Chemistry Chapter 3

Main Ideas

1. Electrochemistry and Redox Reactions

Electrochemistry studies the relationship between chemical change and electrical energy:

• Chemical reactions can produce electricity.
• Electricity can drive chemical reactions that would not occur on their own.

Redox reactions involve electrons moving between substances, accompanied by changes in oxidation numbers:

• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.
• Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain.

2. Galvanic Cells, Corrosion, and Cell Components

A galvanic cell converts chemical energy into electrical energy through a spontaneous redox reaction.

In the zinc–copper example:

• Zinc loses electrons and is oxidized at the anode.
• Copper ions gain electrons and are reduced at the cathode.
• Electrons travel through the external wire from zinc to copper, producing current.

The lecture uses the mnemonic AN OX, RED CAT:

• Anode = Oxidation.
• Reduction = Cathode.

A salt bridge completes the circuit and allows ions to migrate, helping maintain electrical neutrality.

Corrosion is an undesirable redox process. The lecture describes iron reacting in the presence of water, with parts of the iron acting as an anode and losing electrons as rust forms.

In cell notation, the two vertical lines mark the boundary between the half-cells. The lecture identifies the left side with oxidation and the right side with reduction.

3. Oxidation Numbers

The lecture reviews these rules for assigning oxidation numbers:

• An element in its uncombined form has an oxidation number of 0.
• Group 1 elements are usually +1; Group 2 elements are +2.
• Halogens are usually −1, though they can be positive when combined with oxygen.
• Hydrogen is usually +1 with nonmetals and −1 with metals.
• Oxygen is usually −2, or −1 in peroxides.
• Fluorine is −1.
• Oxidation numbers in a neutral compound sum to 0; in an ion, they sum to the ion’s charge.

For example, in dichromate, (\mathrm{Cr_2O_7^{2-}}), oxygen contributes (7(-2)=-14). The two chromium atoms must total (+12), so each chromium has an oxidation number of +6.

4. Balancing Redox Equations in Acidic Solution

The instructor presents the half-reaction method:

1. Assign oxidation numbers and identify which substance is oxidized and which is reduced.
2. Write separate oxidation and reduction half-reactions.
3. Balance atoms other than oxygen and hydrogen.
4. Balance oxygen by adding water.
5. Balance hydrogen by adding (\mathrm{H^+}).
6. Balance charge by adding electrons.
7. Multiply the half-reactions as needed so the electron counts match.
8. Add the half-reactions and cancel electrons and any other species appearing on both sides.

In the permanganate–iron example, permanganate is reduced to (\mathrm{Mn^{2+}}), while (\mathrm{Fe^{2+}}) is oxidized to (\mathrm{Fe^{3+}}). The balanced equation is:

[ \mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}} ]

Therefore, the coefficient of (\mathrm{Fe^{2+}}) is 5.

5. Cell Potential and Spontaneity

Cell potential, measured in volts, describes a cell’s ability to move electrons through a wire. Using standard reduction potentials:

[ E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}} ]

The cathode is the half-reaction with the more positive reduction potential; the other half-reaction is reversed to serve as the oxidation at the anode.

For example, the lecture compares chromium at (-0.73\ \mathrm{V}) and aluminum at (-1.66\ \mathrm{V}). Chromium is assigned as the cathode and aluminum as the anode:

[ E^\circ_{\text{cell}}=(-0.73)-(-1.66)=+0.93\ \mathrm{V} ]

A positive standard cell potential indicates a spontaneous cell reaction and is associated with a negative Gibbs free-energy change. A negative cell potential indicates a nonspontaneous reaction and a positive Gibbs free-energy change.

6. Gibbs Free Energy

The lecture uses:

[ \Delta G^\circ=-nFE^\circ_{\text{cell}} ]

Here, (n) is the number of electrons transferred and (F) is Faraday’s constant. In the example, (n=2), (F=96{,}485\ \mathrm{C/mol}), and (E^\circ=1.10\ \mathrm{V}), producing:

[ \Delta G^\circ\approx-212{,}267\ \mathrm{J} ]

The negative result is consistent with a spontaneous reaction.

7. Nonstandard Conditions and the Nernst Equation

For conditions other than standard conditions, the lecture uses the simplified Nernst equation:

[ E=E^\circ-\frac{0.0592}{n}\log Q ]

To find (Q):

• Write the balanced overall reaction.
• Use product-ion concentrations over reactant-ion concentrations, raised to their stoichiometric coefficients.
• Omit pure solids from (Q).

For the zinc–copper example, (Q=100), (n=2), and (E^\circ=1.10\ \mathrm{V}), giving (E\approx1.0408\ \mathrm{V}).

In a second comparison example, the calculated correction is positive, so the nonstandard potential is greater than the standard potential.

8. Electrolysis and Other Electrochemical Systems

Electrolysis uses an external electrical supply to drive a nonspontaneous chemical change. An electrolytic cell converts electrical energy into chemical change.

The lecture’s examples include separating molten sodium chloride into sodium and chlorine, and splitting water into hydrogen and oxygen. In molten sodium chloride electrolysis:

• Chloride is oxidized, losing electrons.
• Sodium ions are reduced, gaining electrons.
• The external power source drives the reaction.

A hydrogen fuel cell, like a galvanic cell, converts chemical energy into electrical energy. It must be continually supplied with reactants; electricity production stops when the reactants are depleted.

9. Faraday’s Law of Electrolysis

Faraday’s law states that the amount of substance produced or consumed is proportional to the charge passed through the cell. The lecture relates one mole of electrons to one Faraday, approximately (96{,}485\ \mathrm{C}).

For a deposition problem:

1. Convert the deposited mass to moles using molar mass.
2. Use the balanced electrode reaction to relate substance moles to electron moles.
3. Convert electron moles to charge using Faraday’s constant.

Numerical note: For the lecture’s stated 1 g silver example, using (107.87\ \mathrm{g/mol}) and one electron per silver ion gives about (894\ \mathrm{C}). The spoken result in the subtitles, (8{,}194.456\ \mathrm{C}), appears inconsistent with those stated values.

Speakers and Sources Featured

• Speaker: One instructor/lecturer; no other speakers are identifiable in the subtitles.
• Reference materials mentioned: The periodic table and a standard electrode-potential table.