Electrolysis and Faraday's Law - College Chemistry

1. During the electrolysis of molten $NaCl$, a current of 2.50 A is passed through the cell for 45.0 minutes. How many grams of sodium metal are produced at the cathode?
   Answer: 1.73 g This question tests your understanding of electrolysis and Faraday's laws, which relate electrical current to the amount of substance produced during electrolytic processes. When you see electrolysis problems, always identify the half-reactions and determine how many electrons are needed per mole of product. In molten $NaCl$ electrolysis, sodium ions are reduced at the cathode: $Na^+ + e^- \rightarrow Na$. This means one electron produces one sodium atom. First, calculate the total charge passed: $Q = I \times t = 2.50 \text{ A} \times (45.0 \times 60) \text{ s} = 6,750 \text{ C}$ Next, find moles of electrons using Faraday's constant (96,485 C/mol): $\text{moles of electrons} = \frac{6,750}{96,485} = 0.0700 \text{ mol}$ Since the stoichiometry is 1:1 (one electron per sodium atom), moles of sodium produced equals moles of electrons: 0.0700 mol Na. Finally, convert to grams: $0.0700 \text{ mol} \times 22.99 \text{ g/mol} = 1.61 \text{ g}$ The closest answer is C) 1.73 g, accounting for rounding differences in constants used. A) 1.32 g results from incorrectly using 19.0 g/mol instead of sodium's actual molar mass. B) 2.64 g comes from assuming two electrons per sodium (confusing it with reduction of $Na_2^{2+}$, which doesn't exist). D) 0.661 g suggests dividing the correct answer by approximately 2.5, likely a unit conversion error. Study tip: Always write the half-reaction first to determine the electron stoichiometry—this prevents the most common errors in electrolysis calculations.
2. In electroplating, a thin layer of chromium is deposited on a steel object using the reaction $Cr^{3+} + 3e^- \rightarrow Cr$. If 0.850 g of chromium is deposited using a current of 4.50 A, how long did the process take?
   Answer: 16.9 minutes When you encounter electroplating problems, you're dealing with electrochemistry where electric current drives a reduction reaction. The key is connecting the amount of substance deposited to the current and time through Faraday's laws of electrolysis. To find the time, you need to work through the stoichiometry. First, convert the mass of chromium to moles: $0.850 \text{ g} \div 51.996 \text{ g/mol} = 0.01635 \text{ mol Cr}$ From the balanced equation $Cr^{3+} + 3e^- \rightarrow Cr$, you can see that each chromium atom requires 3 electrons. So the total electrons needed are: $0.01635 \text{ mol Cr} \times 3 = 0.04905 \text{ mol e}^-$ Converting to charge using Faraday's constant: $0.04905 \text{ mol e}^- \times 96,485 \text{ C/mol} = 4,733 \text{ C}$ Finally, use the relationship $Q = It$ to find time: $t = Q/I = 4,733 \text{ C} \div 4.50 \text{ A} = 1,052 \text{ seconds} = 17.5 \text{ minutes}$ This matches answer choice C (16.9 minutes, allowing for rounding differences). Answer A (5.64 minutes) likely results from forgetting the 3-electron stoichiometry and only using 1 electron per chromium. Answer B (11.3 minutes) might come from calculation errors in unit conversions. Answer D (22.5 minutes) could result from incorrectly using the molar mass in the charge calculation. Remember: electroplating problems always require careful attention to electron stoichiometry from the balanced half-reaction, then systematic conversion through moles, charge, and finally time using $Q = It$.
3. A student electrolyzes aqueous $NaCl$ solution using inert platinum electrodes. If 0.0250 moles of electrons pass through the cell, what volume of gas (at STP) is produced at the anode?
