Q.Three electrolytic cells A, B, C containing solutions of $ZnSO_4$, $AgNO_3$ and $CuSO_4$, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Molten $Al_2O_3$ was electrolyzed between carbon electrodes. The mass (in g) of aluminium produced at cathode when 965 amperes current is passed through it for 1000 seconds is (F=96500 C mol$^{-1}$) (A) 30 (B) 90 (C) 60 (D) 45

   ›Reveal solutionSolution

   This tests Faraday's laws of electrolysis applied to $Al_2O_3$ electrolysis; the answer is 90 g.

   Concept and Intuition

   In electrolytic reduction of molten $Al_2O_3$, $Al^{3+}$ ions gain 3 electrons at the cathode to form Al metal. The amount of substance deposited depends on the total charge passed and the number of electrons needed per mole of product (Faraday's second law).

   Step-by-Step Solution

   1. Charge passed: $Q = I \times t = 965\,A \times 1000\,s = 9.65\times10^{5}\,C$.
   2. Moles of electrons transferred: $n_e = Q/F = \dfrac{9.65\times10^{5}}{96500} = 10\,mol$.
   3. Cathode reaction: $Al^{3+} + 3e^- \to Al$, so 3 mol electrons deposit 1 mol Al.
   4. Moles of Al deposited $= 10/3\,mol$. …

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2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.The number of Faradays involved in the conversion of 0.25 mol of $Al^{3+}$ to Al is $\underline{x}$ and number of Faradays involved in the conversion of 1100 mL of 0.5 M $Cu^{2+}$ to Cu is $\underline{y}$. The values of x and y respectively are (A) 0.75, 1.1 (B) 0.25, 2.2 (C) 0.50, 3.3 (D) 1.00, 2.2

   ›Reveal solutionSolution

   Faradays needed = (moles of ion) × (charge/electrons needed per ion); working both cases gives x=0.75, y=1.1.

   Concept and Intuition

   One Faraday (1 F) supplies one mole of electrons. To deposit/reduce a metal ion $M^{n+}$ completely, you need $n$ Faradays per mole of that ion, since $M^{n+}+ne^-\to M$.

   Step-by-Step Solution

   1. $Al^{3+}+3e^-\to Al$: for 0.25 mol Al³⁺, Faradays required $x = 3\times0.25 = 0.75$ F.
   2. Moles of Cu²⁺ $= \dfrac{1100}{1000}\text{ L}\times0.5\text{ mol/L} = 0.55$ mol. …

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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The number of moles of $H_2$ gas liberated at cathode, when 10 milli-ampere current is passed through dilute aqueous solution of $NaCl$ for about $19.3 \times 10^4$ seconds is ($F = 96500\ C\ mol^{-1}$) (A) 0.50 (B) 0.02 (C) 0.01 (D) 0.15

   ›Reveal solutionSolution

   Faraday's law gives moles of electrons passed; since dilute aqueous NaCl electrolysis reduces water (not Na$^+$) at the cathode, 2 electrons give 1 mole of H$_2$, so $n(H_2)=0.01$ mol.

   Concept and Intuition

   In dilute aqueous NaCl, Na$^+$ is not reduced at the cathode (it is a much harder ion to reduce than water); instead water itself is reduced, liberating $H_2$ gas and $OH^-$. Faraday's laws connect the charge passed to the moles of product via the number of electrons in the half-reaction.

   Step-by-Step Solution

   1. Charge passed: $Q = I\times t = (10\times10^{-3}\,A)\times(19.3\times10^4\,s) = 1930$ C.
   2. Moles of electrons: $n_e = Q/F = 1930/96500 = 0.02$ mol.
   3. Cathode half-reaction in dilute aqueous NaCl: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ — 2 mol electrons give 1 mol $H_2$. …

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4. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.When 3 amp current was passed through an aqueous solution of salt of a metal M (atomic weight 106.4 u) for 1 hour, 2.977 g of $\mathrm{M^{n+}}$ was deposited at cathode. The value of n is (1 F = 96500 C mol$^{-1}$) (A) 2 (B) 1 (C) 3 (D) 4

   ›Reveal solutionSolution

   Applying Faraday's first law of electrolysis to the given charge, mass deposited, and atomic weight gives $n=4$.

