AP Chemistry Quiz: Electrolysis And Faradays Law

20 questions · exam conditions

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Electrolysis And Faradays LawQuestion 1 of 20

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

1.14 g

0.57 g

2.28 g

4.56 g

0.114 g

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AP Chemistry Quiz

AP Chemistry Quiz: Electrolysis And Faradays Law

Practice Electrolysis And Faradays Law in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Electrolysis And Faradays Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

1.14 g (correct answer)
0.57 g
2.28 g
4.56 g
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, $\text{NaBr}(l)$ , is electrolyzed. The current is $4.82\ \text{A}$ for $2.00\times 10^3\ \text{s}$ . At the anode, bromine is produced according to $2\text{Br}^-\rightarrow \text{Br}_2(g)+2e^-$ . Assuming 100% current efficiency, what amount of $\text{Br}_2(g)$ is produced?

0.200 mol
0.0250 mol
0.0100 mol
0.0500 mol (correct answer)
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 $\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)$ . What total charge is required to produce $0.250\ \text{mol}$ of $\text{Ag}(s)$ , assuming 100% current efficiency?

2.41×10^4 C (correct answer)
4.82×10^4 C
9.65×10^4 C
1.93×10^4 C
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 $2\text{H}_2\text{O}(l)+2e^-\rightarrow \text{H}_2(g)+2\text{OH}^-$ . A current of $9.65\ \text{A}$ is applied for $2.00\times 10^3\ \text{s}$ . Assuming 100% current efficiency, what amount of $\text{H}_2(g)$ is produced?

0.200 mol
0.0100 mol
0.0500 mol
0.100 mol (correct answer)
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 $19.3\ \text{A}$ is applied for $500\ \text{s}$ . At the anode, iodine forms according to $2\text{I}^-\rightarrow \text{I}_2(s)+2e^-$ . Assuming 100% current efficiency, how many moles of $\text{I}_2$ are produced?

0.0250 mol
0.0500 mol (correct answer)
0.100 mol
0.200 mol
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 $\text{Cu}^{2+}$ is electrolyzed with an inert cathode. A total charge of $1.93\times 10^4\ \text{C}$ passes through the circuit. Copper metal plates onto the cathode according to $\text{Cu}^{2+}+2e^-\rightarrow \text{Cu}(s)$ . Assuming 100% current efficiency, how many moles of $\text{Cu}(s)$ are produced?

0.0500 mol
0.100 mol (correct answer)
0.200 mol
0.0100 mol
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 $\text{Ca}^{2+}$ is electrolyzed. Calcium metal forms at the cathode according to $\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}(s)$ . What total charge (in coulombs) is required to produce $0.250\ \text{mol}$ of $\text{Ca}(s)$ , assuming $100\%$ current efficiency? ( $F = 9.65\times 10^4\ \text{C mol}^{-1}\ e^-$ .)

2.41×10^4 C
4.83×10^4 C (correct answer)
9.65×10^4 C
1.93×10^5 C
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 $\text{Cu}^{2+}(aq)$ is electrolyzed with a constant current of $2.00\ \text{A}$ for $40.2\ \text{min}$ . Copper is deposited at the cathode according to $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)$ . Assuming $100\%$ current efficiency, what mass of $\text{Cu}(s)$ is produced? ( $F = 9.65\times 10^4\ \text{C mol}^{-1}\ e^-$ ; molar mass $\text{Cu}=63.5\ \text{g mol}^{-1}$ .)

0.794 g
1.59 g (correct answer)
3.18 g
0.397 g
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 $\text{Fe}^{3+}$ is electrolyzed with an inert cathode. Iron(III) ions are reduced to iron(II) ions according to $\text{Fe}^{3+}+e^-\rightarrow \text{Fe}^{2+}$ . A total charge of $9.65\times 10^3\ \text{C}$ is passed through the cell. Assuming 100% current efficiency, how many moles of $\text{Fe}^{3+}$ are consumed?

0.200 mol
0.0500 mol
0.0100 mol
0.0250 mol
0.100 mol (correct answer)

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 $9.65 \, \text{A}$ for $30.0 \, \text{min}$ . Aluminum metal forms at the cathode according to $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(l)$ . Assuming 100% current efficiency, how many moles of $\text{Al}$ are produced?

0.180 mol
0.0600 mol (correct answer)
0.300 mol
0.0900 mol
0.0200 mol

Explanation: 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.