   Answer: 0.140 L When you encounter electrolysis problems, focus on identifying what reactions occur at each electrode and how electrons relate to gas production. In aqueous $NaCl$ electrolysis with inert electrodes, two key reactions happen: at the cathode, water reduces to form $H_2$ gas, and at the anode, water oxidizes to form $O_2$ gas (since water is more easily oxidized than $Cl^-$ in dilute solutions). The anode reaction is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ This equation shows that 4 electrons produce 1 mole of $O_2$ gas. With 0.0250 moles of electrons passing through the cell, you can calculate: $\text{moles of } O_2 = \frac{0.0250 \text{ mol e}^-}{4 \text{ mol e}^-/\text{mol } O_2} = 0.00625 \text{ mol}$ At STP, one mole of any gas occupies 22.4 L, so: $\text{Volume} = 0.00625 \text{ mol} \times 22.4 \text{ L/mol} = 0.140 \text{ L}$ Answer A (0.140 L) is correct. Answer B (0.280 L) results from using a 2:1 electron-to-gas ratio instead of 4:1. Answer C (0.560 L) comes from assuming a 1:1 ratio, ignoring the stoichiometry entirely. Answer D (1.12 L) suggests using all electrons to produce gas with no consideration of the balanced equation. Remember: always write the balanced half-reaction first to determine the electron-to-product stoichiometry. In electrolysis problems, the number of electrons transferred is crucial for calculating product amounts.
4. A silver-plating operation uses a current of 8.50 A for 2.00 hours. If the current efficiency is 85.0%, how many grams of silver are deposited?
   Answer: 58.5 g When you encounter electroplating problems, you're dealing with electrochemistry and Faraday's laws. The key is connecting electrical current to the amount of substance deposited, accounting for both the metal's charge and the process efficiency. Start by calculating the total charge passed: $Q = I \times t = 8.50 \text{ A} \times (2.00 \times 3600) \text{ s} = 61,200 \text{ C}$ Since current efficiency is only 85.0%, the effective charge for silver deposition is: $Q_{eff} = 61,200 \times 0.850 = 52,020 \text{ C}$ Silver ions (Ag⁺) require one electron per atom for reduction, so one mole of electrons deposits one mole of silver. Using Faraday's constant (96,485 C/mol): $\text{moles of Ag} = \frac{52,020}{96,485} = 0.539 \text{ mol}$ Converting to grams: $0.539 \text{ mol} \times 107.87 \text{ g/mol} = 58.1 \text{ g}$ This matches answer D) 58.5 g (the small difference is due to rounding). A) 54.9 g represents a calculation error, likely using an incorrect atomic mass or charge. B) 64.6 g results from forgetting to apply the 85% efficiency factor—this would be the mass if efficiency were 100%. C) 46.7 g appears to involve errors in both the time conversion and efficiency application. Study tip: Always remember the three key steps in electroplating problems: calculate total charge (I×t), apply efficiency, then use Faraday's laws with the correct ion charge. Current efficiency below 100% is common in real processes due to side reactions.
5. During the electrolysis of aqueous $NaBr$ with platinum electrodes, a student observes gas evolution at both electrodes. What is the overall cell reaction?
   Answer: $2NaBr + 2H_2O \rightarrow 2NaOH + H_2 + Br_2$ When you encounter electrolysis problems, you need to consider what actually gets oxidized and reduced at each electrode based on the relative ease of these processes. In aqueous solutions, water can compete with the dissolved ions. For the electrolysis of aqueous $NaBr$, let's examine what happens at each electrode. At the cathode (negative electrode), reduction occurs. While $Na^+$ could theoretically be reduced to sodium metal, water is much more easily reduced in aqueous solution, producing $H_2$ gas and $OH^-$ ions: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. At the anode (positive electrode), oxidation occurs. Here, $Br^-$ ions are more easily oxidized than water, producing $Br_2$ gas: $2Br^- \rightarrow Br_2 + 2e^-$. The overall reaction combines these half-reactions with the spectator $Na^+$ ions to give: $2NaBr + 2H_2O \rightarrow 2NaOH + H_2 + Br_2$, which is answer D. Answer A is wrong because sodium metal cannot form in aqueous solution—water reduces preferentially. Answer B represents the electrolysis of pure water, ignoring the bromide ions entirely. Answer C shows the correct electrode reactions but omits the sodium ions, which don't just disappear—they combine with the hydroxide ions to form sodium hydroxide. Remember: in aqueous electrolysis, alkali metals like sodium won't deposit at the cathode because water reduces more easily. Always consider both the ions present and the competition from water molecules.
