Electrochemistry and Electrolytic Cells

Exam Information

• Final exam in two weeks.
• Last day for iClicker will be next Monday, followed by a review session on Wednesday.
• Homework due tonight (last Monday homework).
• Final homework assignment due next Wednesday (last day of lecture).
• Extra credit due this Wednesday (exam checkpoint survey).
• Another extra credit due next Wednesday (adaptive quiz for nuclear chemistry).
• Final exam study sheet posted (breakdown of questions, order of concepts, cover sheet format).
• Sample questions for new material posted (from old exams).

Exam Three Scores

• Scores released Friday afternoon.
• Median score: 80.
• Mean score: ~78 (two questions higher than last semester).

Faculty Course Questionnaires (FCQs)

• Open today for at least another week.
• Provide honest feedback.
• Important for tenure and promotion (especially for instructors).
• Nominations for faculty awards also make a difference to departments.

Nernst Equation and Concentration Cells

• Review of concentration cells.
• Concentration cells have the same materials on both sides (e.g., solution, metal).
• Electron flow is due to a concentration gradient.
• Electrons flow from more concentrated to more dilute until equilibrium is reached.
• Highest concentration is desired on the reactant side to have the concentration cell work.
• The concentration cells are in the sample material packet on canvas.

Concentration Cell Problem Example

• Copper two plus and solid copper on both sides.
• Need to solve for the concentration required to achieve a specific voltage (0.046 volts).
• The copper half reaction is the same on both sides.

Steps:

*   Keep one side as a reduction, reverse the other and consider it oxidize.
*   Assign concentrations: higher concentration (1 M) on the reactant side, unknown value (x or y) on the product side.
*   Want Q (reaction quotient) to be small (few products, concentrated reactants).
*   Nernst Equation: $E = E^\circ - \frac{RT}{nF} \ln{Q}$

    *   E = 0.046 V (given voltage).
    *   $E^\circ$ = 0 V (standard cell potential for concentration cell).
    *   R = 8.314 J/(mol*K) (ideal gas constant).
    *   T = 298 K (assumed temperature unless specified).
    *   n = 2 moles of electrons transferred.
    *   F = 96,485 C/mol (Faraday's constant).
    *   Solve for Q, where $Q = \frac{[products]}{[reactants]}$ which is products = x, reactants = 1.

Solution:

1. Evaluate $\frac{RT}{nF}$: $\frac{(8.314 \frac{J}{mol \cdot K}) \cdot (298 K)}{(2 mol) \cdot (96485 \frac{C}{mol})} = 0.012839$
2. Equation becomes: $0.046 = 0 - 0.012839 \ln{Q}$
3. Solve for $\ln{Q}$: $\frac{0.046}{-0.012839} = -3.5827 = \ln{Q}$
4. Solve for Q: $e^{-3.5827} = 0.0278 = Q$
5. Since $Q = \frac{x}{1}$, then x = 0.0278 M (anode concentration).
6. Electrons will flow in this case because reactants are high, and products are low in concentration. Concentrations will eventually settle out and equal each other, and the current will go to zero.
7. Shortcut form of the Nernst equation: $E = E^\circ - \frac{0.0592}{n} \log{Q}$

Concentration Cell Concept Test

• Question: How to increase voltage in a concentration cell?
• Answer: Increase the concentration gradient.
• Reasoning: Higher concentration gradient leads to greater electron flow from high to low concentration.
• Choosing a different half-reaction won't matter because a concentration cell has the same reactions on both sides.
• Selecting a half-reaction with more moles of electrons transferred also won't matter for the same reason.

Electrolytic Cells

• Opposite of spontaneous, voltaic, or galvanic cells.
• Non-spontaneous flow of electrons; requires an external power source.
• Reactants become higher energy.
• Spontaneous/Voltaic/Galvanic vs. Electrolytic:
  • E cell: Positive vs. Negative.
  • Delta G: Negative vs. Positive.
  • K: Big vs. Small.
• Electrolysis: Using electrical energy to drive a non-spontaneous chemical reaction.
• Anode is oxidation, cathode is reduction, definitions stay same.
• Electrons flow in the opposite direction compared to voltaic cells.

