Q.State Faraday's laws of electrolysis.

Showing the 12 most recent of 21 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.Which of the following equations represents the Faraday's first law of electrolysis ?

   (a) mz = c.t

   (b) m = c.z.t

   (c) mc = z.t

   (d) c = m.z.t

   ›Reveal solutionSolution

   Faraday's first law: mass deposited m is proportional to the quantity of charge, m = z x Q = z x c x t (c = current, t = time, z = electrochemical equivalent).

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A) : Reduction of 1 mole of Cu2+ ions requires 2 Faraday of charge. Reason (R) : 1 Faraday is equal to the charge of 1 mole of electrons.

   (a) Both (A) and (R) are true and (R) is the correct explanation of (A)

   (b) Both (A) and (R) are true but (R) is not the correct explanation of (A)

   (c) (A) is true but (R) is false.

   (d) (A) is false but (R) is true.

   ›Reveal solutionSolution

   Cu²⁺ + 2e⁻ → Cu needs 2 moles of electrons, i.e. 2 Faraday of charge, and 1 Faraday is defined as the charge carried by 1 mole of electrons — so the reason directly explains the assertion.

   Assertion: Reduction of 1 mole of Cu²⁺ requires 2 Faraday of charge.

   The reduction half-reaction is:

   $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

   To reduce 1 mole of Cu²⁺ ions, 2 moles of electrons are needed. Since 1 Faraday (F) is exactly the amount of charge carried by 1 mole of electrons (F = N_A × e ≈ 96500 C/mol), 2 moles of electrons corresponds to 2 Faraday of charge. So the assertion is true.

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The charge required to reduce 1 mol of MnO₄⁻ to MnO₂ is-

   (a)

   (i) 1F

   (b)

   (ii) 3F

   (c)

   (iii) 5F

   (d)

   (iv) 6F

   ›Reveal solutionSolution

   $\text{Mn}$ goes from $+7$ (in $\text{MnO}_4^{-}$) to $+4$ (in $\text{MnO}_2$), a gain of 3 electrons per Mn; 1 mole requires 3 F. Correct option: (ii).

   Concept. By Faraday's laws, the charge needed to reduce 1 mole of a species equals (number of electrons gained per ion) $\times$ 1 F, where $1\ \text{F} = 96500\ \text{C}$ is the charge of 1 mole of electrons.

   Steps.

   • Oxidation state of Mn in $\text{MnO}_4^{-}$: $x + 4(-2) = -1 \Rightarrow x = +7$.
   • Oxidation state of Mn in $\text{MnO}_2$: $x + 2(-2) = 0 \Rightarrow x = +4$.
   • Change $= +7 \rightarrow +4$, so each Mn gains $7-4 = 3$ electrons. …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.How much charge is required for the 1 mol Al3+ to Al?

   (a) 1F

   (b) 2F

   (c) 4F

   (d) 3F

   ›Reveal solutionSolution

   Al3+ + 3e- -> Al, so 1 mol Al needs 3 mol electrons = 3F.

   The reduction half-reaction is: Al3+ + 3e- -> Al.

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.If 96500 coulomb of electricity is passed through CuSO4 solution, it will liberate-

   (a) 63.5 gm copper

   (b) 100 gm copper

   (c) 96500 gm copper

   (d) None of the these

   ›Reveal solutionSolution

   96500 C = 1 Faraday = 1 mole of electrons, but Cu2+ needs 2 electrons per atom, so only half a mole (31.75 g) of copper is deposited.

   At the cathode: $Cu^{2+} + 2e^- \to Cu$.

   $96500\ C = 1\ F = 1$ mole of electrons. Since 2 moles of electrons are needed to deposit 1 mole (63.5 g) of copper, 1 mole of electrons (96500 C) deposits only:

   …

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6. CBSE 2025Set D1 markMCQ
   Q.The quantity of electricity required to liberate 32 g of oxygen is

   (a) 1 faraday

   (b) 2 faraday

   (c) 3 faraday

   (d) 4 faraday

   ›Reveal solutionSolution

   32 g O2 = 1 mol; the electrode reaction transfers 4 electrons per O2, so 4 faraday are needed.

   32 g of oxygen (O2, molar mass 32 g/mol) is 1 mole of O2 molecules.

