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Electrolysis and Faraday's Law - AP Chemistry

  1. A solution containing Ag+\text{Ag}^+ ions is electrolyzed using inert electrodes. A constant current of 1.00 A1.00\ \text{A} is applied for 965 s965\ \text{s}. At the cathode, the half-reaction is Ag++e−→Ag(s)\text{Ag}^+ + e^- \rightarrow \text{Ag}(s). Assuming 100% current efficiency, what mass of silver is deposited?
    Answer: 1.08 g This problem uses electrolysis and Faraday's Law to find the mass of silver deposited. Calculate charge: Q = It = (1.00 A)(965 s) = 965 C. Convert to moles of electrons: moles e⁻ = 965 C ÷ 96,500 C/mol = 0.0100 mol e⁻. The half-reaction Ag⁺ + e⁻ → Ag shows a 1:1 ratio between electrons and silver atoms, so 0.0100 mol of Ag is produced. Converting to mass: (0.0100 mol)(107.9 g/mol) = 1.08 g. A common mistake is using the wrong molar mass or forgetting to convert from moles to grams, which might give 0.0100 g. Always use the correct molar mass and complete all unit conversions to get the final mass.
  2. A solution containing Co3+(aq)\text{Co}^{3+}(aq) is electrolyzed, producing cobalt metal at the cathode according to Co3++3e−→Co(s)\text{Co}^{3+} + 3e^- \rightarrow \text{Co}(s). If a constant current of 3.00 A3.00\ \text{A} is applied for 10.0 min10.0\ \text{min}, how many moles of Co(s)\text{Co}(s) are produced? (Assume 100% current efficiency.)
    Answer: 0.00622 mol This problem involves electrolysis and Faraday's Law. Q = I × t gives moles of electrons = Q / F, F ≈ 96500 C/mol. For Co³⁺ + 3e⁻ → Co(s), z=3, moles of Co = (Q / F) / 3. Stoichiometry determines the amount. A tempting distractor is 0.0187 mol, from using one mole of electrons instead of three per mole of product, tripling the moles. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  3. In the electrolysis of molten calcium chloride, calcium metal is produced at the cathode according to Ca2++2e−→Ca(s)\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}(s). If a total charge of 4.83×104 C4.83\times 10^4\ \text{C} passes through the cell, what mass of Ca(s)\text{Ca}(s) is produced? (Assume 100% current efficiency.)
    Answer: 10.0 g This problem involves electrolysis and Faraday's Law. Given Q, moles of electrons = Q / F, F ≈ 96500 C/mol. For Ca²⁺ + 2e⁻ → Ca(s), z=2, mass = [(Q / F) / 2] × molar mass. This relates charge to substance via electrons. A tempting distractor is 20.0 g, from using one mole of electrons instead of two per mole of product, doubling the mass. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  4. A student electrolyzes molten magnesium chloride, MgCl2(l)\text{MgCl}_2(l), using an inert cathode. The current is held constant at 9.65 A9.65\ \text{A} for 1.00×103 s1.00\times 10^3\ \text{s}. At the cathode, magnesium metal is produced according to the half-reaction Mg2++2e−→Mg(s)\text{Mg}^{2+}+2e^-\rightarrow \text{Mg}(s). Assuming 100% current efficiency, what amount of Mg(s)\text{Mg}(s) is produced?
    Answer: 0.0500 mol This problem requires applying electrolysis and Faraday's Law to determine the amount of magnesium produced. First, calculate the total charge: Q = It = (9.65 A)(1.00 × 10³ s) = 9.65 × 10³ C. Next, convert charge to moles of electrons using Faraday's constant: moles of e⁻ = (9.65 × 10³ C)/(96,500 C/mol) = 0.100 mol e⁻. The half-reaction shows Mg²⁺ + 2e⁻ → Mg, meaning 2 moles of electrons produce 1 mole of Mg. Therefore, moles of Mg = (0.100 mol e⁻) × (1 mol Mg/2 mol e⁻) = 0.0500 mol. A common error is forgetting to account for the 2:1 electron stoichiometry, which would incorrectly give 0.100 mol Mg. Always identify the electron-to-product ratio from the balanced half-reaction before calculating the final amount.