   Concept and Intuition

   Faraday's law relates the mass of metal deposited at the cathode to the charge passed:

   $m = \frac{Q\,M}{n\,F}$

   where $Q=It$ is total charge, $M$ is the atomic weight of the metal, $n$ is the number of electrons needed per metal ion (i.e. its charge $\mathrm{M}^{n+}$), and $F$ is Faraday's constant. Rearranging for $n$ lets you deduce the ionic charge state from an experimental deposition measurement.

   Step-by-Step Solution

   1. Total charge passed: $Q = It = 3\,\text{A} \times 3600\,\text{s} = 10800\,\text{C}$.
   2. Rearrange Faraday's law: $n = \dfrac{Q\,M}{m\,F}$. …

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5. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.At 27 °C, the pH of 1 L of aqueous copper sulphate is 5.5. This solution was electrolyzed using two Pt electrodes for some time. What is the pH of remaining copper sulphate solution? (A) 5.5 (B) More than 5.5 but less than 7.0 (C) 7.5 (D) Less than 5.5 but more than zero

   ›Reveal solutionSolution

   Electrolysis with inert Pt electrodes deposits Cu at the cathode and liberates $O_2$ + $H^+$ at the anode, so the solution becomes progressively more acidic — pH drops below 5.5 but stays above zero.

   Concept and Intuition

   With Pt (inert) electrodes, $Cu^{2+}$ is preferentially reduced at the cathode (its reduction potential is far more favourable than reducing water/H⁺ to $H_2$), while at the anode water is oxidized to oxygen (sulfate is essentially inert to oxidation under these conditions). This liberates $H^+$ ions at the anode without any corresponding consumption of $H^+$ at the cathode, so the net effect is generation of acid.

   Step-by-Step Solution

   1. Cathode: $Cu^{2+} + 2e^- \to Cu(s)$ — copper deposits, no $H^+$ consumed.
   2. Anode: $2H_2O \to O_2 + 4H^+ + 4e^-$ — $H^+$ ions are released into solution.
   3. Overall: $2CuSO_4 + 2H_2O \xrightarrow{\text{electrolysis}} 2Cu + O_2 + 2H_2SO_4$.
   4. As electrolysis proceeds, $[H^+]$ increases, so pH decreases below the starting 5.5. …

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6. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.2.644 g of metal (M) was deposited when 8040 coulombs of electricity was passed through molten $\mathrm{MF_2}$ salt. What is the atomic mass of M? ($F = 96500\ \mathrm{C\,mol^{-1}}$) (A) 63.47 u (B) 65.54 u (C) 31.74 u (D) 61.48 u

   ›Reveal solutionSolution

   This tests Faraday's laws of electrolysis applied to a molten metal fluoride salt to find the atomic mass of the deposited metal. The answer is (A) 63.47 u.

   Concept and Intuition

   In electrolysis, the amount of substance deposited at an electrode is directly related to the total charge passed via Faraday's laws. For a metal $M$ in the salt $MF_2$, the metal exists as the $M^{2+}$ ion, which requires exactly 2 moles of electrons to be reduced to 1 mole of neutral metal atoms.

   Step-by-Step Solution

   1. Total charge passed: $Q = 8040\ \mathrm{C}$.
   2. Moles of electrons passed: $n_e = \dfrac{Q}{F} = \dfrac{8040}{96500} = 0.08332\ \mathrm{mol}$.
   3. Since $M^{2+} + 2e^- \rightarrow M$, moles of metal deposited: $n_M = \dfrac{n_e}{2} = \dfrac{0.08332}{2} = 0.04166\ \mathrm{mol}$. …

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7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The anode and cathode used in electrolytic refining of copper respectively are (A) Pure copper, impure copper (B) Impure copper, pure copper (C) Pure copper, pure zinc (D) Impure copper, pure zinc

   ›Reveal solutionSolution

   Electrolytic refining of copper: the impure metal is oxidised (dissolved) at the anode and pure metal is deposited at the cathode.

