9.11 Electrolysis and Faraday’s Law

1. AFaraday's constant, the charge of one mole of electrons Correct
   Faraday's constant is 96,485 C per mole of electrons, so dividing the charge in coulombs by F gives the moles of electrons transferred.
2. BAvogadro's number, the number of electrons in one mole
   A student who treats a charge in coulombs as a count of electrons picks this, intending to divide by Avogadro's number. A coulomb is not the charge of one electron; the charge of a mole of electrons is Faraday's constant.
3. CThe applied voltage, the energy supplied per coulomb
   A student who thinks the applied voltage sets the amount of chemical change picks this. The moles of electrons depend on the charge alone, which is already known from the current and the time.
4. DThe solution's molarity, the moles of ions in each liter
   A student who thinks the concentration of the solution sets the amount of chemical change picks this. The charge that passes fixes the moles of electrons, whatever the concentration.

Working q = I × t gives the charge in coulombs. Faraday's constant is the charge of one mole of electrons, F = 96,485 C/mol e⁻, so moles of electrons = q/F. No other quantity is needed.

CED 9.11.A.1 · Read this in Fix

1. A25.7 min
   A student who thinks one mole of electrons deposits one mole of any metal picks this, using 0.04001 mol of electrons. Each Ni²⁺ ion needs two electrons, so twice the charge and twice the time are required.
2. B12.9 min
   A student who inverts the mole ratio, taking one mole of electrons to deposit two moles of Ni, picks this (0.02000 mol of electrons). Two moles of electrons are needed per mole of Ni, so 0.08001 mol of electrons is required.
3. C51.5 min Correct
   0.04001 mol of Ni requires 0.08001 mol of electrons, because each Ni²⁺ ion gains two electrons. That is 7720 C, which takes 7720 C ÷ 2.50 A = 3088 s = 51.5 min.
4. D19.5 min
   A student who uses the molar mass of NiSO₄ (154.75 g/mol) for the mass gained picks this. The cathode gains nickel metal, so the moles deposited are 2.348 g ÷ 58.69 g/mol.

Working Ni²⁺(aq) + 2e⁻ → Ni(s). Moles of Ni = 2.348 g ÷ 58.69 g/mol = 0.04001 mol. Moles of electrons = 2 × 0.04001 mol = 0.08001 mol. q = (0.08001 mol)(96,485 C/mol) = 7720 C. t = q/I = 7720 C ÷ 2.50 C/s = 3088 s = 51.5 min.

CED 9.11.A.1.iv · Read this in Fix

1. A2.96 g
   A student who thinks one mole of electrons deposits one mole of any metal picks this (0.04664 mol of Cu). Each Cu²⁺ ion needs two electrons, so only 0.02332 mol of Cu forms.
2. B1.48 g Correct
   q = (2.50 A)(1800 s) = 4500 C, which is 0.04664 mol of electrons. Each Cu²⁺ ion gains two electrons, so 0.02332 mol of Cu is deposited: 0.02332 mol × 63.55 g/mol = 1.48 g.
3. C5.93 g
   A student who multiplies the moles of electrons by the charge of the ion picks this (0.09328 mol of Cu). Two electrons are needed for each Cu²⁺ ion, so the moles of electrons are divided by 2.
4. D3.72 g
   A student who uses the molar mass of CuSO₄ (159.61 g/mol) picks this. The deposit is copper metal, formed by Cu²⁺ + 2e⁻ → Cu, so the molar mass of Cu is used.

Working t = 30.0 min × 60 s/min = 1800 s. q = I × t = (2.50 A)(1800 s) = 4500 C. Moles of electrons = 4500 C ÷ 96,485 C/mol = 0.04664 mol. Cu²⁺(aq) + 2e⁻ → Cu(s), so moles of Cu = 0.04664 ÷ 2 = 0.02332 mol. Mass = (0.02332 mol)(63.55 g/mol) = 1.48 g.