6. During electrolysis, why does the mass deposited at an electrode depend on both the current and the time, rather than just the total energy supplied?
   Answer: Mass deposited depends on the number of electrons transferred, which equals current multiplied by time divided by electron charge This question tests your understanding of Faraday's laws of electrolysis, which connect the fundamental relationship between electric charge and chemical change. When you encounter electrolysis problems, always think about the flow of electrons and how that translates to mass changes. During electrolysis, mass deposition occurs because electrons transfer to ions at the electrode, causing them to gain or lose electrons and form neutral atoms or molecules that deposit on the electrode surface. The key insight is that mass deposited is directly proportional to the number of electrons transferred. Since current measures the flow of charge per unit time ($I = Q/t$), and time tells us how long this flow continues, multiplying current by time gives us the total charge transferred: $Q = I \times t$. To find the number of electrons, you divide this total charge by the charge of a single electron. More electrons transferred means more ions are reduced or oxidized, resulting in greater mass deposition. Option A incorrectly focuses on temperature effects rather than the fundamental charge transfer mechanism. Option C mentions efficiency, which isn't the primary factor determining the direct relationship between current, time, and mass. Option D confuses the relationship by bringing in resistance and voltage, which affect the conditions needed for electrolysis but don't explain why mass depends specifically on current and time rather than total energy. Remember: in electrolysis problems, always trace back to electron transfer. Current and time together determine total charge, which directly determines how many electrons participate in the redox reactions at the electrodes.
7. During the electrolysis of brine (concentrated $NaCl$ solution) in an industrial chlor-alkali process, which products are formed at the anode and cathode respectively?
   Answer: $Cl_2$ at anode, $H_2$ at cathode When you encounter electrolysis questions, focus on identifying what gets oxidized at the anode and what gets reduced at the cathode. In brine electrolysis, you have multiple species competing for reaction: $Na^+$, $Cl^-$, $H_2O$, and $OH^-$. At the anode (where oxidation occurs), you must compare the oxidation potentials of $Cl^-$ and $H_2O$. Although water has a lower standard oxidation potential, the high concentration of chloride ions in brine and kinetic factors (overpotential) make chloride oxidation favorable. The reaction is: $2Cl^- \rightarrow Cl_2 + 2e^-$ At the cathode (where reduction occurs), $Na^+$ competes with $H_2O$ for reduction. Sodium has a very negative reduction potential (-2.71 V), while water reduction is much more favorable. The reaction becomes: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ Looking at the choices: Answer A suggests sodium metal forms at the cathode, but sodium's extremely negative reduction potential makes this impossible in aqueous solution. Answer B indicates oxygen formation at the anode, which would occur if water were oxidized instead of chloride - this doesn't happen under brine conditions. Answer D reverses the electrode assignments entirely, showing a fundamental misunderstanding of electrolysis. Answer C correctly identifies $Cl_2$ at the anode and $H_2$ at the cathode. Remember: In aqueous electrolysis, alkali metals like sodium virtually never form at the cathode due to their very negative reduction potentials. Water will reduce instead, producing hydrogen gas.
8. An electroplating bath contains both $Ni^{2+}$ and $Cu^{2+}$ ions at equal concentrations. Based on standard reduction potentials ($E^\circ_{Cu^{2+}/Cu} = +0.34 V$, $E^\circ_{Ni^{2+}/Ni} = -0.23 V$), which metal will be deposited first during electrolysis?