  5. Molten aluminum oxide is electrolyzed with a constant current of 9.65 A9.65 \, \text{A} for 30.0 min30.0 \, \text{min}. Aluminum metal forms at the cathode according to Al3++3e−→Al(l)\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(l). Assuming 100% current efficiency, how many moles of Al\text{Al} are produced?
    Answer: 0.0600 mol This problem involves electrolysis and Faraday's Law. The charge Q=I×tQ = I \times t determines moles of electrons as Q/FQ / F, with F≈96500 C/molF \approx 96500 \, \text{C/mol}. The half-reaction Al3++3e−→Al(l)\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(l) indicates 3 moles of electrons per mole of Al, so moles of Al=Q/F3\text{moles of Al} = \frac{Q / F}{3}. This stoichiometry ensures the amount of Al produced scales with the electron transfer ratio. A tempting distractor is 0.180 mol, from the misconception of using one mole of electrons instead of three per mole of product, tripling the moles calculated. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  6. In an electrolytic cell, nickel is deposited on a metal object from Ni2+(aq)\text{Ni}^{2+}(aq) according to Ni2++2e−→Ni(s)\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}(s). What total charge (in coulombs) is required to deposit 0.100 mol0.100\ \text{mol} of Ni(s)\text{Ni}(s)? (Assume 100% current efficiency.)
    Answer: 1.93×10410^4 C This problem involves electrolysis and Faraday's Law. To find Q for a given moles of substance, use moles of electrons = moles of substance × z, where z is electrons per unit, then Q = (moles of e⁻) × F, F ≈ 96500 C/mol. For Ni²⁺ + 2e⁻ → Ni(s), z=2 for 1 mole Ni. This relates substance to charge via electron count. A tempting distractor is 9.65×10310^3 C, from using one mole of electrons instead of two per mole of product, halving the charge. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  7. An aqueous solution containing Zn2+(aq)\text{Zn}^{2+}(aq) is electrolyzed so that zinc metal forms at the cathode according to Zn2++2e−→Zn(s)\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s). If a constant current of 1.50 A1.50\ \text{A} is applied for 40.0 min40.0\ \text{min}, what mass of Zn(s)\text{Zn}(s) is produced? (Assume 100% current efficiency.)
    Answer: 1.22 g This problem involves electrolysis and Faraday's Law. Charge Q = I × t gives moles of electrons = Q / F, F ≈ 96500 C/mol. For Zn²⁺ + 2e⁻ → Zn(s), 2 moles of electrons yield 1 mole of Zn, so mass = [(Q / F) / 2] × molar mass. Electron stoichiometry determines the scaling. A tempting distractor is 2.44 g, from using one mole of electrons instead of two per mole of product, doubling the mass. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  8. A concentrated aqueous solution of potassium iodide is electrolyzed using an inert anode. A current of 19.3 A19.3\ \text{A} is applied for 500 s500\ \text{s}. At the anode, iodine forms according to 2I−→I2(s)+2e−2\text{I}^-\rightarrow \text{I}_2(s)+2e^-. Assuming 100% current efficiency, how many moles of I2\text{I}_2 are produced?
    Answer: 0.0500 mol This problem applies electrolysis and Faraday's Law to iodine production. Calculate charge: Q = It = (19.3 A)(500 s) = 9.65 × 10³ C. Convert to moles of electrons: moles of e⁻ = (9.65 × 10³ C)/(96,500 C/mol) = 0.100 mol e⁻. The half-reaction 2I⁻ → I₂ + 2e⁻ indicates that 2 moles of electrons are produced per mole of I₂ formed. Thus, moles of I₂ = (0.100 mol e⁻) × (1 mol I₂/2 mol e⁻) = 0.0500 mol. A common misconception is using one electron per iodine atom instead of recognizing that I₂ is the product, leading to an incorrect answer of 0.100 mol. Always write out the complete half-reaction to identify the actual product species and its electron stoichiometry.
  9. A molten salt containing Ca2+\text{Ca}^{2+} is electrolyzed. Calcium metal forms at the cathode according to Ca2++2e−→Ca(s)\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}(s). What total charge (in coulombs) is required to produce 0.250 mol0.250\ \text{mol} of Ca(s)\text{Ca}(s), assuming 100%100\% current efficiency? (F=9.65×104 C mol−1 e−F = 9.65\times 10^4\ \text{C mol}^{-1}\ e^-.)
