AP Chemistry 9.11 Electrolysis and Faraday’s Law Study Notes - New Syllabus Effective fall 2024
AP Chemistry 9.11 Electrolysis and Faraday’s Law Study Notes- New syllabus
AP Chemistry 9.11 Electrolysis and Faraday’s Law Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.
LEARNING OBJECTIVE
Calculate the amount of chargeflowbasedon changes in the amounts of reactants and products in an electrochemical cell.
Key Concepts:
- Quantitative Relationships in Electrochemical Cells (Faraday’s Laws)
Quantitative Relationships in Electrochemical Cells (Faraday’s Laws)
Faraday’s Laws of Electrolysis describe the quantitative relationship between the amount of electric charge passed through an electrochemical cell and the amount of chemical change that occurs at the electrodes.
These relationships connect the flow of electric charge (current) with the stoichiometry of the redox process — allowing us to calculate quantities such as the mass of a substance deposited, moles of electrons transferred, or time required for a given reaction.
Key Variables and Units
- \( \mathrm{I} \): current (amperes, A = C/s)
- \( \mathrm{q} \): charge (coulombs, C)
- \( \mathrm{t} \): time (seconds, s)
- \( \mathrm{F} \): Faraday’s constant = \( \mathrm{96,485\ C/mol\ e^-} \)
- \( \mathrm{n} \): moles of electrons transferred per mole of reactant
Fundamental Relationships
1. \( \mathrm{q = It} \)
2. \( \mathrm{q = n_e F} \)
Where:
- \( \mathrm{q} \): total charge passed (Coulombs)
- \( \mathrm{n_e} \): moles of electrons transferred
- \( \mathrm{F} \): Faraday constant
Combined Expression:
\( \mathrm{It = n_e F} \)
Relating Charge to Mass Change at Electrodes
From stoichiometry, the mass of metal deposited or oxidized at an electrode is proportional to the number of moles of electrons passed.
\( \mathrm{m = \dfrac{M I t}{n F}} \)
- \( \mathrm{m} \): mass of substance (g)
- \( \mathrm{M} \): molar mass (g/mol)
- \( \mathrm{I} \): current (A)
- \( \mathrm{t} \): time (s)
- \( \mathrm{n} \): number of electrons per ion (charge of the ion)
Applications of Faraday’s Laws
- Electroplating (mass of metal deposited)
- Electrolysis (amount of gas evolved or ion reduced)
- Battery discharge and charge capacity
- Determining total charge transferred during redox reactions
Key Idea: Faraday’s laws link electrical quantities (I, t, q) with chemical change in redox processes:
- \( \mathrm{q = It} \) connects current and time to total charge.
- \( \mathrm{q = n_e F} \) relates charge to moles of electrons.
- \( \mathrm{m = \dfrac{MIt}{nF}} \) connects charge to the mass of material deposited or oxidized.
These relationships are essential in analyzing batteries, electroplating, and electrolysis quantitatively.
Example :
A current of 2.50 A passes through an electrolytic cell for 45.0 minutes. Calculate (a) the total charge transferred and (b) the number of moles of electrons.
▶️ Answer / Explanation
Step 1: Convert time to seconds.
\( \mathrm{t = 45.0\ min \times 60 = 2700\ s} \)
Step 2: Calculate total charge using \( \mathrm{q = It} \).
\( \mathrm{q = (2.50)(2700) = 6750\ C} \)
Step 3: Calculate moles of electrons using \( \mathrm{q = n_e F} \).
\( \mathrm{n_e = \dfrac{q}{F} = \dfrac{6750}{96485} = 0.0700\ mol\ e^-} \)
Final Answer:
- Total charge = \( \mathrm{6750\ C} \)
- Moles of electrons = \( \mathrm{0.0700\ mol\ e^-} \)
Example :
How many grams of copper will be deposited on the cathode when a 3.00 A current runs through a \( \mathrm{CuSO_4(aq)} \) solution for 2.00 hours? (Given: \( \mathrm{Cu^{2+} + 2e^- \rightarrow Cu(s)} \), \( \mathrm{M_{Cu} = 63.55\ g/mol} \))
▶️ Answer / Explanation
Step 1: Calculate total charge.
\( \mathrm{t = 2.00\ h \times 3600 = 7200\ s} \)
\( \mathrm{q = It = (3.00)(7200) = 21,600\ C} \)
Step 2: Calculate moles of electrons.
\( \mathrm{n_e = \dfrac{q}{F} = \dfrac{21600}{96485} = 0.224\ mol\ e^-} \)
Step 3: Use reaction stoichiometry (\( \mathrm{2\ mol\ e^- \rightarrow 1\ mol\ Cu} \)).
\( \mathrm{n_{Cu} = \dfrac{0.224}{2} = 0.112\ mol\ Cu} \)
Step 4: Calculate mass of copper deposited.
\( \mathrm{m = nM = (0.112)(63.55) = 7.12\ g} \)
Final Answer: \( \mathrm{7.12\ g} \) of copper will be deposited on the cathode after 2.00 hours.