Open by holding up a piece of costume jewelry or a chrome-plated object (a shiny bolt, a plated spoon, or a photo of a gold-plated iPhone connector). Tell students: 'A factory needs to gold-plate 10,000 connectors per hour. The plant manager needs to know exactly how many amps to run and for how long. Guess too low — the coating peels. Guess too high — you waste gold at $80/gram. Today you become the plant engineer.' Ask students to predict: if we double the current, what happens to the mass of gold deposited? If we switch from Ag⁺ to Cu²⁺ at the same current and time, does the same mass deposit? Take 2-3 predictions, DO NOT resolve — tell them we'll answer both by the end of class. Targets SP 6 (making a predictive claim they'll defend with evidence).

1. 6m

   Charge, Current, and Time: Q = I·t

   Content

   Electrolysis problems always start by counting how much charge flowed through the wire. Current (I), measured in amperes, is the rate of charge flow: 1 A = 1 coulomb per second. Total charge Q in coulombs equals current times time in seconds: Q = I·t. If a cell runs at 2.50 A for 15.0 minutes, first convert time to seconds (15.0 × 60 = 900. s), then Q = 2.50 A × 900. s = 2250 C. Units check: A · s = (C/s)·s = C. The ampere is a rate; the coulomb is an amount — students routinely mix these up.

   Delivery

   Anchor the ampere-vs-coulomb distinction with a water analogy: amperes are gallons per second, coulombs are gallons. Emphasize that time MUST be in seconds — this is the #1 arithmetic error. Ask a cold-call check: 'A cell runs at 0.500 A for 2.00 hours. What is Q?' (Expected: 0.500 × 7200 = 3600 C.) The slide's flowchart highlights the first arrow: I, t → Q. Targets SP 5.

2. 7m

   The Faraday Constant: Charge → Moles of Electrons

   Content

   One mole of electrons carries a fixed amount of charge: the Faraday constant, F = 96,485 C/mol e⁻. This is just Avogadro's number times the elementary charge (6.022 × 10²³ × 1.602 × 10⁻¹⁹ C). To convert charge to moles of electrons, divide: n(e⁻) = Q/F. Continuing the previous example: n(e⁻) = 2250 C ÷ 96,485 C/mol = 0.02332 mol e⁻. F is given on the AP equations sheet — students do not memorize it, but they must recognize what it does: it is the unit conversion between the electrical world (coulombs) and the chemical world (moles).

   Delivery

   Stress that F is a bridge, not a multiplier you apply blindly. Ask: 'Why divide by F and not multiply?' — because F has units of C per mole, so to cancel coulombs you divide. Pre-empt the misconception that coulombs, amperes, and F are interchangeable — put the three units side by side and have students state what each measures (amount of charge, rate of charge, charge per mole). Targets SP 1 and SP 5.

3. 8m

   Electrons → Moles of Metal → Mass: Why Cu²⁺ ≠ Ag⁺

   Content

   The half-reaction tells you how many electrons reduce one ion. For silver: Ag⁺ + e⁻ → Ag, so 1 mol e⁻ makes 1 mol Ag. For copper: Cu²⁺ + 2 e⁻ → Cu, so 2 mol e⁻ makes only 1 mol Cu. Divide moles of electrons by the electron count n to get moles of metal, then multiply by molar mass. Full worked example — 2.50 A for 15.0 min through a Cu²⁺ solution: n(e⁻) = 0.02332 mol (from previous beat). n(Cu) = 0.02332 / 2 = 0.01166 mol. Mass Cu = 0.01166 × 63.55 g/mol = 0.741 g. Now the same charge through AgNO₃: n(Ag) = 0.02332 / 1 = 0.02332 mol. Mass Ag = 0.02332 × 107.87 = 2.515 g. Same current, same time, same coulombs — but 3.4× more silver deposits than copper because silver only needs one electron per atom, AND silver has a larger molar mass. The comparison graphic on the slide shows equal packets of electrons arriving at two cathodes: the Ag⁺ side plates one atom per electron, the Cu²⁺ side plates one atom per TWO electrons.