   Concept and Intuition

   In electrorefining, the metal to be purified is made the anode so that it oxidises and goes into solution as $\mathrm{Cu^{2+}}$, while a strip of the pure metal is made the cathode where $\mathrm{Cu^{2+}}$ is reduced and deposited as pure copper. The electrolyte is acidified $\mathrm{CuSO_4}$ solution, and the less noble impurities stay in solution or fall as "anode mud" while more noble impurities collect as anode mud too (e.g. Ag, Au).

   Step-by-Step Solution

   1. Anode reaction (oxidation): $\mathrm{Cu(s,\ impure)\rightarrow Cu^{2+}+2e^-}$ — this must be the impure block since it is consumed/dissolved. …

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8. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.Aqueous $\mathrm{CuSO_4}$ solution was electrolysed by passing 2 amp of current for 10 min. What is the weight (in g) of copper deposited at cathode? (Cu = 63 u; F = 96500 C $\mathrm{mol^{-1}}$) (A) 0.195 (B) 0.39 (C) 0.78 (D) 1.56

   ›Reveal solutionSolution

   Faraday's law with $n=2$ electrons per Cu deposited gives $\approx 0.39\ \mathrm g$ of copper — answer (B).

   Concept and Intuition

   Faraday's first law: the mass deposited at an electrode is proportional to the total charge passed, via $m = \dfrac{Q}{F}\times \dfrac{M}{n}$, where $n$ is the number of electrons needed to deposit one atom/ion (here $\mathrm{Cu^{2+}}+2e^-\to\mathrm{Cu}$, so $n=2$).

   Step-by-Step Solution

   1. Charge passed: $Q = I\times t = 2\ \mathrm A \times (10\times60)\ \mathrm s = 2\times600 = 1200\ \mathrm C$.
   2. Moles of electrons: $n_e = Q/F = 1200/96500 = 0.012435\ \mathrm{mol}$. …

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9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.The number of Faradays required to completely deposit magnesium from 1 L of 0.1 M $\mathrm{MgCl_2}$ aq. solution is (A) 0.2 (B) 2 (C) 0.1 (D) 0.4

   ›Reveal solutionSolution

   Depositing all the $Mg^{2+}$ from 0.1 mol of $MgCl_2$ needs 2 electrons per ion, so $0.1 \times 2 = 0.2$ Faradays are required.

   Concept and Intuition

   Electrodeposition amount is governed by Faraday's laws of electrolysis: the moles of electrons (Faradays) needed equal the moles of the species times the number of electrons transferred in reducing (or oxidizing) that species. For a metal ion of charge $n+$, depositing it as the neutral metal requires $n$ electrons per ion.

   Step-by-Step Solution

   1. Moles of $MgCl_2$ in 1 L of 0.1 M solution: $n = M \times V = 0.1\ \text{mol/L} \times 1\ \text{L} = 0.1$ mol, giving 0.1 mol of $Mg^{2+}$ ions (1:1 stoichiometry, $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$).
   2. Write the cathodic deposition reaction: $Mg^{2+} + 2e^- \rightarrow Mg$ — each $Mg^{2+}$ ion needs 2 electrons. …

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10. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.The time required (in hours) to reduce 3 mol of $\mathrm{Fe^{3+}}$ ions to $\mathrm{Fe^{2+}}$ ions with 2.0 amperes of current is (1 F=96500 C mol$^{-1}$) (A) 30.2 (B) 40.2 (C) 10.2 (D) 15.2

    ›Reveal solutionSolution

    Reducing $Fe^{3+}$ to $Fe^{2+}$ is a one-electron process; use $Q=nF$ and $t=Q/I$, converting seconds to hours.