    Answer: 4.83×10410^4 C This problem involves electrolysis and Faraday's Law. The total charge required is calculated based on the desired moles of product and the electron stoichiometry. The moles of electrons needed is the moles of product multiplied by the number of electrons per mole from the half-reaction. The total charge is then the moles of electrons multiplied by Faraday's constant, relating charge to electron quantity. A tempting distractor is 2.41×10410^4 C, which is incorrect because it used one mole of electrons instead of two per mole of product. Use the half-reaction to count electrons before converting to substance.
  10. A solution containing Ni2+\text{Ni}^{2+} ions is electrolyzed using inert electrodes. At the cathode, the half-reaction is Ni2++2e−→Ni(s)\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}(s). What total charge (in coulombs) must pass through the cell to deposit 0.100 mol0.100\ \text{mol} of Ni(s)\text{Ni}(s), assuming 100% current efficiency? (Use F=96,500 C mol−1 e−F = 96{,}500\ \text{C mol}^{-1}\ e^-.)
    Answer: 19,300 C This problem uses electrolysis and Faraday's Law to find the charge needed to deposit a specific amount of nickel. The half-reaction Ni²⁺ + 2e⁻ → Ni shows that 2 moles of electrons are required per mole of nickel. For 0.100 mol Ni, we need: moles e⁻ = (0.100 mol Ni)(2 mol e⁻/mol Ni) = 0.200 mol e⁻. Converting to charge: Q = (0.200 mol e⁻)(96,500 C/mol e⁻) = 19,300 C. A common mistake is forgetting the 2:1 electron-to-nickel ratio, using only 0.100 mol e⁻, which would give 9,650 C. Always use the half-reaction stoichiometry to determine the moles of electrons before calculating charge.
  11. In an electrolytic cell, oxygen gas is produced at the anode from water according to 2H2O(l)→O2(g)+4H+(aq)+4e−2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-. If a total charge of 3.86×104 C3.86\times 10^4\ \text{C} passes through the cell, how many moles of O2(g)\text{O}_2(g) are produced? (Assume 100% current efficiency.)
    Answer: 0.100 mol This problem involves electrolysis and Faraday's Law. Given Q, moles of electrons = Q / F, F ≈ 96500 C/mol. The half-reaction 2H₂O → O₂ + 4H⁺ + 4e⁻ requires 4 moles of electrons for 1 mole of O₂, so moles of O₂ = (Q / F) / 4. This stoichiometry scales the product. A tempting distractor is 0.400 mol, from using one mole of electrons instead of four per mole of product, quadrupling the moles. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  12. Molten calcium chloride is electrolyzed to produce calcium metal at the cathode according to Ca2++2e−→Ca(s)\text{Ca}^{2+}+2e^-\rightarrow \text{Ca}(s). If a current of 19.3 A19.3\ \text{A} is applied for 500 s500\ \text{s}, assuming 100% current efficiency, how many moles of Ca(s)\text{Ca}(s) are produced?
    Answer: 0.0500 mol This question tests electrolysis and Faraday's Law for calcium production. Calculate charge: Q = It = (19.3 A)(500 s) = 9.65 × 10³ C. Convert to moles of electrons: moles of e⁻ = (9.65 × 10³ C)/(96,500 C/mol) = 0.100 mol e⁻. The half-reaction Ca²⁺ + 2e⁻ → Ca shows that 2 moles of electrons produce 1 mole of Ca. Therefore, moles of Ca = (0.100 mol e⁻) × (1 mol Ca/2 mol e⁻) = 0.0500 mol. A typical error is forgetting the 2:1 electron stoichiometry and calculating 0.100 mol Ca instead. Always use the balanced half-reaction to establish the electron-to-product ratio before performing calculations.
  13. A solution containing Cu2+(aq)\text{Cu}^{2+}(aq) is electrolyzed with a constant current of 2.00 A2.00\ \text{A} for 40.2 min40.2\ \text{min}. Copper is deposited at the cathode according to Cu2++2e−→Cu(s)\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s). Assuming 100%100\% current efficiency, what mass of Cu(s)\text{Cu}(s) is produced? (F=9.65×104 C mol−1 e−F = 9.65\times 10^4\ \text{C mol}^{-1}\ e^-; molar mass Cu=63.5 g mol−1\text{Cu}=63.5\ \text{g mol}^{-1}.)