   Delivery

   This is the beat where the standard's core misconception lives. Walk both calculations in parallel columns so students see the ONLY difference is the '/n' step. Ask: 'If I want equal MOLES of Ag and Cu deposited, do I need equal charge?' (No — Cu needs twice the charge.) Then ask: 'If I want equal MASS?' (Depends on molar mass too — targets SP 4, model analysis.) Pre-empt the misconception that mass depends only on time: it depends on I, t, n, AND M jointly.

1. 25m

   Copper Electroplating Lab: Predict, Measure, ReconcileLab

   Targets SP 2 (method), SP 3 (data), SP 5 (calculation), SP 6 (argument from evidence). Groups of 3. Teacher pre-cleans copper strips with fine sandpaper and rinses with acetone the day before. Distribute the handout below. Walk around and verify (a) the ammeter is in SERIES not parallel, (b) both electrodes are copper (this is electrorefining-style — the anode dissolves to replenish Cu²⁺, so current efficiency approaches 100%), (c) students record initial cathode mass BEFORE any current flows, (d) current stays roughly constant — if it drifts, students should record an average. Expect actual mass to come in 90-100% of predicted (side reactions and rinsing loss). If a group gets >100%, likely didn't dry cathode fully — have them re-dry and reweigh. If <85%, check for loose plating that flaked off during rinse. In the argument step, students must distinguish between systematic error (drying, rinsing) and a broken model. Student handout: Copper Electroplating & Faraday's Law Goal: Predict the mass of copper deposited on the cathode from the current and time you run, then measure it and defend your model. Half-reactions: - Cathode: Cu²⁺(aq) + 2 e⁻ → Cu(s) - Anode: Cu(s) → Cu²⁺(aq) + 2 e⁻ Part 1 — Setup 1. Put on goggles and gloves. Do not touch electrode tips with bare fingers — skin oils prevent plating. 2. Sand both copper strips lightly, rinse with acetone, and dry. 3. Label one strip C (cathode) and one A (anode). Mass the cathode on the analytical balance. - Initial cathode mass = ______ g 4. Fill the 250 mL beaker with ~100 mL of 0.5 M CuSO₄. 5. Clip the cathode to the NEGATIVE terminal and the anode to the POSITIVE terminal of the power supply. Wire the ammeter in series. 6. Suspend both strips in the solution without touching each other. Part 2 — Predict BEFORE turning on the power You will run the cell at approximately 0.40 A for 8.0 minutes. Predict the mass of copper deposited on the cathode. - t (s) = ______ - Q = I × t = ______ C - n(e⁻) = Q / 96,485 = ______ mol - n(Cu) = n(e⁻) / 2 = ______ mol - Predicted mass Cu = n(Cu) × 63.55 = ______ g Part 3 — Run the cell 1. Turn on the power supply and adjust to ~0.40 A. Start the timer. 2. Record current every 2 min. If it drifts, average the readings. - t = 0 min: I = ______ A - t = 2 min: I = ______ A - t = 4 min: I = ______ A - t = 6 min: I = ______ A - t = 8 min: I = ______ A (turn off) - Average I = ______ A 3. Carefully lift the cathode with tongs, rinse gently with distilled water, blot dry, and air-dry 2 min. 4. Mass the cathode. - Final cathode mass = ______ g - Measured mass deposited = ______ g Part 4 — Recalculate with your ACTUAL average current - Q_actual = I_avg × 480 s = ______ C - n(e⁻)_actual = ______ mol - Recalculated predicted mass = ______ g Part 5 — Current efficiency - Current efficiency = (measured mass / recalculated predicted mass) × 100% = ______ % Part 6 — Argue from evidence (SP 6) Answer in 3-4 sentences on the back: 1. Does your data support Faraday's law? Cite your % efficiency. 2. If efficiency < 100%, identify ONE specific physical source of loss (not 'human error') and explain how it lowers measured mass. 3. A classmate claims 'if we double the time, we double the mass.' Under what condition is this claim true? When would it fail?