Question 11

An electrolytic cell is used to generate hydrogen gas at the cathode from water according to $2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)$ . A constant current of $3.00\ \text{A}$ is applied for $10.0\ \text{min}$ . Assuming 100% current efficiency, how many moles of $\text{H}_2(g)$ are produced?

0.0187 mol
0.00933 mol (correct answer)
0.0373 mol
0.00467 mol
0.0933 mol

Explanation: 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.

Question 12

A solution containing $\text{Ag}^+$ ions is electrolyzed using inert electrodes. A constant current of $1.00\ \text{A}$ is applied for $965\ \text{s}$ . At the cathode, the half-reaction is $\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)$ . Assuming 100% current efficiency, what mass of silver is deposited?

1.08 g (correct answer)
0.108 g
10.8 g
0.0108 g
2.16 g

Explanation: This problem uses electrolysis and Faraday's Law to find the mass of silver deposited. Calculate charge: Q = It = (1.00 A)(965 s) = 965 C. Convert to moles of electrons: moles e⁻ = 965 C ÷ 96,500 C/mol = 0.0100 mol e⁻. The half-reaction Ag⁺ + e⁻ → Ag shows a 1:1 ratio between electrons and silver atoms, so 0.0100 mol of Ag is produced. Converting to mass: (0.0100 mol)(107.9 g/mol) = 1.08 g. A common mistake is using the wrong molar mass or forgetting to convert from moles to grams, which might give 0.0100 g. Always use the correct molar mass and complete all unit conversions to get the final mass.

Question 13

A solution containing $\text{Ni}^{2+}$ ions is electrolyzed using inert electrodes. At the cathode, the half-reaction is $\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}(s)$ . What total charge (in coulombs) must pass through the cell to deposit $0.100\ \text{mol}$ of $\text{Ni}(s)$ , assuming 100% current efficiency? (Use $F = 96{,}500\ \text{C mol}^{-1}\ e^-$ .)

4,825 C
9,650 C
19,300 C (correct answer)
965 C
38,600 C

Explanation: This problem uses electrolysis and Faraday's Law to find the charge needed to deposit a specific amount of nickel. The half-reaction Ni²⁺ + 2e⁻ → Ni shows that 2 moles of electrons are required per mole of nickel. For 0.100 mol Ni, we need: moles e⁻ = (0.100 mol Ni)(2 mol e⁻/mol Ni) = 0.200 mol e⁻. Converting to charge: Q = (0.200 mol e⁻)(96,500 C/mol e⁻) = 19,300 C. A common mistake is forgetting the 2:1 electron-to-nickel ratio, using only 0.100 mol e⁻, which would give 9,650 C. Always use the half-reaction stoichiometry to determine the moles of electrons before calculating charge.

Question 14

Molten calcium chloride is electrolyzed to produce calcium metal at the cathode according to $\text{Ca}^{2+}+2e^-\rightarrow \text{Ca}(s)$ . If a current of $19.3\ \text{A}$ is applied for $500\ \text{s}$ , assuming 100% current efficiency, how many moles of $\text{Ca}(s)$ are produced?

0.100 mol
0.0100 mol
0.200 mol
0.0250 mol
0.0500 mol (correct answer)

Explanation: This question tests electrolysis and Faraday's Law for calcium 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 Ca²⁺ + 2e⁻ → Ca shows that 2 moles of electrons produce 1 mole of Ca. Therefore, moles of Ca = (0.100 mol e⁻) × (1 mol Ca/2 mol e⁻) = 0.0500 mol. A typical error is forgetting the 2:1 electron stoichiometry and calculating 0.100 mol Ca instead. Always use the balanced half-reaction to establish the electron-to-product ratio before performing calculations.

Question 15

In an electrolytic cell, oxygen gas is produced at the anode from water according to $2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$ . If a total charge of $3.86\times 10^4\ \text{C}$ passes through the cell, how many moles of $\text{O}_2(g)$ are produced? (Assume 100% current efficiency.)

0.400 mol
0.100 mol (correct answer)
0.200 mol
0.0500 mol
0.0250 mol

Explanation: This problem involves electrolysis and Faraday's Law. Given Q, moles of electrons = Q / F, F ≈ 96500 C/mol. The half-reaction 2H₂O → O₂ + 4H⁺ + 4e⁻ requires 4 moles of electrons for 1 mole of O₂, so moles of O₂ = (Q / F) / 4. This stoichiometry scales the product. A tempting distractor is 0.400 mol, from using one mole of electrons instead of four per mole of product, quadrupling the moles. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.