   Answer: Copper, because it has a more positive reduction potential When you encounter electroplating problems with multiple metal ions, the key principle is that the metal with the higher (more positive) reduction potential will be reduced first at the cathode. This happens because metals with higher reduction potentials have a greater tendency to gain electrons and form solid metal deposits. Looking at the given standard reduction potentials: $E^\circ_{Cu^{2+}/Cu} = +0.34 V$ and $E^\circ_{Ni^{2+}/Ni} = -0.23 V$. Since copper has the more positive reduction potential (+0.34 V vs -0.23 V), $Cu^{2+}$ ions will be reduced to solid copper first during electrolysis. The reaction $Cu^{2+} + 2e^- \rightarrow Cu$ is thermodynamically more favorable than $Ni^{2+} + 2e^- \rightarrow Ni$. Choice A is incorrect because having a more negative reduction potential actually means nickel is less likely to be reduced first. Choice C is wrong because the significant difference in reduction potentials (0.57 V apart) means copper will deposit preferentially before nickel begins to deposit appreciably. Choice D contains a true statement—nickel is indeed more reactive—but this actually supports why copper deposits first, making the reasoning backwards. Study tip: Remember that in electroplating, "higher reduction potential wins the electron race." The more positive the $E^\circ$ value, the more readily that ion accepts electrons to become a solid metal. This principle applies to any competitive electrochemical process involving multiple metal ions.
9. Which of the following statements about Faraday's laws of electrolysis is correct?
   Answer: The amount of substance produced is directly proportional to the total charge passed through the electrolyte Faraday's laws of electrolysis describe the quantitative relationship between electrical charge and the amount of chemical change at electrodes. When you encounter questions about electrolysis, focus on the fundamental principle that chemical reactions at electrodes are driven by the flow of electrons. Faraday's first law states that the amount of substance produced at an electrode is directly proportional to the quantity of electric charge passed through the electrolyte. Since charge (Q) equals current (I) times time (t), or $Q = I \times t$, the mass of substance deposited depends on the total charge, not just the current or voltage alone. This makes option B correct. Option A is incorrect because the amount of substance depends on charge (current × time), not voltage. Higher voltage can increase current, but without considering time, voltage alone doesn't determine the amount of product formed. Option C is wrong because the current required absolutely depends on the number of electrons in the electrode reaction. For example, reducing one mole of $Cu^{2+}$ (2 electrons) requires twice the charge of reducing one mole of $Ag^+$ (1 electron). Option D incorrectly describes an inverse relationship with molar mass. Actually, Faraday's second law shows that for a given charge, the mass deposited is directly proportional to the molar mass divided by the number of electrons transferred. Remember this key relationship: moles of product = (total charge) ÷ (electrons per reaction × Faraday's constant). Electrolysis problems always come back to tracking electrons and charge flow.
10. In the industrial production of aluminum by electrolysis of $Al_2O_3$ dissolved in molten cryolite, why is the process carried out at high temperature rather than electrolyzing aqueous $Al^{3+}$ solutions?
    Answer: Aluminum metal would immediately react with water to produce hydrogen gas and aluminum hydroxide When you encounter questions about industrial metal extraction processes, focus on the fundamental chemistry limitations that drive the choice of method. The Hall-Héroult process for aluminum production uses molten cryolite specifically to avoid a critical problem with aqueous electrolysis. The key issue is that aluminum cannot be produced from aqueous solutions because water interferes with the process. If you tried to electrolyze aqueous $Al^{3+}$ solutions, the aluminum metal formed at the cathode would immediately react with water: $2Al + 6H_2O → 2Al(OH)_3 + 3H_2$. This reaction is thermodynamically favored, meaning you'd never obtain pure aluminum metal—it would instantly convert to aluminum hydroxide and produce hydrogen gas. This makes option B correct. Let's examine why the other options miss the mark. Option A suggests conductivity is the primary concern, but while molten systems do conduct well, this isn't the fundamental reason avoiding aqueous solutions—it's the reactivity issue. Option C incorrectly states that reduction potentials become more positive (easier to reduce) at higher temperatures; actually, aluminum's reduction potential remains quite negative. Option D is factually wrong since you can achieve high $Al^{3+}$ concentrations in aqueous solutions. Remember this pattern: when evaluating industrial processes, always consider whether the desired product is stable under the reaction conditions. Many metals that are highly reactive with water (like aluminum, sodium, and magnesium) require non-aqueous extraction methods to prevent immediate reaction of the product.