CED 9.11.A.1.ii · Read this in Fix

1. A30 min
   A student who thinks the same charge deposits the same number of moles of any metal picks this. Each Au³⁺ ion needs three electrons, so 0.010 mol of Au needs three times the charge that 0.010 mol of Ag needs.
2. B10 min
   A student who thinks an ion with three times the charge is deposited three times as fast picks this. A larger charge on the ion means more electrons are needed per atom, so the time is longer, not shorter.
3. C55 min
   A student who thinks a given charge deposits a given mass of any metal picks this, scaling the time by the molar masses (196.97 ÷ 107.87). The charge needed depends on the moles of metal and the charge of its ion, not on the mass.
4. D90 min Correct
   Each Au³⁺ ion gains three electrons, while each Ag⁺ ion gains one. The same number of moles of metal needs three times the charge, and at the same current that takes three times as long.

Working Ag⁺ + e⁻ → Ag needs 1 mol of electrons per mole of Ag; Au³⁺ + 3e⁻ → Au needs 3 mol of electrons per mole of Au. The same number of moles of Au therefore needs three times the charge. At the same current, q = I × t, so the time is three times as long: 3 × 30 min = 90 min.

CED 9.11.A.1.v · Read this in Fix

1. A2.28 A
   A student who thinks one mole of electrons deposits one mole of any metal picks this, using 0.01888 mol of electrons. Each Cu²⁺ ion gains two electrons, so the charge and the current are twice as large.
2. B80.5 A
   A student who reads the height of the final point, 21.20 g, as the mass deposited picks this. The vertical axis shows the total mass of the cathode; the copper deposited is the change in mass, 1.20 g.
3. C4.55 A Correct
   The cathode gains 1.20 g (0.01888 mol of Cu) in 800 s. Two electrons are needed per Cu²⁺ ion, so 0.03777 mol of electrons (3644 C) passes, and I = 3644 C ÷ 800 s = 4.55 A.
4. D1.81 A
   A student who uses the molar mass of CuSO₄ (159.61 g/mol) for the mass gained picks this. The cathode gains copper metal, so the moles deposited are 1.20 g ÷ 63.55 g/mol.

Working From the graph, the cathode gains 21.20 g − 20.00 g = 1.20 g in 800 s. Moles of Cu = 1.20 g ÷ 63.55 g/mol = 0.01888 mol. Cu²⁺ + 2e⁻ → Cu, so moles of electrons = 0.03777 mol. q = (0.03777 mol)(96,485 C/mol) = 3644 C. I = q/t = 3644 C ÷ 800 s = 4.55 A.

CED 9.11.A.1.iii · Read this in Fix

1. AThe concentration of the chromium ions in the solution used
   A student who thinks a more concentrated solution deposits more metal picks this. With the current and the time known, the charge that passes fixes the amount deposited.
2. BThe charge of the chromium ions that are in the solution Correct
   Current × time ÷ F gives the moles of electrons. The charge of the ion gives the number of electrons needed per chromium atom deposited, and so the moles of Cr.
3. CThe molar mass of the chromium compound used for the solution
   A student who thinks the molar mass of the dissolved compound is used for the deposit picks this. The deposit is chromium metal, whose molar mass is already available.
4. DThe voltage applied across the two electrodes of the cell
   A student who thinks the applied voltage sets the amount deposited picks this. The amount depends on the charge, which is already fixed by the known current and time.

Working q = I × t gives the charge, and q/F gives the moles of electrons. To convert moles of electrons to moles of Cr, the number of electrons gained by each chromium ion is needed, which is the charge of the ion. Moles of Cr × molar mass of Cr then gives the mass.

CED 9.11.A.1.v · Read this in Fix

1. AFour
   A student who thinks one electron is transferred for each particle of product, whatever the charge of the ion, picks this from the four Al atoms. Each Al³⁺ ion must gain three electrons to become an atom.
2. BSix
   A student who assigns two electrons, the charge of one oxide ion, to each of the three O₂ molecules picks this. Each O₂ molecule forms from two O²⁻ ions, which lose four electrons in all.
3. CThree
   A student who reports the number of electrons written in the half-reaction Al³⁺ + 3e⁻ → Al picks this. That is the number gained by one ion; four Al³⁺ ions gain 4 × 3 = 12 electrons.
4. DTwelve Correct
   Four Al³⁺ ions each gain three electrons at the cathode (4 × 3). The same electrons were released at the anode, where six O²⁻ ions each lose two electrons to form three O₂ molecules (6 × 2).

