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Electrochemistry
Faraday's Laws of Electrolysis
5 questions from 5 papers, each with its answer.
- questions
- 5
- papers
- 5
- years asked, 2017–2026
- 2 of 10
- of Electrochemistry
- 6.76%
Appears in
Concepts tested
- faraday-time-chlorine
- Mass deposited
- Number of Faradays required for redox conversions
- Second law of electrolysis
All Questions
2026 1 question
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A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: [Given: Molar mass of Cu = 63 g $mol^{-1}$; 1 F = 96487 C $mol^{-1}$]$w = \frac{E}{96500} \times I \times t$ $w = \frac{63}{2 \times 96500} \times 1.5 \times 10 \times 60$ $w = 0.293$ g
2024 1 question
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Match List-I with List-II regarding the number of Faraday required for the given conversions.
Pick the matching item from Column II for each row.
Column I
- A. 1 mol of $H_2O$ to $O_2$
- B. 1 mol of $MnO_4^-$ to $Mn^{2+}$
- C. 1.5 mol of $Ca$ from molten $CaCl_2$
- D. 1 mol of $FeO$ to $Fe_2O_3$
Column II
- I. 3F
- II. 2F
- III. 1F
- IV. 5F
Correct answer: A → II, B → IV, C → I, D → III
(A) $H_2O \rightarrow 2H^+ + \frac{1}{2}O_2 + 2e^-$; Q = nF = 2F (B) $MnO_4^- + 5e^- \rightarrow Mn^{2+}$; Q = 5F (C) $Ca^{2+} + 2e^- \rightarrow Ca$; 1 mol Ca needs $2e^-$, so 1.5 mol needs $3e^-$; Q = 3F (D) Oxidation number of Fe in FeO is +2 and in $Fe_2O_3$ is +3; 1 electron per Fe is required, so Q = 1F Correct match: A-II, B-IV, C-I, D-III.
2016 Phase II 1 question
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During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is
0.1 mol Cl₂ needs 0.2 F of charge.
$2Cl^- \rightarrow Cl_2(g) + 2e^-$
$W = \frac{E}{96500}\times it$
$0.1\times71 = \frac{35.5}{96500}\times3\times t(sec)$
t = 6433.33 s = 107.22 min ≈ 110 min
2009 AIPMT 1 question
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$Al_2O_3$ is reduced by electrolysis at low potentials and high currents. If $4.0\times10^4$ amperes of current is passed through molten $Al_2O_3$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, at. mass of Al = 27 g $mol^{-1}$)Charge passed: $Q = It = 4.0\times10^4\times6\times3600 = 8.64\times10^8$ C
Moles of electrons = $\frac{Q}{F} = \frac{8.64\times10^8}{96500} = 8953$ mol
Cathode reaction: $Al^{3+} + 3e^- \rightarrow Al$, so 3 mol of electrons deposit 1 mol (27 g) of Al.
Moles of Al = $\frac{8953}{3} = 2984$ mol
Mass of Al = $2984\times27 = 8.06\times10^4$ g
$\approx 8.1\times10^4$ g
2005 AIPMT 1 question
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4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from $Al^{3+}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^+$ ions in solution by the same quantity of electric charge will be -By Faraday's second law, the same quantity of electricity liberates the same number of equivalents of different substances.
$Al^{3+} + 3e^- \rightarrow Al$, so the equivalent mass of Al $= \dfrac{27}{3} = 9$ g
Equivalents of Al deposited $= \dfrac{4.5}{9} = 0.5$
So the same charge (0.5 F) liberates 0.5 equivalent of hydrogen.
$2H^+ + 2e^- \rightarrow H_2$: 1 mole of $H_2$ = 2 equivalents, so 0.5 equivalent = 0.25 mol of $H_2$.
Volume at STP $= 0.25 \times 22.4 = 5.6$ L