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AP Chemistry

AP Chemistry Help: Electrolysis And Faradays Law

Review real example questions for Electrolysis And Faradays Law in AP Chemistry.

Question 1

Molten sodium chloride is electrolyzed, producing sodium metal at the cathode according to Na++e−→Na(l)\text{Na}^+ + e^- \rightarrow \text{Na}(l)Na++e−→Na(l). If a constant current of 4.00 A4.00\ \text{A}4.00 A is applied for 20.0 min20.0\ \text{min}20.0 min, what mass of Na\text{Na}Na is produced? (Assume 100% current efficiency.)

  1. 1.14 g
  2. 0.57 g
  3. 2.28 g
  4. 4.56 g
  5. 0.114 g
Explanation: 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.

Question 2

Molten sodium bromide, NaBr(l)\text{NaBr}(l)NaBr(l), is electrolyzed. The current is 4.82 A4.82\ \text{A}4.82 A for 2.00×103 s2.00\times 10^3\ \text{s}2.00×103 s. At the anode, bromine is produced according to 2Br−→Br2(g)+2e−2\text{Br}^-\rightarrow \text{Br}_2(g)+2e^-2Br−→Br2​(g)+2e−. Assuming 100% current efficiency, what amount of Br2(g)\text{Br}_2(g)Br2​(g) is produced?

  1. 0.200 mol
  2. 0.0250 mol
  3. 0.0100 mol
  4. 0.0500 mol
  5. 0.100 mol
Explanation: 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.

Question 3

Silver metal is deposited during electrolysis according to the cathode half-reaction Ag++e−→Ag(s)\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)Ag++e−→Ag(s). What total charge is required to produce 0.250 mol0.250\ \text{mol}0.250 mol of Ag(s)\text{Ag}(s)Ag(s), assuming 100% current efficiency?

  1. 2.41×10^4 C
  2. 4.82×10^4 C
  3. 9.65×10^4 C
  4. 1.93×10^4 C
  5. 9.65×10^3 C
Explanation: This question applies electrolysis and Faraday's Law to find the charge needed for silver deposition. The half-reaction Ag⁺ + e⁻ → Ag shows a 1:1 ratio between electrons and silver atoms. To produce 0.250 mol Ag, we need 0.250 mol e⁻. Using Faraday's constant: Q = (0.250 mol e⁻) × (96,500 C/mol) = 2.41 × 10⁴ C. A common mistake is assuming silver requires 2 electrons (like many transition metals), which would double the charge to 4.82 × 10⁴ C. Always verify the oxidation state in the given half-reaction rather than assuming based on common patterns.

Question 4

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}^-2H2​O(l)+2e−→H2​(g)+2OH−. A current of 9.65 A9.65\ \text{A}9.65 A is applied for 2.00×103 s2.00\times 10^3\ \text{s}2.00×103 s. Assuming 100% current efficiency, what amount of H2(g)\text{H}_2(g)H2​(g) is produced?

  1. 0.200 mol
  2. 0.0100 mol
  3. 0.0500 mol
  4. 0.100 mol
  5. 0.0250 mol
Explanation: 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.

Question 5

A concentrated aqueous solution of potassium iodide is electrolyzed using an inert anode. A current of 19.3 A19.3\ \text{A}19.3 A is applied for 500 s500\ \text{s}500 s. At the anode, iodine forms according to 2I−→I2(s)+2e−2\text{I}^-\rightarrow \text{I}_2(s)+2e^-2I−→I2​(s)+2e−. Assuming 100% current efficiency, how many moles of I2\text{I}_2I2​ are produced?

  1. 0.0250 mol
  2. 0.0500 mol
  3. 0.100 mol
  4. 0.200 mol
  5. 0.0100 mol
Explanation: 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.

Question 6

An aqueous solution containing Cu2+\text{Cu}^{2+}Cu2+ is electrolyzed with an inert cathode. A total charge of 1.93×104 C1.93\times 10^4\ \text{C}1.93×104 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)Cu2++2e−→Cu(s). Assuming 100% current efficiency, how many moles of Cu(s)\text{Cu}(s)Cu(s) are produced?