Question 16

In the electrolysis of molten calcium chloride, calcium metal is produced at the cathode according to $\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}(s)$ . If a total charge of $4.83\times 10^4\ \text{C}$ passes through the cell, what mass of $\text{Ca}(s)$ is produced? (Assume 100% current efficiency.)

20.0 g
10.0 g (correct answer)
40.0 g
5.00 g
2.50 g

Explanation: This problem involves electrolysis and Faraday's Law. Given Q, moles of electrons = Q / F, F ≈ 96500 C/mol. For Ca²⁺ + 2e⁻ → Ca(s), z=2, mass = [(Q / F) / 2] × molar mass. This relates charge to substance via electrons. A tempting distractor is 20.0 g, from using one mole of electrons instead of two per mole of product, doubling the mass. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.

Question 17

In an electrolytic cell, nickel is deposited on a metal object from $\text{Ni}^{2+}(aq)$ according to $\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}(s)$ . What total charge (in coulombs) is required to deposit $0.100\ \text{mol}$ of $\text{Ni}(s)$ ? (Assume 100% current efficiency.)

9.65×10^3 C
4.83×10^4 C
1.93×10^4 C (correct answer)
9.65×10^4 C
1.93×10^3 C

Explanation: This problem involves electrolysis and Faraday's Law. To find Q for a given moles of substance, use moles of electrons = moles of substance × z, where z is electrons per unit, then Q = (moles of e⁻) × F, F ≈ 96500 C/mol. For Ni²⁺ + 2e⁻ → Ni(s), z=2 for 1 mole Ni. This relates substance to charge via electron count. A tempting distractor is 9.65×10^3 C, from using one mole of electrons instead of two per mole of product, halving the charge. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.

Question 18

A solution containing $\text{Fe}^{3+}(aq)$ is electrolyzed, and iron metal is produced at the cathode according to $\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}(s)$ . If a total charge of $2.895\times 10^4\ \text{C}$ passes through the cell, how many moles of $\text{Fe}(s)$ are produced? (Assume 100% current efficiency.)

0.300 mol
0.100 mol (correct answer)
0.200 mol
0.0500 mol
0.0100 mol

Explanation: This problem involves electrolysis and Faraday's Law. Given Q, moles of electrons = Q / F, F ≈ 96500 C/mol. For Fe³⁺ + 3e⁻ → Fe(s), z=3, so moles of Fe = (Q / F) / 3. Stoichiometry sets the ratio. A tempting distractor is 0.300 mol, from using one mole of electrons instead of three per mole of product, tripling the moles. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.

Question 19

An aqueous solution containing $\text{Zn}^{2+}(aq)$ is electrolyzed so that zinc metal forms at the cathode according to $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s)$ . If a constant current of $1.50\ \text{A}$ is applied for $40.0\ \text{min}$ , what mass of $\text{Zn}(s)$ is produced? (Assume 100% current efficiency.)

1.22 g (correct answer)
2.44 g
0.611 g
0.305 g
12.2 g

Explanation: This problem involves electrolysis and Faraday's Law. Charge Q = I × t gives moles of electrons = Q / F, F ≈ 96500 C/mol. For Zn²⁺ + 2e⁻ → Zn(s), 2 moles of electrons yield 1 mole of Zn, so mass = [(Q / F) / 2] × molar mass. Electron stoichiometry determines the scaling. A tempting distractor is 2.44 g, from using one mole of electrons instead of two per mole of product, doubling the mass. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.

Question 20

A solution containing $\text{Co}^{3+}(aq)$ is electrolyzed, producing cobalt metal at the cathode according to $\text{Co}^{3+} + 3e^- \rightarrow \text{Co}(s)$ . If a constant current of $3.00\ \text{A}$ is applied for $10.0\ \text{min}$ , how many moles of $\text{Co}(s)$ are produced? (Assume 100% current efficiency.)

0.00311 mol
0.0622 mol
0.00622 mol (correct answer)
0.0187 mol
0.00933 mol

Explanation: This problem involves electrolysis and Faraday's Law. Q = I × t gives moles of electrons = Q / F, F ≈ 96500 C/mol. For Co³⁺ + 3e⁻ → Co(s), z=3, moles of Co = (Q / F) / 3. Stoichiometry determines the amount. A tempting distractor is 0.0187 mol, from using one mole of electrons instead of three per mole of product, tripling the moles. Always use the half-reaction to count the number of electrons transferred per mole of substance before applying Faraday's Law.