CED 9.11.A.1.i · Read this in Fix

1. ADoubling the length of time the current passes Correct
   At a constant current, q = I × t, so doubling the time doubles the charge, the moles of electrons and the moles of copper deposited.
2. BDoubling the concentration of the CuSO₄(aq) solution
   A student who thinks a more concentrated solution deposits more metal picks this. The current and the time are unchanged, so the same charge passes and the same mass of copper is deposited.
3. CDoubling the area of the cathode in the solution used
   A student who thinks a larger electrode gains more mass picks this. The same charge passes, so the same mass of copper is deposited, as a thinner layer over a larger area.
4. DDoubling the voltage setting on the power supply
   A student who thinks the applied voltage sets the amount deposited picks this. With the current held at 0.50 A for the same time, the charge and the mass deposited are unchanged.

CED 9.11.A.1.ii · Read this in Fix

1. ACell 2 deposits the same number of moles, as the same charge passes through it
   A student who thinks the same charge deposits the same number of moles of any metal picks this. The same charge does pass, but each Cu²⁺ ion needs two electrons and each Ag⁺ ion needs one.
2. BCell 2 deposits twice as many moles of metal, as an Ag⁺ ion gains half as many electrons Correct
   The cells are in series, so the same charge passes through each. Each Cu²⁺ ion gains two electrons and each Ag⁺ ion gains one, so the same moles of electrons deposit twice as many moles of Ag as of Cu.
3. CCell 2 deposits half as many moles of metal, as each Ag⁺ ion has half the charge
   A student who thinks an ion with a larger charge gives more metal per mole of electrons picks this. An ion with a smaller charge needs fewer electrons, so a given charge deposits more moles of Ag, not fewer.
4. DCell 2 deposits fewer moles of metal, as some current is used up in cell 1
   A student who thinks current is used up as it passes through each cell picks this. Charge is conserved: the same current passes through both cells, and cell 2 deposits more moles because each Ag⁺ ion needs one electron.

CED 9.11.A.1.i · Read this in Fix

1. A22.46 g
   A student who thinks one mole of electrons reacts with one mole of any metal picks this, removing 0.04001 mol of Cu (2.54 g). Two electrons are released for each Cu atom oxidized, so 0.02000 mol (1.27 g) is removed.
2. B25.00 g
   A student who thinks the anode only completes the circuit, and that the copper deposited comes only from ions already in the solution, picks this. The copper anode is oxidized, Cu → Cu²⁺ + 2e⁻, and loses mass.
3. C26.27 g
   A student who places the deposition of metal at the anode picks this, adding 1.27 g. Cu²⁺ ions are reduced at the cathode, which gains mass; copper is oxidized at the anode, which loses mass.
4. D23.73 g Correct
   3860 C is 0.04001 mol of electrons. Copper atoms of the anode are oxidized, Cu → Cu²⁺ + 2e⁻, so 0.02000 mol of Cu (1.27 g) is removed and 25.00 g − 1.27 g = 23.73 g remains.

Working q = (1.930 A)(2000 s) = 3860 C. Moles of electrons = 3860 C ÷ 96,485 C/mol = 0.04001 mol. At the copper anode, Cu(s) → Cu²⁺(aq) + 2e⁻, so 0.02000 mol of Cu is oxidized: 0.02000 mol × 63.55 g/mol = 1.27 g removed. Mass of anode = 25.00 g − 1.27 g = 23.73 g.

