Cell Potential Under Nonstandard Conditions; Electrolysis and Faraday's Law (9.10, 9.11)

Introduction to Nonstandard Cells and Electrolysis

Welcome to the final stretch of Unit 9! So far, you have learned how to calculate the standard cell potential (\(E^\circ_{cell}\)) when everything is "perfect"—concentrations are exactly \(1.0 \text{ M}\) and gases are at \(1.0 \text{ atm}\). But in the real world (and in your phone's battery), concentrations change as the reaction proceeds. In this chapter, we will explore what happens to the voltage when conditions aren't standard and how we can use electricity to force non-spontaneous reactions to happen through electrolysis. Don't worry if this seems like a lot of math at first; we will break it down into simple, logical steps!

9.10: Cell Potential Under Nonstandard Conditions

In previous chapters (like 9.8 and 9.9), we focused on the standard cell potential (\(E^\circ_{cell}\)). However, as a battery runs, the reactants are consumed and products are formed. This changes the concentrations, which in turn changes the voltage the battery can provide.

1. The Qualitative Approach (Le Chatelier’s Principle)

You can often predict how the cell potential (\(E_{cell}\)) will change without doing heavy math by using Le Chatelier's Principle. Think of the cell potential as the "driving force" of the reaction.

• If you increase the concentration of a reactant (or decrease a product), the "driving force" increases, and \(E_{cell} > E^\circ_{cell}\).
• If you increase the concentration of a product (or decrease a reactant), the reaction is closer to equilibrium, the "driving force" decreases, and \(E_{cell} < E^\circ_{cell}\).
• When the system reaches equilibrium, the battery is "dead." At this point, \(Q = K\) and \(E_{cell} = 0\).

2. The Quantitative Approach (The Nernst Equation)

To calculate the exact voltage under nonstandard conditions, we use the Nernst Equation, which is provided on your AP Equation Sheet:

\(E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q\)

Where:

• \(E_{cell}\) = Potential under nonstandard conditions (V)
• \(E^\circ_{cell}\) = Standard cell potential (V)
• \(R\) = Gas constant (\(8.314 \text{ J/mol} \cdot \text{K}\))
• \(T\) = Temperature in Kelvin (usually \(298 \text{ K}\))
• \(n\) = Number of moles of electrons transferred in the balanced redox equation
• \(F\) = Faraday’s constant (\(96,485 \text{ C/mol } e^-\))
• \(Q\) = Reaction quotient (calculated using \(\frac{[\text{products}]}{[\text{reactants}]}\))

Analogy: Think of \(E^\circ_{cell}\) as the "starting price" of an item. The \(\frac{RT}{nF} \ln Q\) part is like a discount or a surcharge based on how much "stuff" you currently have in your reaction container.

3. Concentration Cells

A concentration cell is a special type of galvanic cell where both the anode and cathode compartments contain the same substances, but at different concentrations.
Example: A cell with \(Cu(s)\) electrodes in both sides, but one side is \(0.1 \text{ M } Cu^{2+}\) and the other is \(1.0 \text{ M } Cu^{2+}\).
Because the substances are the same, \(E^\circ_{cell}\) is always \(0\). The only reason a voltage exists is because the system wants to reach equilibrium by equalizing the concentrations!

Quick Review:
- If \(Q < 1\), then \(E_{cell} > E^\circ_{cell}\) (Reactant-heavy)
- If \(Q > 1\), then \(E_{cell} < E^\circ_{cell}\) (Product-heavy)
- If \(Q = 1\), then \(E_{cell} = E^\circ_{cell}\) (Standard conditions)
- If \(Q = K\), then \(E_{cell} = 0\) (Dead battery/Equilibrium)

9.11: Electrolysis and Faraday's Law

While galvanic cells produce electricity, electrolytic cells use electricity to force a non-spontaneous reaction (\(\Delta G > 0\)) to occur. This process is called electrolysis.

1. Understanding Current and Charge

To do calculations for electrolysis, you need to understand how we measure the "amount" of electricity moving through a wire. Two equations on your reference sheet are key here:

\(I = \frac{q}{t}\)

Where:

• \(I\) = Current, measured in Amperes (\(A\)). Note: \(1 \text{ Ampere} = 1 \text{ Coulomb per second} (C/s)\).
• \(q\) = Charge, measured in Coulombs (\(C\)).
• \(t\) = Time, measured in seconds (\(s\)).

2. Faraday's Law

Faraday’s Law tells us that the amount of substance produced or consumed at an electrode is proportional to the amount of electric charge that passed through the cell. To solve these problems, you usually follow a "unit conversion" path.

The "Stoichiometry of Electricity" Pathway:
Current & Time (\(I \times t\)) \(\rightarrow\) Coulombs (\(q\)) \(\rightarrow\) Moles of Electrons (\(e^-\)) \(\rightarrow\) Moles of Substance \(\rightarrow\) Grams of Substance

Key Conversion Factor: Faraday's Constant (\(F = 96,485 \text{ C/mol } e^-\)). This is the "bridge" between the physics of electricity and the chemistry of moles.

3. Step-by-Step Example

Question: How many grams of \(Cu(s)\) are plated from a \(Cu^{2+}\) solution if a current of \(2.0 \text{ A}\) is applied for \(30 \text{ minutes}\)?

Step 1: Convert time to seconds.
\(30 \text{ min} \times 60 \text{ s/min} = 1800 \text{ s}\)

Step 2: Calculate total charge (\(q = I \times t\)).
\(2.0 \text{ C/s} \times 1800 \text{ s} = 3600 \text{ C}\)

Step 3: Convert Coulombs to moles of electrons.
\(3600 \text{ C} \div 96,485 \text{ C/mol } e^- \approx 0.0373 \text{ mol } e^-\)

Step 4: Convert moles of electrons to moles of metal.
Since the ion is \(Cu^{2+}\), the half-reaction is \(Cu^{2+} + 2e^- \rightarrow Cu(s)\).
\(0.0373 \text{ mol } e^- \times \frac{1 \text{ mol Cu}}{2 \text{ mol } e^-} = 0.0187 \text{ mol Cu}\)

Step 5: Convert moles to grams.
\(0.0187 \text{ mol Cu} \times 63.55 \text{ g/mol} \approx 1.19 \text{ g Cu}\)

Common Pitfalls to Avoid

• Time Units: Always convert time to seconds. If the problem gives you hours, multiply by \(3600\).
• Mole Ratios: Don't forget the charge of the ion! \(Al^{3+}\) requires \(3\) moles of electrons per mole of metal, while \(Ag^+\) only requires \(1\).
• Standard vs. Nonstandard: Use the Nernst equation only when the concentrations are not \(1.0 \text{ M}\). If they are \(1.0 \text{ M}\), \(Q = 1\), \(\ln(1) = 0\), and \(E_{cell} = E^\circ_{cell}\).

Summary Table

Concept: Cell Potential (\(E_{cell}\))
Equation/Relation: \(E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q\)
Key Takeaway: Potential changes as concentrations change. \(E\) drops as the reaction moves toward equilibrium.

Concept: Electrolysis
Equation/Relation: \(I = \frac{q}{t}\)
Key Takeaway: Non-spontaneous reactions driven by outside current. Use Faraday's constant (\(96,485 \text{ C/mol } e^-\)) to link charge to mass.

Did you know? Electroplating, the process used to coat jewelry in gold or silver, is a direct application of electrolysis and Faraday's Law! By controlling the current and time, jewelers know exactly how thick the layer of gold will be.