   Materials

   • 0.5 M CuSO₄ solution (~100 mL per group)
   • 2 copper strips per group (one anode, one cathode; ~2 × 6 cm)
   • DC power supply (0-6 V, variable) or 6 V lantern battery
   • Ammeter or multimeter (in series)
   • Alligator clip leads (2 per group)
   • 250 mL beaker
   • Stopwatch or phone timer
   • Analytical balance (0.001 g)
   • Acetone or ethanol for cleaning strips
   • Paper towels, tongs
   • Safety goggles and gloves

   Example outputs

   • Predicted mass at 0.40 A for 480 s: Q = 192 C; n(e⁻) = 1.99 × 10⁻³ mol; n(Cu) = 9.95 × 10⁻⁴ mol; mass = 0.0632 g
   • Measured mass = 0.061 g; current efficiency ≈ 96.5%; loss attributed to a thin layer of loosely bound Cu rinsing off during the water rinse.
   • SP 6 answer: 'Doubling time doubles mass ONLY if current stays constant AND the same half-reaction dominates. If current drops as ion concentration falls, or if H₂ evolution begins competing at the cathode, the linear relationship breaks.'

1. A current of 1.50 A is passed through molten AlCl₃ for 30.0 minutes. What mass of aluminum is deposited at the cathode? (Al³⁺ + 3 e⁻ → Al; M(Al) = 26.98 g/mol)

   calculationTargets SP 5. 1. t = 30.0 × 60 = 1800 s 2. Q = 1.50 × 1800 = 2700 C 3. n(e⁻) = 2700 / 96,485 = 0.02798 mol 4. n(Al) = 0.02798 / 3 = 9.328 × 10⁻³ mol 5. mass Al = 9.328 × 10⁻³ × 26.98 = 0.252 g

2. Equal amounts of charge are passed through separate cells containing AgNO₃, CuSO₄, and AuCl₃. Which cell deposits the GREATEST number of MOLES of metal, and which deposits the greatest MASS? (Ag⁺, Cu²⁺, Au³⁺; molar masses 107.87, 63.55, 196.97 g/mol)

   multiple choiceTargets SP 4 and SP 6. For equal charge Q, n(metal) = (Q/F) / n_electrons. - Ag: divides by 1 → most moles - Cu: divides by 2 - Au: divides by 3 → fewest moles Greatest MOLES: Ag (fewest electrons per ion). Greatest MASS: compare (M/n) — Ag: 107.87/1 = 107.87; Cu: 63.55/2 = 31.78; Au: 196.97/3 = 65.66. Greatest mass: Ag also wins. Correct answer: Ag for both.

3. A student electroplates nickel from Ni²⁺ solution at 2.00 A for 20.0 minutes and measures 0.680 g deposited. The theoretical mass is 0.730 g. Calculate the current efficiency and give ONE chemically specific reason (not 'human error') for the shortfall. (M(Ni) = 58.69 g/mol)

   short answerTargets SP 5 and SP 6. Efficiency = 0.680 / 0.730 × 100% = 93.2%. A valid chemical reason: some of the current reduced H⁺ or H₂O to H₂ gas at the cathode (a competing reduction) instead of Ni²⁺, so not every electron went into depositing nickel. Accept also: loose Ni flaked off during rinsing, or trace impurity ions in solution were reduced preferentially.

4. To deposit 5.00 g of silver from AgNO₃ at 3.00 A, how many minutes must the cell run?

   calculationTargets SP 5. 1. n(Ag) = 5.00 / 107.87 = 0.04635 mol 2. n(e⁻) = 0.04635 mol (since Ag⁺ + e⁻ → Ag) 3. Q = n(e⁻) × F = 0.04635 × 96,485 = 4472 C 4. t = Q / I = 4472 / 3.00 = 1491 s 5. t = 1491 / 60 = 24.8 min

• ☐0.5 M CuSO₄ solution (~100 mL per group)
• ☐Copper strips, 2 per group, pre-sanded and rinsed with acetone
• ☐DC power supply (0-6 V variable) or 6 V lantern battery
• ☐Ammeter or multimeter per group
• ☐Alligator clip leads
• ☐250 mL beakers
• ☐Analytical balance (0.001 g)
• ☐Stopwatch/timer
• ☐Acetone or ethanol for pre-cleaning
• ☐Paper towels and tongs
• ☐Safety goggles and nitrile gloves
• ☐Student handout (printed, one per student)
• ☐AP equation sheet or F = 96,485 C/mol posted