AP Chemistry: Electrochemistry (Galvanic/Voltaic Cells, E°, Nernst Equation)

AP Chemistry – Electrochemistry (Galvanic/Voltaic Cells, E°, Nernst Equation)

AP Chemistry: Electrochemistry Study Guide

(Galvanic/Voltaic Cells, Standard Reduction Potentials, Nernst Equation)

What This Is

Electrochemistry is the study of how chemical reactions produce or consume electrical energy. On the AP exam, you’ll need to predict cell voltages, balance redox reactions, and calculate non-standard cell potentials using the Nernst equation. Real-world example: The first battery (Voltaic pile, 1800) was made by stacking zinc and copper discs separated by saltwater-soaked cloth—this is the basis for modern galvanic cells, like the AA batteries powering your remote.

Key Terms & Concepts

• Galvanic (Voltaic) Cell: A spontaneous electrochemical cell that converts chemical energy into electrical energy. Key components: Anode (oxidation), cathode (reduction), salt bridge (ion flow), and external circuit (electron flow).
• Standard Reduction Potential (E°): The voltage associated with a half-reaction under standard conditions (1 M, 1 atm, 25°C). Example: Cu²? + 2e?-Cu(s) has E° = +0.34 V.
• Cell Potential (E°cell): E°cell = E°cathode – E°anode. Positive E°cell = spontaneous reaction.
• Oxidation vs. Reduction:
• Oxidation: Loss of electrons (anode). Example: Zn(s)-Zn²? + 2e?.
• Reduction: Gain of electrons (cathode). Example: Cu²? + 2e?-Cu(s).
• Salt Bridge: Allows ion flow to maintain charge balance in the cell. Typically contains KNO? or Na?SO?.
• Nernst Equation: Calculates cell potential under non-standard conditions: E = E° – (RT/nF) ln(Q) or E = E° – (0.0592/n) log(Q) at 25°C.
• Variables: E = cell potential, E° = standard cell potential, R = gas constant (8.314 J/mol·K), T = temperature (K), n = moles of electrons, F = Faraday’s constant (96,485 C/mol), Q = reaction quotient.
• Faraday’s Constant (F): Charge of 1 mole of electrons (96,485 C/mol).
• Gibbs Free Energy (?G°): ?G° = –nFE°. Negative ?G° = spontaneous reaction.
• Line Notation: Shorthand for galvanic cells. Example: Zn(s) | Zn²?(1 M) || Cu²?(1 M) | Cu(s).
• Rules: Anode on left, cathode on right; single line = phase boundary; double line = salt bridge.
• Concentration Cells: A galvanic cell where both half-cells have the same species but different concentrations. E°cell = 0, but Ecell-0 due to Q-1.

Step-by-Step: Solving Electrochemistry Problems

1. Predicting Spontaneity & Calculating E°cell

Problem: Will Ag? oxidize Cu(s) to Cu²? under standard conditions? Steps:
1. Write half-reactions and E° values: - Ag? + e?-Ag(s) E° = +0.80 V - Cu²? + 2e?-Cu(s) E° = +0.34 V
2. Identify anode (oxidation) and cathode (reduction): - Cu(s) must be oxidized (anode), so reverse its half-reaction: Cu(s)-Cu²? + 2e? (E° = –0.34 V). - Ag? is reduced (cathode).
3. Balance electrons and add E° values: - Multiply Ag? half-reaction by 2 to balance electrons: 2Ag? + 2e?-2Ag(s). - E°cell = E°cathode – E°anode = 0.80 V – 0.34 V = +0.46 V.
4. Conclusion: E°cell > 0, so the reaction is spontaneous.

2. Using the Nernst Equation

Problem: Calculate the cell potential for Zn(s) | Zn²?(0.10 M) || Cu²?(1.0 M) | Cu(s) at 25°C. Steps:
1. Write the balanced reaction and E°cell: - Zn(s) + Cu²?-Zn²? + Cu(s) - E°cell = E°Cu – E°Zn = 0.34 V – (–0.76 V) = 1.10 V.
2. Determine Q (reaction quotient): - Q = [Zn²?]/[Cu²?] = 0.10 / 1.0 = 0.10.
3. Plug into Nernst equation (simplified for 25°C): - E = E° – (0.0592/n) log(Q) - n = 2 (electrons transferred). - E = 1.10 V – (0.0592/2) log(0.10) = 1.10 V – (0.0296)(–1) = 1.13 V.