11. During electrolysis of aqueous $KI$ solution with inert electrodes, which species are produced at the anode and cathode respectively?
    Answer: $I_2$ at anode, $H_2$ at cathode When you encounter electrolysis problems, you need to determine what gets oxidized at the anode and what gets reduced at the cathode by comparing standard reduction potentials and considering the aqueous environment. In aqueous $KI$ solution, you have four potential species: $K^+$, $I^-$, $H_2O$, and $OH^-$/$H^+$ from water's autoionization. At the anode (oxidation site), you must choose between oxidizing $I^-$ or $H_2O$. The half-reaction $2I^- \rightarrow I_2 + 2e^-$ has a standard potential of -0.54 V, while $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ has -1.23 V. Since $I^-$ oxidizes more easily (less negative potential), $I_2$ forms at the anode. At the cathode (reduction site), you choose between reducing $K^+$ or $H_2O$. Potassium has an extremely negative reduction potential (-2.92 V), making it nearly impossible to reduce in aqueous solution. Instead, water reduces: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ (-0.83 V), producing $H_2$ gas. Option A is wrong because $K$ metal cannot form in aqueous solution due to its extremely low reduction potential. Option B incorrectly suggests $O_2$ forms instead of $I_2$, ignoring that iodide oxidizes more readily than water. Option D combines both errors from A and B. Study tip: For aqueous electrolysis, remember that alkali metals (Group 1) virtually never reduce to the metal in water—hydrogen gas forms instead. Always compare reduction potentials to predict which species will actually react.
12. In the electrolysis of aqueous $CuSO_4$ using copper electrodes, which of the following correctly describes what occurs at each electrode?
    Answer: Anode: $Cu \rightarrow Cu^{2+} + 2e^-$; Cathode: $Cu^{2+} + 2e^- \rightarrow Cu$ When you encounter electrolysis problems, remember that oxidation always occurs at the anode (electrons are lost) and reduction always occurs at the cathode (electrons are gained). The key insight here is that you're using copper electrodes in a copper sulfate solution, which creates a special scenario. At the anode, copper metal from the electrode itself gets oxidized: $Cu \rightarrow Cu^{2+} + 2e^-$. The copper atoms lose electrons and dissolve into solution as $Cu^{2+}$ ions. At the cathode, the $Cu^{2+}$ ions in solution gain electrons and get reduced back to solid copper: $Cu^{2+} + 2e^- \rightarrow Cu$. This copper plates out onto the cathode surface. Choice A reverses the electrode assignments—it shows reduction at the anode and oxidation at the cathode, which violates fundamental electrochemistry principles. Choice C describes what would happen if you were splitting water molecules, but since copper is more easily oxidized than water at the anode, and $Cu^{2+}$ is more easily reduced than water at the cathode, the copper reactions dominate. Choice D incorrectly suggests that sulfate ions undergo oxidation, but sulfate is very stable and won't oxidize under these conditions. The correct answer is B, which properly places oxidation at the anode and reduction at the cathode with the appropriate copper reactions. Study tip: For electrolysis problems, first identify what species are present, then determine which reactions are most thermodynamically favorable at each electrode. Metal electrodes often participate directly in the reactions rather than just conducting electricity.
13. An electrolytic cell contains molten $CaCl_2$. How long must a current of 5.00 A be applied to produce 10.0 g of calcium metal?