    Answer: 1.59 g This problem involves electrolysis and Faraday's Law. The total charge passed is the current multiplied by the time converted to seconds, giving the quantity of electricity in coulombs. The moles of electrons transferred is this charge divided by Faraday's constant, which represents the charge per mole of electrons. The electron stoichiometry from the half-reaction determines how many moles of the substance are formed per mole of electrons, so the moles of substance is moles of electrons divided by the number of electrons in the half-reaction. A tempting distractor is 3.18 g, which is incorrect because it used one mole of electrons instead of two per mole of product. Use the half-reaction to count electrons before converting to substance.
  14. During the electrolysis of molten aluminum oxide, aluminum metal is produced at the cathode according to Al3++3e−→Al(l)\text{Al}^{3+}+3e^-\rightarrow \text{Al}(l). If a total charge of 2.90×104 C2.90\times 10^4\ \text{C} passes through the cell, assuming 100% current efficiency, how many moles of Al\text{Al} are produced?
    Answer: 0.100 mol This question involves electrolysis and Faraday's Law for aluminum production. Given charge of 2.90 × 10⁴ C, convert to moles of electrons: moles of e⁻ = (2.90 × 10⁴ C)/(96,500 C/mol) = 0.300 mol e⁻. The half-reaction Al³⁺ + 3e⁻ → Al shows that 3 moles of electrons are required per mole of Al produced. Therefore, moles of Al = (0.300 mol e⁻) × (1 mol Al/3 mol e⁻) = 0.100 mol. Students often err by using 2 electrons per Al (confusing with common 2+ ions) or forgetting the electron stoichiometry entirely, yielding 0.300 mol Al. Always check the charge on the ion in the half-reaction to determine the correct number of electrons transferred.
  15. Molten sodium chloride is electrolyzed, producing sodium metal at the cathode according to Na++e−→Na(l)\text{Na}^+ + e^- \rightarrow \text{Na}(l). If a constant current of 4.00 A4.00\ \text{A} is applied for 20.0 min20.0\ \text{min}, what mass of Na\text{Na} is produced? (Assume 100% current efficiency.)
    Answer: 1.14 g This problem involves electrolysis and Faraday's Law. Total charge Q = I × t yields moles of electrons = Q / F, F ≈ 96500 C/mol. For Na⁺ + e⁻ → Na(l), 1 mole of electrons produces 1 mole of Na, directly tying charge to mass via molar mass. Stoichiometry here is 1:1 for electrons to Na. A tempting distractor is 0.57 g, from using two moles of electrons instead of one per mole of product, halving the mass. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  16. An electrolytic cell is used to produce hydrogen gas at the cathode from water according to 2H2O(l)+2e−→H2(g)+2OH−2\text{H}_2\text{O}(l)+2e^-\rightarrow \text{H}_2(g)+2\text{OH}^-. A current of 9.65 A9.65\ \text{A} is applied for 2.00×103 s2.00\times 10^3\ \text{s}. Assuming 100% current efficiency, what amount of H2(g)\text{H}_2(g) is produced?
    Answer: 0.100 mol This problem tests electrolysis and Faraday's Law for hydrogen gas production. Calculate charge: Q = It = (9.65 A)(2.00 × 10³ s) = 1.93 × 10⁴ C. Convert to moles of electrons: moles of e⁻ = (1.93 × 10⁴ C)/(96,500 C/mol) = 0.200 mol e⁻. The half-reaction 2H₂O + 2e⁻ → H₂ + 2OH⁻ shows that 2 moles of electrons produce 1 mole of H₂. Thus, moles of H₂ = (0.200 mol e⁻) × (1 mol H₂/2 mol e⁻) = 0.100 mol. A frequent error is confusing the coefficient of water (2) with the product ratio, incorrectly calculating 0.0500 mol H₂. Focus on the electron-to-H₂ ratio in the balanced equation, not the water coefficient.
  17. Molten sodium bromide, NaBr(l)\text{NaBr}(l), is electrolyzed. The current is 4.82 A4.82\ \text{A} for 2.00×103 s2.00\times 10^3\ \text{s}. At the anode, bromine is produced according to 2Br−→Br2(g)+2e−2\text{Br}^-\rightarrow \text{Br}_2(g)+2e^-. Assuming 100% current efficiency, what amount of Br2(g)\text{Br}_2(g) is produced?
