#Electrochemistry: Your Ultimate Review ⚡

Hey there! Let's dive into electrochemistry, where we explore how redox reactions create electrical energy. This is a crucial area, so let's make sure you're feeling confident!

Electrochemistry is a high-value topic, often accounting for a significant portion of both multiple-choice and free-response questions. Mastering this unit can greatly boost your score.

# Review of Redox Reactions

#What are Redox Reactions?

• Redox reactions (oxidation-reduction reactions) involve the transfer of electrons. 🔄
• A reducing agent loses electrons and is oxidized. ⬆️
• An oxidizing agent gains electrons and is reduced. ⬇️

#Example:

Consider the reaction: 2AgNO₃ + Cu → Cu(NO₃)₂ + 2Ag

• Copper (Cu) is oxidized (0 → +2 oxidation state).
• Silver (Ag⁺) is reduced (+1 → 0 oxidation state).
• Electrons move from copper to silver.

Image From SciencePhoto

#Half-Reactions:

• Oxidation: Cu → Cu²⁺ + 2e⁻
• Reduction: 2Ag⁺ + 2e⁻ → 2Ag

# Reduction Potentials

#What are Reduction Potentials?

• Electromotive force (EMF) is the force that pushes electrons in a redox reaction. It's measured in volts (V). ⚡
• Standard reduction potentials are the voltages of reduction half-reactions. They're provided in a table (like this one).
• Negative voltage means energy is needed for the reaction to occur.

#Calculating Cell Potential (E°)

Let's use the reaction: Zn(s) + Pb²⁺(aq) → Zn²⁺(aq) + Pb(s)

1. Half-Reactions:
   • Oxidation: Zn → Zn²⁺ + 2e⁻
   • Reduction: Pb²⁺ + 2e⁻ → Pb
2. Reduction Potentials (from the table):
   • Pb²⁺ reduction: -0.13 V
   • Zn²⁺ reduction: -0.76 V
3. Flip the sign for the oxidation half-reaction (Zn → Zn²⁺ + 2e⁻): +0.76 V
4. Calculate E°:
   • E° = E°(reduction) + E°(oxidation) = -0.13 V + 0.76 V = +0.63 V

#Alternative Formula

You can also use: E°cell = E°(cathode) - E°(anode). This is the same as E° = E°red - E°ox. We will use this in the next section.

# Galvanic/Voltaic Cells

#How do they work?

• Galvanic (voltaic) cells use spontaneous redox reactions to generate electricity. 🔋
• Electrons travel through a wire from the anode (where oxidation occurs) to the cathode (where reduction occurs). ➡️

#Example:

Consider the reaction: Cu + 2Ag⁺ → 2Ag + Cu²⁺

Image From BCCampus

• Anode (oxidation): Copper (Cu) electrode shrinks as Cu → Cu²⁺ + 2e⁻.
• Cathode (reduction): Silver (Ag) electrode grows as Ag⁺ + e⁻ → Ag.
• Voltmeter measures the cell potential (E°cell), which is +0.46 V in this case.
• Salt bridge maintains charge neutrality by allowing ions to flow. It contains spectator ions (e.g., NaNO₃).

#Practice

Try to write out the half-reactions and calculate the cell potential for the reaction above. You should get E°cell = +0.46 V. Also, think about why we can use E°cell = E°cathode - E°anode.

# Electrolytic Cells 🔋

#What are Electrolytic Cells?

• Electrolytic cells use electrical energy to drive nonspontaneous redox reactions. ⚡
• E°cell is negative for these reactions.

#Example: Electrolysis of NaCl

Consider the reaction: 2NaCl → 2Na + Cl₂

1. Half-Reactions:
   • Reduction: 2Na⁺ + 2e⁻ → 2Na (E° = -2.71 V)
   • Oxidation: 2Cl⁻ → Cl₂ + 2e⁻ (E° = -1.36 V)
2. Calculate E°cell:
   • E°cell = E°cathode - E°anode = -2.71 V - (-1.36 V) = -1.35 V

Image From TropicSU

• A battery (or other power source) is needed to push electrons from anode to cathode.
• Inert electrodes are used to collect products (e.g., Cl₂ gas at the anode and Na metal at the cathode).
• The battery voltage must be greater than or equal to the absolute value of E°cell (1.35 V in this case).