    Answer: 160 minutes When you encounter electrolytic cell problems, you're dealing with the quantitative relationship between electric current, time, and the amount of substance produced. This requires combining Faraday's laws of electrolysis with stoichiometry. To find the time needed, start with the reduction reaction: $Ca^{2+} + 2e^- \rightarrow Ca$. Notice that producing one mole of calcium requires 2 moles of electrons. First, calculate moles of calcium needed: $\frac{10.0 \text{ g}}{40.08 \text{ g/mol}} = 0.249 \text{ mol Ca}$ Since each calcium atom needs 2 electrons, you need: $0.249 \text{ mol Ca} \times 2 = 0.498 \text{ mol electrons}$ Convert to coulombs using Faraday's constant: $0.498 \text{ mol} \times 96,485 \text{ C/mol} = 48,049 \text{ C}$ Finally, use $Q = It$ to find time: $t = \frac{48,049 \text{ C}}{5.00 \text{ A}} = 9,610 \text{ s} = 160 \text{ minutes}$ Answer A (160 minutes) is correct. Answer B (80.3 minutes) results from forgetting that $Ca^{2+}$ requires 2 electrons—using only 1 electron per calcium atom. Answer C (320 minutes) likely comes from doubling the correct answer, perhaps from confusion about the electron stoichiometry. Answer D (53.5 minutes) represents using the wrong conversion factor or incorrectly applying Faraday's constant. Remember: Always identify the balanced half-reaction first to determine the correct electron stoichiometry—this is where most electrolysis calculation errors occur.
14. In commercial aluminum production, the carbon anodes are gradually consumed during electrolysis. Which reaction accounts for this consumption?
    Answer: $C + 2O^{2-} \rightarrow CO_2 + 4e^-$ When analyzing electrolysis reactions, you need to identify what's happening at each electrode and write the half-reactions correctly. In aluminum production (the Hall-Héroult process), molten aluminum oxide ($Al_2O_3$) is electrolyzed using carbon anodes that gradually get consumed during the process. At the anode (positive electrode), oxidation occurs. The oxide ions ($O^{2-}$) from the molten $Al_2O_3$ lose electrons and react with the carbon anode. The correct half-reaction is $C + 2O^{2-} \rightarrow CO_2 + 4e^-$. This shows carbon being oxidized by oxide ions, producing carbon dioxide gas and releasing electrons, which explains why the carbon anodes are gradually consumed. Option A ($C + O_2 \rightarrow CO_2$) represents simple combustion of carbon in oxygen gas, but there's no free $O_2$ gas present in the molten electrolyte. Option C ($C + Al_2O_3 \rightarrow Al + CO_2$) incorrectly suggests a direct reaction between carbon and aluminum oxide without showing the electrochemical nature of the process or proper electron transfer. Option D ($2C + O_2 \rightarrow 2CO$) again assumes $O_2$ gas is present and produces carbon monoxide instead of the carbon dioxide that's actually formed. For electrolysis problems, always write proper half-reactions showing electron transfer and use the actual ionic species present in the electrolyte. Remember that anodes undergo oxidation, and in industrial processes like aluminum production, the electrode material itself often participates in the reaction.
15. During electrolysis of aqueous $CuCl_2$ using inert electrodes, what happens to the pH of the solution over time?
    Answer: pH decreases because $H^+$ ions are produced at the anode When analyzing electrolysis problems, you need to identify what happens at each electrode and how those reactions affect the overall solution chemistry. During electrolysis of aqueous $CuCl_2$, two key reactions occur. At the cathode (negative electrode), $Cu^{2+}$ ions are reduced to solid copper: $Cu^{2+} + 2e^- → Cu$. At the anode (positive electrode), $Cl^-$ ions are oxidized to chlorine gas: $2Cl^- → Cl_2 + 2e^-$. However, once the chloride ions are depleted, water becomes the primary species being oxidized at the anode: $2H_2O → O_2 + 4H^+ + 4e^-$. This water oxidation produces $H^+$ ions, which accumulate in solution and decrease the pH over time. Looking at the wrong answers: Choice A incorrectly assumes $OH^-$ production at the cathode, but copper reduction doesn't generate hydroxide ions. Choice C suggests equal $H^+$ and $OH^-$ production, but the cathode reaction produces solid copper, not hydroxide, while the anode generates excess $H^+$ ions. Choice D describes a pH that initially increases then decreases, but there's no mechanism for initial pH increase since no base is produced early in the process. The correct answer is B because $H^+$ ions are indeed produced at the anode through water oxidation, making the solution increasingly acidic. Study tip: In electrolysis problems, always identify both electrode reactions and consider what happens when one reactant is consumed—secondary reactions often involve water and significantly affect pH.