  1. 0.0500 mol
  2. 0.100 mol
  3. 0.200 mol
  4. 0.0100 mol
  5. 0.400 mol
Explanation: 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.

Question 7

A molten salt containing Ca2+\text{Ca}^{2+}Ca2+ is electrolyzed. Calcium metal forms at the cathode according to Ca2++2e−→Ca(s)\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}(s)Ca2++2e−→Ca(s). What total charge (in coulombs) is required to produce 0.250 mol0.250\ \text{mol}0.250 mol of Ca(s)\text{Ca}(s)Ca(s), assuming 100%100\%100% current efficiency? (F=9.65×104 C mol−1 e−F = 9.65\times 10^4\ \text{C mol}^{-1}\ e^-F=9.65×104 C mol−1 e−.)

  1. 2.41×10^4 C
  2. 4.83×10^4 C
  3. 9.65×10^4 C
  4. 1.93×10^5 C
  5. 7.24×10^4 C
Explanation: 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 2.41×10^4 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.

Question 8

A solution containing Cu2+(aq)\text{Cu}^{2+}(aq)Cu2+(aq) is electrolyzed with a constant current of 2.00 A2.00\ \text{A}2.00 A for 40.2 min40.2\ \text{min}40.2 min. Copper is deposited at the cathode according to Cu2++2e−→Cu(s)\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)Cu2++2e−→Cu(s). Assuming 100%100\%100% current efficiency, what mass of Cu(s)\text{Cu}(s)Cu(s) is produced? (F=9.65×104 C mol−1 e−F = 9.65\times 10^4\ \text{C mol}^{-1}\ e^-F=9.65×104 C mol−1 e−; molar mass Cu=63.5 g mol−1\text{Cu}=63.5\ \text{g mol}^{-1}Cu=63.5 g mol−1.)

  1. 0.794 g
  2. 1.59 g
  3. 3.18 g
  4. 0.397 g
  5. 6.35 g
Explanation: 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.

Question 9

An aqueous solution containing Fe3+\text{Fe}^{3+}Fe3+ is electrolyzed with an inert cathode. Iron(III) ions are reduced to iron(II) ions according to Fe3++e−→Fe2+\text{Fe}^{3+}+e^-\rightarrow \text{Fe}^{2+}Fe3++e−→Fe2+. A total charge of 9.65×103 C9.65\times 10^3\ \text{C}9.65×103 C is passed through the cell. Assuming 100% current efficiency, how many moles of Fe3+\text{Fe}^{3+}Fe3+ are consumed?

  1. 0.200 mol
  2. 0.0500 mol
  3. 0.0100 mol
  4. 0.0250 mol
  5. 0.100 mol
Explanation: This problem involves electrolysis and Faraday's Law for the reduction of Fe³⁺ to Fe²⁺. Given charge of 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 Fe³⁺ + e⁻ → Fe²⁺ shows a 1:1 ratio between electrons and Fe³⁺ consumed. Therefore, moles of Fe³⁺ consumed = 0.100 mol. Students might mistakenly think 3 electrons are involved because of the 3+ charge, but the reaction only involves a one-electron reduction from Fe³⁺ to Fe²⁺. Always read the half-reaction carefully to determine the actual electron transfer, not the total charge on the ion.

Question 10

Molten aluminum oxide is electrolyzed with a constant current of 9.65 A9.65 \, \text{A}9.65A for 30.0 min30.0 \, \text{min}30.0min. Aluminum metal forms at the cathode according to Al3++3e−→Al(l)\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(l)Al3++3e−→Al(l). Assuming 100% current efficiency, how many moles of Al\text{Al}Al are produced?

  1. 0.180 mol
  2. 0.0600 mol
  3. 0.300 mol
  4. 0.0900 mol
  5. 0.0200 mol
Explanation: This problem involves electrolysis and Faraday's Law. The charge Q=I×tQ = I \times tQ=I×t determines moles of electrons as Q/FQ / FQ/F, with F≈96500 C/molF \approx 96500 \, \text{C/mol}F≈96500C/mol. The half-reaction Al3++3e−→Al(l)\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(l)Al3++3e−→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}moles of Al=3Q/F​. 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.