Faraday's Law

• Amount of substance produced is directly proportional to the quantity of charge flowing through the cell.
• Longer something is plugged in, the more charge builds up, the more reactants can be created.
• Electrolysis calculations are stoichiometric calculations.
• Conversion factors needed:
  • Faraday's constant: 96,485 coulombs per mole of electrons.
  • Current: Amps (A) = Coulombs per second (C/s) [Given on the cover sheet].
  • $1 A = 1 \frac{C}{s}$
• Cover sheet also defines a joule. E cell can be given, current or potential.

Electrolysis Problem Example: Copper Deposition

• Calculate the mass of copper metal deposited by passing a 10 amp current through a solution of copper(II) sulfate for 3 hours.

Steps:

1. Reaction: $Cu^{2+} + 2e^- \rightarrow Cu(s)$
2. Convert amps to coulombs per second: 10 amps = 10 C/s.
3. Start with the simplest unit: 3 hours.
4. Convert hours to seconds: $3 hours \cdot \frac{60 minutes}{1 hour} \cdot \frac{60 seconds}{1 minute} = 10800 seconds$
5. Multiply by current: $10800 seconds \cdot \frac{10 C}{1 second} = 108000 C$
6. Use Faraday's constant: $108000 C \cdot \frac{1 mol e^-}{96485 C} = 1.119 mol e^-$
7. Use stoichiometry: For every one mole of solid copper, there are two moles of electrons transferred, $1.119 mol e^- \cdot \frac{1 mol Cu}{2 mol e^-} = 0.5595 mol Cu$
8. Convert moles to grams: $0.5595 mol Cu \cdot \frac{63.5 g Cu}{1 mol Cu} = 35.6 g Cu$

• Result: 35.6 grams of solid copper will be deposited.

Electrolysis Clicker Question: Chromium Deposition

• Simplified version of the previous example.
• Given: Coulombs of charge, find mass.
• Hint of the Chromium three plus ion from chromium nitrate.
• Reduction Equation: $Cr^{3+} + 3e^- \rightarrow Cr(s)$
• 3 moles of electrons transferred.
• Convert coulombs to grams.

Electrolysis of Water

• Reactions:
  • Reduction (Cathode): $2H<em>2O(l) + 2e^- \rightarrow H</em>2(g) + 2OH^-(aq)$
  • Oxidation (Anode): $2H<em>2O(l) \rightarrow O</em>2(g) + 4H^+(aq) + 4e^-$
• Electrolysis of water would need to be scaled up.
• Standard cell potential is negative. $E^\circ$ = Negative.
• Costs more energy to make this happen than would get back from hydrogen fuel.

Electrolysis Problem: Hydrogen Gas Production

• Calculate the time required to produce 2.5 grams of hydrogen gas by the electrolysis of water. The current is 5 amps.
• The electrolye is not important.

Steps:

1. H redox Equation: 2 electrons for every 1 mol of $H_2$ gas for every 2 moles of water that we need.
2. Start with 2.5 grams of hydrogen gas.
3. $2.5 g \cdot \frac{1 mol \; H<em>2}{2.02 g} \cdot \frac{2 mol e^-}{1 mol H</em>2} \cdot \frac{96485 C}{1 mol e^-} \cdot \frac{1 second}{5 C}$
4. Result: Enormous amount of seconds can be converted into minutes or hours.

Electrolysis Practice Problem

• I like to solve for two things separately in these types of problems.
  • Like to go from mass to coulombs.
  • Solve for time in seconds then divide the two values separately.
• You will be solving for Current. For me, this is most challenging.
• Answers are in the slides.
• The most the test will expect you to do is go from a charged metal to a neutral metal and count the electrons.

Oscar Statues

• Made using electrolysis.
• An iron statue is placed inside a solution of gold cyanide.
• Electricity is pumped through the solution.
• The gold deposits on the iron (the electrode).