3. Drawing a Galvanic Cell

Problem: Sketch the cell for Mg(s) | Mg²?(1 M) || Ag?(1 M) | Ag(s). Steps:
1. Label anode (oxidation) and cathode (reduction): - Anode: Mg(s)-Mg²? + 2e?. - Cathode: Ag? + e?-Ag(s).
2. Draw two half-cells: - Left: Mg(s) in Mg²? solution. - Right: Ag(s) in Ag? solution.
3. Add salt bridge (e.g., KNO?) and external wire: - Salt bridge connects the two solutions. - Wire connects Mg(s) to Ag(s) with a voltmeter.
4. Indicate electron flow: - Electrons flow from Mg (anode) to Ag (cathode).

Common Mistakes

• Mistake: Forgetting to reverse the anode’s E° value when calculating E°cell. Correction: E°cell = E°cathode – E°anode. The anode’s half-reaction is reversed, so its E° sign flips.

• Mistake: Using the wrong sign for ?G° in ?G° = –nFE°. Correction: Negative ?G° = spontaneous. If E°cell is positive, ?G° is negative.

• Mistake: Misapplying the Nernst equation (e.g., using ln instead of log or forgetting to divide by n). Correction: At 25°C, use E = E° – (0.0592/n) log(Q). Check units: log(Q), not ln(Q).

• Mistake: Confusing Q for concentration cells (e.g., using [anode]/[cathode] instead of [dilute]/[concentrated]). Correction: For concentration cells, Q = [dilute]/[concentrated]. Electrons flow from low to high concentration.

• Mistake: Ignoring coefficients when balancing half-reactions for E°cell. Correction: Multiply half-reactions to balance electrons, but do not multiply E° values. E° is intensive (doesn’t depend on amount).

AP Exam Insights

• Tricky Distinction: E°cell vs. Ecell
• E°cell = standard conditions (1 M, 1 atm).

• Ecell = non-standard conditions (use Nernst equation).

• FRQ Favorite: Designing a galvanic cell (line notation, labeling anode/cathode, calculating E°cell).

• Example: "Write the line notation for a cell with Al(s) and Ni²?(aq). Calculate E°cell."

• Multiple-Choice Trap: Sign of E° for non-spontaneous reactions.

• If E°cell < 0, the reaction is non-spontaneous (but the reverse reaction is spontaneous).

• Lab-Based Question: Predicting voltage changes when concentrations are altered.

• Example: "If [Cu²?] decreases, what happens to Ecell?" (Answer: Decreases, because Q increases.)

Quick Check Questions

1. Multiple Choice

A galvanic cell is constructed with the following half-reactions: - Pb²? + 2e?-Pb(s) E° = –0.13 V - Fe³? + e?-Fe²? E° = +0.77 V

What is the standard cell potential (E°cell) for this cell? (A) –0.90 V (B) +0.64 V (C) +0.90 V (D) +1.54 V

Answer: (C) +0.90 V. Explanation: E°cell = E°cathode – E°anode = 0.77 V – (–0.13 V) = +0.90 V.

2. Short FRQ

A concentration cell is made with Ag?(0.010 M) in one half-cell and Ag?(1.0 M) in the other at 25°C. (a) Write the line notation for this cell. (b) Calculate the cell potential (Ecell).

Answer: (a) Ag(s) | Ag?(0.010 M) || Ag?(1.0 M) | Ag(s). (b) Ecell = 0 – (0.0592/1) log(0.010/1.0) = 0.118 V. Explanation: For concentration cells, E°cell = 0, and Q = [dilute]/[concentrated].

Last-Minute Cram Sheet

1. Galvanic cell: Spontaneous (E°cell > 0), anode = oxidation, cathode = reduction.
2. E°cell = E°cathode – E°anode ( Don’t multiply E° values when balancing electrons!).
3. ?G° = –nFE° ( Negative ?G° = spontaneous).
4. Nernst equation: E = E° – (0.0592/n) log(Q) at 25°C.
5. Q for cells: [products]/[reactants], but for concentration cells, Q = [dilute]/[concentrated].
6. Salt bridge: Maintains charge balance ( Not part of the external circuit!).
7. Line notation: Anode | anode solution || cathode solution | cathode.
8. Faraday’s constant (F): 96,485 C/mol (charge of 1 mole of electrons).
9. Concentration cells: E°cell = 0, but Ecell-0 if [ion] differs.
10. Common trap: Forgetting to reverse the anode’s E° sign when calculating E°cell.