Q.If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

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 $= 10/3\,mol$.
   5. Mass of Al $= \dfrac{10}{3}\times 27\,g/mol = 90\,g$.

   Common Mistakes

   • Forgetting the 3-electron requirement per Al atom and dividing by 1 or 2 instead of 3.
   • Mixing up current × time with moles of electrons directly (must divide by F).

   ✓Final answer

   The correct option is (B) — 90 g of aluminium is deposited.

   ANSWER: B

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.
   3. $Cu^{2+}+2e^-\to Cu$: Faradays required $y = 2\times0.55 = 1.1$ F.

   Common Mistakes

   • Forgetting to convert 1100 mL to 1.1 L before multiplying by molarity.
   • Using the wrong number of electrons (using 2 for Al or 3 for Cu).

   ✓Final answer

   The correct option is (A) — 0.75, 1.1.

   ANSWER: A

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 $H_2$.
   4. Moles of $H_2$ liberated $= n_e/2 = 0.02/2 = 0.01$ mol.

   Common Mistakes

   • Assuming Na$^+$ is discharged at the cathode instead of water (Na is far more electropositive; water is preferentially reduced in aqueous solution).
   • Forgetting to divide by 2 (the stoichiometric electron count for $H_2$ formation).

   ✓Final answer

   The correct option is (C) — 0.01.

   ANSWER: C

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: $n = \dfrac{Q\,M}{m\,F}$.
   3. Substitute: $n = \dfrac{10800 \times 106.4}{2.977 \times 96500} = \dfrac{1{,}149{,}120}{287{,}290.5} \approx 4.0$.
   4. So the metal ion is $\mathrm{M^{4+}}$, $n=4$.

   Common Mistakes

   • Forgetting to convert the time from hours to seconds before computing $Q=It$.
   • Inverting the Faraday-law rearrangement (solving for $M$ instead of $n$, or vice versa).

   ✓Final answer

   The correct option is (D) — 4.

   ANSWER: D

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.
   5. Since only a limited quantity of $Cu^{2+}$ is present ("for some time", not complete electrolysis), the pH falls but remains a positive, finite value greater than zero.

   Common Mistakes

   • Assuming the solution stays neutral or basic because $SO_4^{2-}$ is "spectator" — the anode reaction (water oxidation) is what actually drives the pH down.

   ✓Final answer

   The correct option is (D) — Less than 5.5 but more than zero.

   ANSWER: D

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: $n_M = \dfrac{n_e}{2} = \dfrac{0.08332}{2} = 0.04166\ \mathrm{mol}$.
   4. Atomic mass of $M$: $\dfrac{\text{mass deposited}}{\text{moles deposited}} = \dfrac{2.644\ \mathrm{g}}{0.04166\ \mathrm{mol}} \approx 63.47\ \mathrm{u}$.

   Common Mistakes

   • Forgetting the factor of 2 electrons per M(2+) ion (treating it as a monovalent metal), which would halve the correct atomic mass answer.

   ✓Final answer

   The correct option is (A) — 63.47 u.

   ANSWER: A

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.
   2. Cathode reaction (reduction): $\mathrm{Cu^{2+}+2e^-\rightarrow Cu(s,\ pure)}$ — pure copper is deposited here, so the cathode is thin pure copper (as a seed for deposition).
   3. Hence anode = impure copper, cathode = pure copper.

   Common Mistakes

   • Swapping anode and cathode roles (remembering that in electrolytic refining, the impure metal is always the anode being dissolved, not the cathode).

   ✓Final answer

   The correct option is (B) — Impure copper, pure copper.

   ANSWER: B

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: $n_e = Q/F = 1200/96500 = 0.012435\ \mathrm{mol}$.
   3. Half-reaction: $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$, so moles of Cu deposited $= n_e/2 = 0.012435/2 = 0.0062176\ \mathrm{mol}$.
   4. Mass of Cu $= 0.0062176 \times 63 = 0.3917\ \mathrm g \approx 0.39\ \mathrm g$.

   Common Mistakes

   • Forgetting to convert minutes to seconds before computing $Q=It$.
   • Forgetting the factor of 2 (electrons per Cu²⁺) and reporting double the correct mass.

   ✓Final answer

   The correct option is (B) — 0.39.

   ANSWER: B

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.
   3. Total Faradays (moles of electrons) needed $= 0.1\ \text{mol} \times 2 = 0.2\ F$.

   Common Mistakes

   • Forgetting the "2" electrons needed for a divalent ion and answering 0.1 F instead (that would be for a monovalent ion).
   • Confusing "1 L of 0.1 M" with "0.1 mol/L of Cl⁻" and using the wrong ion's concentration.

   ✓Final answer

   The correct option is (A) — 0.2.

   ANSWER: A

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: $t = Q/I = 289500/2.0 = 144750\ \text{s}$.
    4. Convert to hours: $144750/3600 = 40.208\ \text{h}\approx40.2\ \text{h}$.

    Common Mistakes

    • Using the wrong number of electrons (e.g. assuming 3 electrons per Fe as if it were $Fe^{3+}\to Fe^0$).
    • Forgetting to convert seconds to hours at the end.

    ✓Final answer

    The correct option is (B) — 40.2.

    ANSWER: B

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 $= 0.02\ \mathrm{mol}\times63.54\ \mathrm{g\,mol^{-1}} = 1.2708\ \mathrm{g}\approx1.27\ \mathrm{g}$.
    5. Anode: $2\mathrm{H_2O}\to\mathrm{O_2}+4\mathrm{H^+}+4e^-$, so moles of $\mathrm{O_2} = 0.04/4 = 0.01\ \mathrm{mol}$.
    6. Volume at STP $= 0.01\times22400\ \mathrm{mL} = 224\ \mathrm{mL} = 0.224\ \mathrm{L}$.

    Common Mistakes

    • Forgetting Cu$^{2+}$ needs 2 electrons per atom (halving the moles of Cu, not equating directly to moles of electrons).
    • Assuming $\mathrm{SO_4^{2-}}$ is discharged at the anode instead of water — with Pt electrodes and a sulphate solution, O$_2$ (from water) is liberated, not a sulphur oxide.

    ✓Final answer

    The correct option is (C) — 1.27 g, 0.224 L.

    ANSWER: C

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: $n_e = Q/F = 9650/96500 = 0.1\ \text{mol}$.
    3. Cathode reaction: $\mathrm{Al^{3+} + 3e^- \rightarrow Al}$. So moles of Al deposited $= n_e/3 = 0.1/3 = 0.0333\ \text{mol}$.
    4. Mass of Al deposited $= 0.0333 \times 27 = 0.9\ \text{g}$.

    Common Mistakes

    • Forgetting to divide by the valency factor (3 for $\mathrm{Al^{3+}}$) and directly multiplying moles of electrons by atomic weight.
    • Using an incorrect value of the Faraday constant instead of $96500\ \text{C mol}^{-1}$.

    ✓Final answer

    The correct option is (A) — 0.90 g.

    ANSWER: A