# Learning Summary

#Key Takeaways

• Electrochemistry studies redox reactions and their relationship to electrical energy. ⚡
• Oxidation occurs at the anode, and reduction occurs at the cathode. ➡️
• Galvanic cells use spontaneous reactions (E°cell > 0) to generate electricity.
• Electrolytic cells use electrical energy to drive nonspontaneous reactions (E°cell < 0).
• Half-reactions and cell diagrams are essential tools for analysis.

# Final Exam Focus

#High-Priority Topics

• Redox reactions: Identifying oxidation and reduction, balancing equations.
• Cell potentials: Calculating E°cell using standard reduction potentials.
• Galvanic vs. Electrolytic cells: Understanding the differences and applications.
• Cell diagrams: Interpreting and drawing cell diagrams.

#Common Question Types

• Multiple Choice: Identifying oxidizing/reducing agents, calculating cell potentials, predicting spontaneity.
• Free Response: Designing electrochemical cells, writing half-reactions, explaining the role of the salt bridge, calculating cell potential and relating it to Gibbs Free Energy.

#Last-Minute Tips

• Time Management: Quickly identify the type of cell and the required calculations.
• Common Pitfalls: Double-check the signs of reduction potentials, balance electrons correctly, and remember that multiplying a half-reaction does not affect the potential.
• Strategies: Start with the basics, write out half-reactions, and use the formulas correctly. If you're stuck, try to identify the knowns and unknowns and work from there.

# Practice Questions

Practice Question

#Multiple Choice Questions

1. Which of the following is the strongest oxidizing agent? (A) Li⁺ (B) F⁻ (C) F₂ (D) Li

2. A voltaic cell is set up using the following reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) Given the standard reduction potentials: E°(Zn²⁺/Zn) = -0.76 V and E°(Cu²⁺/Cu) = +0.34 V. What is the standard cell potential (E°cell) for this reaction? (A) -1.10 V (B) -0.42 V (C) +0.42 V (D) +1.10 V

3. In an electrolytic cell, which of the following is true? (A) Oxidation occurs at the cathode. (B) Reduction occurs at the anode. (C) Electrons flow from the cathode to the anode. (D) An external power source is required.

#Free Response Question

A voltaic cell is constructed using a standard hydrogen electrode (SHE) and a silver electrode in a solution of silver nitrate. The overall reaction is:

2Ag⁺(aq) + H₂(g) → 2Ag(s) + 2H⁺(aq)

(a) Write the balanced half-reactions for the oxidation and reduction processes.

(b) Draw a diagram of the voltaic cell, labeling the anode, cathode, direction of electron flow, and the salt bridge.

(c) Calculate the standard cell potential (E°cell) for this reaction, given that the standard reduction potential for Ag⁺/Ag is +0.80 V and for H⁺/H₂ is 0.00 V.

(d) If the concentration of Ag⁺ is decreased, how will it affect the cell potential? Explain using Le Chatelier’s principle.

#FRQ Scoring Breakdown

(a) Half-Reactions (2 points)

• Oxidation: H₂(g) → 2H⁺(aq) + 2e⁻ (1 point)
• Reduction: 2Ag⁺(aq) + 2e⁻ → 2Ag(s) (1 point)

(b) Cell Diagram (3 points)

• Correctly labeled anode (H₂ electrode) (1 point)
• Correctly labeled cathode (Ag electrode) (1 point)
• Correctly labeled direction of electron flow (anode to cathode) and salt bridge (1 point)

(c) Cell Potential Calculation (2 points)

• E°cell = E°cathode - E°anode = +0.80 V - 0.00 V = +0.80 V (2 points)

(d) Effect of Concentration (2 points)

• Decreasing [Ag⁺] will decrease the cell potential (1 point)
• Explanation using Le Chatelier’s principle: Decreasing [Ag⁺] shifts the equilibrium to the left, reducing the driving force of the reaction and thus decreasing the cell potential (1 point)

Good luck! You've got this! 💪