   At the anode, oxygen is liberated by:

   2H2O → O2 + 4H+ + 4e- (or 4OH- → O2 + 2H2O + 4e-)

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The number of electrons with one coulomb of charge will be:

   (a) 6.29 x 10^11

   (b) 1.6 x 10^19

   (c) 6.24 x 10^18

   (d) 5.46 x 10^29

   ›Reveal solutionSolution

   Since the charge on one electron is 1.6 × 10^-19 C, the number of electrons carrying 1 C of charge is 1 / (1.6 × 10^-19) ≈ 6.24 × 10^18.

   The charge on a single electron is e = 1.6 × 10^-19 coulomb (this value itself is option b, which is a distractor — it is the charge of ONE electron, not the count of electrons).

   To find how many electrons (n) are needed to make up a total charge of 1 C:

   n = Total charge / charge per electron = 1 / (1.6 × 10^-19) = 6.25 × 10^18, which rounds to the listed 6.24 × 10^18.

   …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The amount of electricity required to deposit 1 mol of aluminium from a solution of $AlCl_3$ will be

   (a) 0.33 faraday

   (b) 1 faraday

   (c) 3 faraday

   (d) 1 ampere

   ›Reveal solutionSolution

   Aluminium exists as $Al^{3+}$ in $AlCl_3$, so depositing one mole of Al at the cathode needs 3 moles of electrons, i.e. 3 faradays.

   Reasoning

   The cathodic reduction half-reaction is:

   $Al^{3+} + 3e^- \rightarrow Al$

   By Faraday's first law, the quantity of electricity needed to deposit 1 mole of a substance equals (number of electrons transferred per ion) $\times F$, where $F = 96500\ \text{C mol}^{-1}$.

   Here $n = 3$, so the charge required is:

   $Q = nF = 3F$

   …

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9. CBSE 2024Set D1 markMCQ
   Q.A charge of 96500 coulomb liberates .............. from the solution of CuSO4.

   (a) 63.5 gm copper

   (b) 31.76 gm copper

   (c) 96500 gm copper

   (d) 100 gm copper

   ›Reveal solutionSolution

   96500 C = 1 faraday = 1 mole of electrons. Depositing Cu requires 2 electrons per Cu atom, so 1 F deposits 63.5/2 = 31.76 g Cu.

   Electrode reaction: Cu2+ + 2e- -> Cu.

   To deposit 1 mole of copper (63.5 g) you need 2 moles of electrons = 2 x 96500 C = 193000 C.

   Therefore the charge passed here, 96500 C (1 faraday, i.e. 1 mole of electrons), deposits half a mole of copper: …

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10. CBSE 2024Set B1 mark
    Q.Fill in the blank: One Faraday electricity equals to ______ coulomb.

    ›Reveal solutionSolution

    1 Faraday = charge carried by one mole of electrons = 96,500 C (more precisely 96,487 C, usually rounded to 96,500 C).

    One Faraday (F) is defined as the quantity of electric charge carried by one mole (Avogadro's number, 6.022x10^23) of electrons:

    $1F = N_A \times e = 6.022\times10^{23} \times 1.602\times10^{-19}\text{ C} \approx 96{,}500\text{ C mol}^{-1}$

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11. CBSE 2024Set ANNUAL1 mark
    Q.State Faraday's first law of Electrolysis.

    ›Reveal solutionSolution

    More charge passed through an electrolytic cell means proportionally more substance deposited at the electrode.

    Faraday's first law of electrolysis states: the mass (w) of a substance produced (deposited or liberated) at an electrode during electrolysis is directly proportional to the quantity of electricity (charge, Q) that passes through the electrolyte.

    w ∝ Q

    Since charge Q = current (I) × time (t), this can be written:

    w = Z × I × t

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12. CBSE 2023Set ANNUAL1 mark
    Q.A solution of $MgSO_4$ is electrolysed to carry out a deposition of 24.3 g of magnesium at cathode. How many electrons pass through the solution during the process ?

    ›Reveal solutionSolution

    Depositing 1 mole of Mg²⁺ needs 2 moles of electrons (per Faraday's law), i.e. $2N_A$ electrons.

    Magnesium is deposited at the cathode by the two-electron reduction:

    $Mg^{2+} + 2e^- \rightarrow Mg(s)$

    Moles of Mg deposited $= \dfrac{24.3\ g}{24.3\ g/mol} = 1\ mol$

    Since each mole of Mg requires 2 moles of electrons:

    $\text{moles of electrons} = 1 \times 2 = 2\ mol$

    …

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