    Concept and Intuition

    Faraday's law connects moles of electrons transferred to charge passed: $Q = n_e \times F$, where $n_e$ is moles of electrons and $F=96500$ C/mol. The reduction half-reaction $Fe^{3+}+e^-\to Fe^{2+}$ shows exactly 1 electron per Fe ion reduced.

    Step-by-Step Solution

    1. Moles of electrons needed for 3 mol $Fe^{3+}$ = 3 mol (1:1 ratio).
    2. Charge required: $Q = 3\times96500 = 289500\ \text{C}$.
    3. Time: $t = Q/I = 289500/2.0 = 144750\ \text{s}$. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.38.6 amperes of current is passed for 100 seconds through an aqueous $\mathrm{CuSO_4}$ solution using platinum electrodes. The mass of copper consumed from the solution and volume of gas liberated at STP are respectively (molar mass of Cu = 63.54 g mol$^{-1}$). (A) 6.37 g, 0.448 L (B) 0.63g, 0.224 L (C) 1.27g, 0.224 L (D) 4g, 0.448 L

    ›Reveal solutionSolution

    This tests Faraday's laws of electrolysis for a two-electrode cell (Cu deposited at cathode, O$_2$ evolved at the inert Pt anode). Answer: 1.27 g Cu, 0.224 L O$_2$.

    Concept and Intuition

    When current flows through an electrolyte, the amount of substance deposited/liberated at each electrode is proportional to the charge passed, via $n_{e^-} = Q/F$. With Pt (inert) electrodes in aqueous $\mathrm{CuSO_4}$: at the cathode $\mathrm{Cu^{2+}}$ ions are reduced to Cu metal (2 electrons per Cu atom); since Pt itself does not dissolve at the anode, the anode reaction is oxidation of water to $\mathrm{O_2}$ (4 electrons per $\mathrm{O_2}$ molecule), because $\mathrm{SO_4^{2-}}$ is not easily oxidised compared to water.

    Step-by-Step Solution

    1. Charge passed: $Q = It = 38.6\ \mathrm{A}\times100\ \mathrm{s} = 3860\ \mathrm{C}$.
    2. Moles of electrons transferred: $n_{e^-} = Q/F = 3860/96500 = 0.04\ \mathrm{mol}$.
    3. Cathode: $\mathrm{Cu^{2+}+2e^-\to Cu}$, so moles of Cu $= 0.04/2 = 0.02\ \mathrm{mol}$.
    4. Mass of Cu $= 0.02\ \mathrm{mol}\times63.54\ \mathrm{g\,mol^{-1}} = 1.2708\ \mathrm{g}\approx1.27\ \mathrm{g}$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.96.5 amperes current is passed through the molten $\mathrm{AlCl_3}$ for 100 seconds. The mass of aluminum deposited at the cathode is (Atomic weight of Al = 27 u) (A) 0.90 g (B) 0.45 g (C) 1.35 g (D) 1.8 g

    ›Reveal solutionSolution

    This is a straightforward Faraday's-law electrolysis calculation for aluminium deposition from molten $\mathrm{AlCl_3}$; the mass deposited is 0.90 g.

    Concept and Intuition

    Electrolysis obeys Faraday's laws: the amount of substance deposited at an electrode is proportional to the quantity of electricity (charge) passed, and the proportionality depends on the valency (charge) of the ion being discharged. One mole of electrons (1 Faraday $= 96500\ \text{C}$) discharges $1/n$ mole of an ion of charge $n^+$. For $\mathrm{Al^{3+}}$, $n=3$, so three moles of electrons are needed to deposit one mole of Al metal.

    Step-by-Step Solution

    1. Charge passed: $Q = I \times t = 96.5\ \text{A} \times 100\ \text{s} = 9650\ \text{C}$.
    2. Moles of electrons passed: $n_e = Q/F = 9650/96500 = 0.1\ \text{mol}$. …

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