \(Cr_2O_7^{2-}\) into \(Cr^{3+} \)?
1. 3
2. 6
3. 4
4. 2
1. 3
2. 6
3. 4
4. 2
Subtopic: Faraday’s Law of Electrolysis |
64%
Level 2: 60%+
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Electrolysis of Fe₂(SO₄)₃ solution is carried out for 'x' minutes with a current of 1.5 A, depositing 0.3482 g of Fe. What is the value of x to the nearest integer?
[Given: \(1 ~\text{F}=96500 ~\text{C} ~\text{mol}^{-1}\)
Atomic mass of \(\text{Fe}=56~ \text{g}~ \text{mol}^{-1}\)]
1. 20
2. 25
3. 18
4. 17
[Given: \(1 ~\text{F}=96500 ~\text{C} ~\text{mol}^{-1}\)
Atomic mass of \(\text{Fe}=56~ \text{g}~ \text{mol}^{-1}\)]
1. 20
2. 25
3. 18
4. 17
Subtopic: Faraday’s Law of Electrolysis |
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Level 3: 35%-60%
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A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. The number of moles of Ni that will be deposited at the cathode are:
| 1. | 0.10 | 2. | 0.05 |
| 3. | 0.20 | 4. | 0.15 |
Subtopic: Faraday’s Law of Electrolysis |
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Level 2: 60%+
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Aluminium oxide may be electrolysed at 1000 °C to furnish aluminium metal.
The cathode reaction is Al3+ + 3e- → Al.
To prepare 5.12 kg of aluminium metal by this method would require:
(Atomic mass = 27 amu; 1 faraday=96,500 Coulombs)
1. 5.49×101 C of electricity
2. 5.49×104 C of electricity
3. 1.83×107 C of electricity
4. 5.49×107 C of electricity
Subtopic: Faraday’s Law of Electrolysis |
63%
Level 2: 60%+
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