CED 9.11.A.1.ii · Read this in Fix

1. AMoles of metal deposited are directly proportional to the ion's charge
   A student who thinks an ion with a larger charge gives more metal per mole of electrons picks this. The moles deposited fall from 0.0100 (Ag⁺) to 0.00500 (Cu²⁺) to 0.00334 (Au³⁺) as the charge of the ion rises.
2. BMoles of metal deposited are equal for the three ions, as the charge passed is equal
   A student who thinks the same charge deposits the same number of moles of any metal picks this. The masses correspond to 0.0100 mol, 0.00500 mol and 0.00334 mol, which are not equal.
3. CMoles of metal deposited are inversely proportional to the ion's charge Correct
   Dividing each mass by the molar mass gives 0.0100 mol of Ag, 0.00500 mol of Cu and 0.00334 mol of Au: in the ratio 1 : 1/2 : 1/3 for ions of charge 1+, 2+ and 3+.
4. DMass of metal deposited is fixed by the charge passed, whichever ion is used
   A student who thinks a given charge deposits a given mass of any metal picks this. The three masses, 1.08 g, 0.318 g and 0.657 g, are different although the charge passed is the same.

CED 9.11.A.1.v · Read this in Fix

1. A483 C
   A student who thinks one mole of electrons deposits one mole of any metal picks this, taking 0.00500 mol of electrons. Each Zn²⁺ ion needs 2 electrons, so 0.0100 mol of electrons, 965 C, passed.
2. B241 C
   A student who thinks a 2+ ion gives twice as much metal per mole of electrons picks this, dividing the moles of Zn by 2 to get 0.00250 mol of electrons. The moles of Zn are multiplied by 2, because each Zn²⁺ ion gains 2 electrons.
3. C391 C
   A student who divides the mass gained by the molar mass of ZnSO₄, 161.44 g/mol, picks this. Only zinc is deposited on the cathode, so the mass is divided by the molar mass of Zn, 65.38 g/mol.
4. D965 C Correct
   0.327 g of Zn is 0.00500 mol. Each Zn²⁺ ion gains 2 electrons, so 0.0100 mol of electrons passed, and q = 0.0100 mol × 96,485 C/mol = 965 C.

Working Moles of Zn deposited = 0.327 g ÷ 65.38 g/mol = 0.00500 mol. Zn²⁺(aq) + 2 e⁻ → Zn(s), so moles of electrons = 2 × 0.00500 mol = 0.0100 mol. q = 0.0100 mol × 96,485 C/mol = 965 C.

CED 9.11.A.1.i · Read this in Fix

1. A0.304 g Correct
   The charge is found for each period of constant current: (1.50 A)(400 s) + (0.50 A)(800 s) = 1000 C, which is 0.01036 mol of electrons. Each Ni²⁺ ion gains 2 electrons, so 0.005182 mol of Ni, 0.304 g, is deposited.
2. B0.608 g
   A student who thinks one mole of electrons deposits one mole of any metal picks this, taking 0.01036 mol of Ni. Each Ni²⁺ ion gains 2 electrons, so 0.005182 mol of Ni is deposited.
3. C0.802 g
   A student who multiplies the 0.005182 mol by the molar mass of NiSO₄, 154.75 g/mol, picks this. Only nickel is deposited on the cathode, so the molar mass of Ni, 58.69 g/mol, is used.
4. D0.547 g
   A student who multiplies the starting current by the total time picks this: (1.50 A)(1200 s) = 1800 C. The current was 1.50 A for only 400 s and 0.50 A for the remaining 800 s, so the charge is 600 C + 400 C = 1000 C.

Working From the graph, the current is 1.50 A for 400 s and then 0.50 A for 800 s. q = (1.50 A)(400 s) + (0.50 A)(800 s) = 600 C + 400 C = 1000 C. Moles of electrons = 1000 C ÷ 96,485 C/mol = 0.01036 mol. Ni²⁺(aq) + 2 e⁻ → Ni(s), so moles of Ni = 0.005182 mol, and mass = 0.005182 mol × 58.69 g/mol = 0.304 g.

CED 9.11.A.1 · Read this in Fix