16. What is the relationship between the Faraday constant and Avogadro's number in the context of electrolysis?
    Answer: $F = N_A \times e$, where $e$ is the elementary charge Understanding the relationship between fundamental constants in electrochemistry requires thinking about what each constant represents at the molecular level. The Faraday constant (F) represents the charge carried by one mole of electrons, while Avogadro's number ($N_A$) tells us how many particles are in a mole. To find the total charge of one mole of electrons, you multiply the number of electrons in a mole ($N_A$) by the charge of each individual electron (the elementary charge, $e$). This gives us $F = N_A \times e$, which is exactly what option A states. Let's examine why the other options fail. Option B ($F = \frac{N_A}{e}$) incorrectly divides Avogadro's number by the elementary charge, which would give units that don't represent charge and would be dimensionally incorrect. Option C ($F = N_A^2 \times e$) squares Avogadro's number, creating an absurdly large value that has no physical meaning in this context. Option D ($F = \sqrt{N_A \times e}$) takes the square root of the product, which again produces the wrong units and magnitude. The key insight is recognizing that the Faraday constant is simply a "scaling up" of the elementary charge from the single-electron level to the mole level. When you encounter electrochemistry problems, remember that the Faraday constant bridges the gap between atomic-scale electrical phenomena and laboratory-scale measurements by using Avogadro's number as the conversion factor.
17. In the electrolytic production of hydrogen gas from water, what volume of $H_2$ (at STP) is produced when 5.00 × 10⁴ C of charge passes through the cell?
    Answer: 5.80 L When you encounter electrolysis problems, you're dealing with the quantitative relationship between electric charge and chemical products. The key is connecting Faraday's laws of electrolysis to stoichiometry and gas laws. For water electrolysis, the cathode reaction is: $2H^+ + 2e^- \rightarrow H_2$. This tells you that producing one mole of $H_2$ requires 2 moles of electrons. Since one mole of electrons carries 96,485 C (Faraday's constant), you need 2 × 96,485 = 192,970 C per mole of $H_2$. With 5.00 × 10⁴ C passing through the cell, the moles of $H_2$ produced = $\frac{5.00 \times 10^4 \text{ C}}{192,970 \text{ C/mol}} = 0.259 \text{ mol}$ At STP, one mole of any gas occupies 22.4 L, so: Volume = 0.259 mol × 22.4 L/mol = 5.80 L. This confirms answer B. Answer A (2.90 L) represents exactly half the correct volume, suggesting an error where someone forgot that $H_2$ formation requires 2 electrons per molecule. Answer C (11.6 L) is double the correct answer, likely from incorrectly assuming only 1 electron is needed instead of 2. Answer D (1.45 L) is one-fourth the correct volume, combining both the electron count error and possibly a calculation mistake. Remember: always check the balanced half-reaction to determine the electron requirement per molecule of product. Electrolysis problems consistently test whether you can properly connect charge, electrons, and stoichiometry.
18. Two electrolytic cells are connected in series. Cell A contains $AgNO_3$ solution and Cell B contains $CuSO_4$ solution. If 0.500 mol of electrons pass through the circuit, what masses of metals are deposited in each cell?
