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.

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.

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.

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.

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.

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.

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.

This problem involves electrolysis and Faraday's Law. The charge $Q = I \times t$ determines moles of electrons as $Q / F$, with $F \approx 96500 , \text{C/mol}$. The half-reaction $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(l)$ indicates 3 moles of electrons per mole of Al, so $\text{moles of Al} = \frac{Q / F}{3}$. 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.

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.

This problem involves electrolysis and Faraday's Law. Charge Q = I × t gives moles of electrons = Q / F, F ≈ 96500 C/mol. The half-reaction 2H₂O + 2e⁻ → H₂ + 2OH⁻ means 2 moles of electrons produce 1 mole of H₂, so moles of H₂ = (Q / F) / 2. Electron stoichiometry dictates the product amount from charge. A tempting distractor is 0.0187 mol, from using one mole of electrons instead of two per mole of product, doubling the moles. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.