#🎉 Congratulations on Reaching the Final Stretch! 🎉

You've conquered so much in AP Chemistry, from acids and bases to equilibrium and thermodynamics. Now, let's tackle the last topic: nonspontaneous redox reactions and electrolytic cells. This is where we use external power to make the impossible happen!

#Review of Electrolytic Cells 🔋

Electrolytic cells use an external power source (like a battery) to drive **nonspontaneous redox reactions**. This is the opposite of galvanic cells, which generate electricity from spontaneous reactions. Think of it like pushing a ball uphill—it requires energy input.

Let's look at a classic example:

Caption: In this electrolytic cell, electrons are forced from Cu to Zn, causing Cu to oxidize and Zn to reduce, which is the reverse of the spontaneous reaction.

Normally, copper wouldn't oxidize to form Cu²⁺, and Zn²⁺ wouldn't reduce to form Zn metal. However, by applying external voltage, we can make this happen! The reaction is: **Cu + Zn²⁺ → Cu²⁺ + Zn**

Calculating Cell Potential (E_cell):

Half-reactions:

Cu → Cu²⁺ + 2e^- (E = -0.34 V)

Zn²⁺ + 2e^- → Zn (E = -0.76 V)

E_cell = -0.34 V + (-0.76 V) = -1.10 V

The negative E_cell confirms that this reaction is nonspontaneous, and we need a battery with voltage greater than 1.10V to drive it.

In this setup, the mass of the copper electrode will decrease, and the mass of the zinc electrode will increase, as copper is oxidized and zinc is reduced.

#Comparing Electrolytic and Galvanic Cells

Let's compare galvanic (voltaic) and electrolytic cells side-by-side:

Caption: Notice the key differences: Galvanic cells generate electricity spontaneously, while electrolytic cells require an external power source.

Galvanic Cell (Left):

• Anode: Oxidation occurs (e.g., Cd → Cd²⁺ + 2e^-)
• Cathode: Reduction occurs (e.g., Cu²⁺ + 2e^- → Cu)
• Electrons flow from anode to cathode through a wire
• Voltmeter measures the electromotive force (voltage)
• Salt bridge maintains neutrality

Galvanic Cell Example

Caption: In this galvanic cell, the copper electrode grows, and the zinc electrode shrinks as the reaction proceeds spontaneously.

Electrolytic Cell (Right):

• Anode: Oxidation occurs (e.g., Cu → Cu²⁺ + 2e^-)
• Cathode: Reduction occurs (e.g., Cd²⁺ + 2e^- → Cd)
• Electrons are *forced* to flow from anode to cathode by a power supply
• Power supply provides the required voltage
• Salt bridge maintains neutrality

Key Similarity: In *both* cell types, **oxidation always occurs at the anode and reduction always occurs at the cathode**. This is a fundamental rule!

Here’s a table summarizing the key differences:

#Faraday’s Law and Electrolysis Problems ⚡

Faraday's Law allows us to calculate the mass of a substance deposited or consumed during electrolysis. The key is to use a chain of conversions:

Key Relationships:

• 1 Amp (A) = 1 Coulomb/second (C/s)
• Faraday's Constant: 1 mole of electrons = 96485 Coulombs

Let’s tackle an example:

Problem: Determine the mass of chromium produced when a solution of Cr(NO₃)₂ is electrolyzed for 60 minutes with a current of 15 amps.

Solution:

1. Convert time to seconds and calculate total charge:

60 min * 60 s/min * 15 C/s = 54000 Coulombs

2. Convert Coulombs to moles of electrons using Faraday's constant:

54000 C * (1 mol e^- / 96485 C) = 0.559 mol e^-

3. Use stoichiometry to convert from mol e^- to moles of chromium:

0.559 mol e^- * (1 mol Cr / 2 mol e^-) = 0.2795 mol Cr

4. Convert moles of Cr to grams using its molar mass (52.0 g/mol):

0.2795 mol Cr * (52 g/mol) = **14.55 g Cr**

We can summarize this in one long chain:

Important Note: You can manipulate this chain to solve for any unknown: mass, time, current, or charge. For example, if you are given mass and need to find time, start with mass and work backward.

Unit Cancellation is Key!

#Final Exam Focus 🎯

Here’s what to focus on for the exam:

• Electrolytic vs. Galvanic Cells: Know the differences in setup, spontaneity, and electron flow.
• Anode and Cathode: Remember that oxidation *always* occurs at the anode, and reduction *always* occurs at the cathode.
• Faraday's Law: Practice converting between charge, moles of electrons, and mass of substances.
• Dimensional Analysis: Master unit cancellation to avoid errors in calculations.

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Practice Question

Practice Questions

Multiple Choice Questions

1. In an electrolytic cell, which of the following is true regarding the flow of electrons? (A) Electrons flow from the cathode to the anode through the external circuit. (B) Electrons flow from the anode to the cathode through the external circuit. (C) Electrons flow from the cathode to the anode through the salt bridge. (D) Electrons flow from the anode to the cathode through the salt bridge.

2. A solution of AgNO₃ is electrolyzed. What mass of silver will be deposited at the cathode if a current of 2.00 A is passed through the solution for 30.0 minutes? (A) 0.00112 g (B) 0.00224 g (C) 4.02 g (D) 10.0 g

3. Which of the following statements is true regarding the standard cell potential, E°, for a galvanic cell and an electrolytic cell? (A) E° is always positive for both galvanic and electrolytic cells. (B) E° is always negative for both galvanic and electrolytic cells. (C) E° is positive for a galvanic cell and negative for an electrolytic cell. (D) E° is negative for a galvanic cell and positive for an electrolytic cell.

Free Response Question

A voltaic cell is constructed using a standard hydrogen electrode (SHE) and a standard copper electrode. The cell operates at 298 K. The overall reaction is:

Cu²⁺(aq) + H₂(g) ⇌ 2H⁺(aq) + Cu(s)

(a) Write the half-reaction that occurs at the anode.

(b) Write the half-reaction that occurs at the cathode.

(c) Calculate the standard cell potential, E°, for the cell.

(d) Calculate the standard free energy change, ΔG°, for the reaction.

(e) If the cell is set up as an electrolytic cell, what is the minimum voltage required to drive the reaction in the reverse direction? Explain.

Scoring Guide

(a) H₂(g) → 2H⁺(aq) + 2e^- (1 point)

(b) Cu²⁺(aq) + 2e^- → Cu(s) (1 point)

(c) E°_cell = E°_cathode - E°_anode = 0.34 V - 0.00 V = 0.34 V (1 point)

(d) ΔG° = -nFE° = -(2 mol)(96485 C/mol)(0.34 V) = -65610 J = -65.6 kJ (2 points)

(e) The minimum voltage is 0.34 V. In an electrolytic cell, the external voltage must overcome the non-spontaneity of the reverse reaction, which is equal to the absolute value of the standard cell potential. (2 points)