    Answer: 0.0500 mol This question tests electrolysis and Faraday's Law for bromine production at the anode. Calculate total charge: Q = It = (4.82 A)(2.00 × 10³ s) = 9.64 × 10³ C. Convert to moles of electrons: moles of e⁻ = (9.64 × 10³ C)/(96,500 C/mol) = 0.100 mol e⁻. The half-reaction 2Br⁻ → Br₂ + 2e⁻ shows that 2 moles of electrons are released when 1 mole of Br₂ forms. Therefore, moles of Br₂ = (0.100 mol e⁻) × (1 mol Br₂/2 mol e⁻) = 0.0500 mol. Students often mistakenly think 2 moles of Br⁻ means 2 moles of product, but the coefficient refers to bromide ions, not Br₂ molecules. Always focus on the electron-to-product ratio, not the reactant coefficients.
  18. A solution containing Fe3+(aq)\text{Fe}^{3+}(aq) is electrolyzed, and iron metal is produced at the cathode according to Fe3++3e−→Fe(s)\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}(s). If a total charge of 2.895×104 C2.895\times 10^4\ \text{C} passes through the cell, how many moles of Fe(s)\text{Fe}(s) are produced? (Assume 100% current efficiency.)
    Answer: 0.100 mol This problem involves electrolysis and Faraday's Law. Given Q, moles of electrons = Q / F, F ≈ 96500 C/mol. For Fe³⁺ + 3e⁻ → Fe(s), z=3, so moles of Fe = (Q / F) / 3. Stoichiometry sets the ratio. A tempting distractor is 0.300 mol, from using one mole of electrons instead of three per mole of product, tripling the moles. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.
  19. During electrolysis of an aqueous solution, oxygen gas is produced at the anode according to 2H2O(l)→O2(g)+4H+(aq)+4e−2\text{H}_2\text{O}(l)\rightarrow \text{O}_2(g)+4\text{H}^+(aq)+4e^-. A total charge of 3.86×104 C3.86\times 10^4\ \text{C} passes through the cell. Assuming 100% current efficiency, what amount of O2(g)\text{O}_2(g) is produced?
    Answer: 0.100 mol This problem applies electrolysis and Faraday's Law to oxygen gas production. Given charge of 3.86 × 10⁴ C, convert to moles of electrons: moles of e⁻ = (3.86 × 10⁴ C)/(96,500 C/mol) = 0.400 mol e⁻. The half-reaction 2H₂O → O₂ + 4H⁺ + 4e⁻ shows that 4 moles of electrons are produced per mole of O₂ formed. Thus, moles of O₂ = (0.400 mol e⁻) × (1 mol O₂/4 mol e⁻) = 0.100 mol. Students often use 2 electrons per O₂ (thinking of the O₂²⁻ to O₂ conversion), yielding an incorrect 0.200 mol O₂. Always use the complete half-reaction as written to determine the correct electron stoichiometry.
  20. An aqueous solution containing Cu2+\text{Cu}^{2+} is electrolyzed with an inert cathode. A total charge of 1.93×104 C1.93\times 10^4\ \text{C} passes through the circuit. Copper metal plates onto the cathode according to Cu2++2e−→Cu(s)\text{Cu}^{2+}+2e^-\rightarrow \text{Cu}(s). Assuming 100% current efficiency, how many moles of Cu(s)\text{Cu}(s) are produced?
    Answer: 0.100 mol This problem involves electrolysis and Faraday's Law to calculate copper production. Given a total charge of 1.93 × 10⁴ C, first convert to moles of electrons: moles of e⁻ = (1.93 × 10⁴ C)/(96,500 C/mol) = 0.200 mol e⁻. The half-reaction Cu²⁺ + 2e⁻ → Cu shows that 2 moles of electrons produce 1 mole of Cu. Therefore, moles of Cu = (0.200 mol e⁻) × (1 mol Cu/2 mol e⁻) = 0.100 mol. A tempting error is to use a 1:1 ratio between electrons and copper, which would incorrectly yield 0.200 mol Cu. Remember to always examine the half-reaction carefully to determine how many electrons are required per mole of product formed.

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