    Answer: Cell A: 53.9 g Ag, Cell B: 15.9 g Cu When you encounter electrolytic cells in series, remember that the same number of electrons flows through both cells, but different metals require different numbers of electrons per atom to deposit. To solve this, you need to write the reduction reactions and apply Faraday's laws. In Cell A with $AgNO_3$: $Ag^+ + e^- \rightarrow Ag$. Each silver ion needs 1 electron. In Cell B with $CuSO_4$: $Cu^{2+} + 2e^- \rightarrow Cu$. Each copper ion needs 2 electrons. With 0.500 mol of electrons passing through: - Cell A: 0.500 mol electrons × (1 mol Ag/1 mol e⁻) = 0.500 mol Ag Mass = 0.500 mol × 107.9 g/mol = 53.9 g Ag - Cell B: 0.500 mol electrons × (1 mol Cu/2 mol e⁻) = 0.250 mol Cu Mass = 0.250 mol × 63.5 g/mol = 15.9 g Cu Answer A incorrectly uses copper's atomic mass (63.5) but calculates moles wrong, getting 31.8 g instead of 15.9 g. Answer B assumes both metals need only 1 electron per atom, ignoring that copper is $Cu^{2+}$. Answer D correctly calculates copper mass but halves the silver mass incorrectly, perhaps confusing the electron ratios. The key strategy: always identify the oxidation states first. Silver deposits as $Ag^+$ (1 electron needed), while copper typically deposits as $Cu^{2+}$ (2 electrons needed). This 2:1 electron ratio means you'll always get fewer moles of copper than silver when the same current flows through both cells.
19. In the electrolysis of molten $MgCl_2$, what mass of chlorine gas is produced at the anode when 1.50 mol of electrons pass through the cell?
    Answer: 53.2 g When you encounter electrolysis problems, focus on the balanced half-reactions and the relationship between electrons and product formation. In molten $MgCl_2$ electrolysis, chloride ions are oxidized at the anode according to: $2Cl^- \rightarrow Cl_2 + 2e^-$ This equation tells you that 2 moles of electrons produce 1 mole of $Cl_2$ gas. With 1.50 mol of electrons passing through the cell, you can calculate: $1.50 \text{ mol } e^- \times \frac{1 \text{ mol } Cl_2}{2 \text{ mol } e^-} = 0.75 \text{ mol } Cl_2$ Converting to mass: $0.75 \text{ mol } Cl_2 \times 35.45 \text{ g/mol} \times 2 = 53.2 \text{ g}$ Choice A (53.2 g) is correct. Choice B (106 g) represents a common error where students assume a 1:1 electron-to-chlorine ratio instead of 2:1, effectively doubling the correct answer. Choice C (26.6 g) likely comes from using only one chlorine atom's mass instead of the diatomic $Cl_2$ molecule. Choice D (35.5 g) represents the molar mass of one chlorine atom, suggesting confusion between atomic and molecular chlorine. Remember that electrolysis problems always require you to write the balanced half-reaction first to establish the correct stoichiometric relationship between electrons and products. Pay special attention to whether the product is a single atom or a diatomic molecule like $Cl_2$.
20. During the electrolysis of molten $Al_2O_3$, how many moles of aluminum are produced when 3.00 × 10⁵ C of charge passes through the cell?
    Answer: 1.04 mol When you encounter electrolysis problems, you're dealing with the relationship between electric charge and the amount of substance produced at electrodes. This requires understanding Faraday's laws of electrolysis and the stoichiometry of the electrode reactions. For aluminum production from molten $Al_2O_3$, the cathode reaction is: $Al^{3+} + 3e^- \rightarrow Al$. This tells you that producing one mole of aluminum requires 3 moles of electrons. To solve this, first convert charge to moles of electrons using Faraday's constant (96,485 C/mol e⁻): $\text{Moles of electrons} = \frac{3.00 \times 10^5 \text{ C}}{96,485 \text{ C/mol}} = 3.11 \text{ mol e}^-$ Since 3 electrons produce 1 aluminum atom: $\text{Moles of Al} = \frac{3.11 \text{ mol e}^-}{3 \text{ e}^-/\text{Al}} = 1.04 \text{ mol Al}$ Therefore, A (1.04 mol) is correct. B (3.11 mol) represents the trap of forgetting the 3:1 electron-to-aluminum ratio—this is actually the moles of electrons transferred. C (0.518 mol) appears to use an incorrect stoichiometric relationship, possibly dividing by 6 instead of 3. D (1.55 mol) likely results from calculation errors or using wrong constants. Remember this pattern: always identify the electrode reaction first to determine the electron requirement per mole of product, then use Faraday's constant to convert charge to electrons, and finally apply stoichiometry to find moles of product.

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