Nernst Equation and Concentration Effects

📉 Non-Standard Conditions — The Nernst Equation

Part 1 of 7 — Beyond Standard Potentials

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

• Understanding the core concepts covered in Part 1
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

🔋 Deriving the Nernst Equation

Starting Point — Free Energy and Equilibrium

We know from thermodynamics:

$\Delta G = \Delta G° + RT\ln Q$

And from electrochemistry:

$\Delta G = -nFE \qquad \text{and} \qquad \Delta G° = -nFE°$

The Derivation

Substituting both into the free energy equation:

$-nFE = -nFE° + RT\ln Q$

Dividing every term by $-nF$:

$\boxed{E = E° - \frac{RT}{nF}\ln Q}$

💡 This is the general form of the Nernst equation — valid at any temperature.

📊 Key Variables

🧮 At 25°C (298 K) — The Simplified Form

At room temperature, the constants combine to give:

$E = E° - \frac{0.0257}{n}\ln Q \qquad \text{or} \qquad E = E° - \frac{0.0592}{n}\log Q$

The second form uses $\log$ (base 10) instead of $\ln$ — both are commonly seen on exams.

🧭 Interpreting the Nernst Equation

$E = E° - \frac{RT}{nF}\ln Q$

The correction term $\frac{RT}{nF}\ln Q$ shifts $E$ up or down from $E°$ depending on the value of $Q$.

How Q Affects E

🔑 Key Insight — Why Batteries Die

As a galvanic cell operates:

1. Reactants are consumed → $Q$ increases
2. $E$ decreases as $Q$ approaches $K$
3. When $Q = K$: $E = 0$ — the battery is "dead"

💡 Tip: A "dead" battery is simply a cell that has reached equilibrium — there is no longer any thermodynamic driving force for the reaction.

🧪 Worked Example — Daniell Cell

Problem: For the Daniell cell: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$, with $E° = +1.10$ V, $n = 2$, $T = 298$ K. Find $E$ when $[\text{Zn}^{2+}] = 0.10$ M and $[\text{Cu}^{2+}] = 2.0$ M.

Solution:

Step 1 — Write the Reaction Quotient

Remember: solids are excluded from $Q$!

$Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.10}{2.0} = 0.050$

Step 2 — Apply the Nernst Equation

$E = E° - \frac{RT}{nF}\ln Q$

$E = 1.10 - \frac{(8.314)(298)}{(2)(96{,}485)}\ln(0.050)$

$E = 1.10 - \frac{2478}{192{,}970}(-3.00)$

$E = 1.10 - (0.01284)(-3.00) = 1.10 + 0.039$

$\boxed{E = 1.14 \text{ V}}$

Step 3 — Check the Result

✅ $E > E°$ because $Q < 1$ — there are excess reactants ($\text{Cu}^{2+}$ is high), which drives a higher voltage than standard conditions.

Nernst Equation Concept Quiz 🎯

Nernst Equation Calculations 🧮

For a cell with $E° = +0.80$ V and $n = 2$ at 298 K:

1) If $Q = 1$, what is $E$? (in V)

2) If $Q = 100$, is $E$ greater than or less than $E°$? (type "greater" or "less")

3) If $Q = K$, what is $E$? (in V)

Round all answers to 3 significant figures.

Nernst Equation Concepts 🔽

Exit Quiz — Nernst Equation ✅

🔢 Simplified Nernst at 25°C

Part 2 of 7 — E = E° − (0.0592/n) log Q

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

• Understanding the core concepts covered in Part 2
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📌 The Simplified Form

Starting from: $E = E° - \frac{RT}{nF}\ln Q$

At $T = 298$ K:

$\frac{RT}{F} = \frac{(8.314)(298)}{96{,}485} = 0.02569 \text{ V}$

Converting $\ln$ to $\log$: $\ln Q = 2.303 \log Q$

$E = E° - \frac{(0.02569)(2.303)}{n}\log Q$

$\boxed{E = E° - \frac{0.0592}{n}\log Q} \quad \text{(at 25°C)}$

Why This Form Is Useful

• $\log$ (base 10) is easier to compute mentally than $\ln$
• The constant $0.0592$ is easy to remember
• Most AP problems are at 25°C

Example

Problem: For a 2-electron cell with $E° = 1.10$ V and $Q = 100$:

Solution:

$E = 1.10 - \frac{0.0592}{2}\log(100) = 1.10 - (0.0296)(2) = 1.10 - 0.059 = 1.04 \text{ V}$

🔋 Applications of the Simplified Nernst

Effect of 10-Fold Concentration Change

For each 10-fold change in $Q$:

$\Delta E = \frac{0.0592}{n} \cdot 1 = \frac{0.0592}{n} \text{ V per decade}$

For a 2-electron process: each 10× change in $Q$ shifts $E$ by $0.0296$ V

🔑 Key Concept: When $Q < 1$ (excess reactants), $E > E°$ — the cell produces more voltage. When $Q > 1$ (excess products), $E < E°$ — voltage decreases toward zero.

Common Q Expressions

Remember: solids and pure liquids are excluded from Q!

⚠️ Warning: Always write $Q$ as products over reactants, and never include solids or pure liquids. Getting $Q$ upside down flips the sign of the correction term!

Simplified Nernst Quiz 🎯

Simplified Nernst Calculations 🧮

All at 25°C. Use $E = E° - (0.0592/n)\log Q$.

1) $E° = 0.46$ V, $n = 2$, $Q = 0.01$. Calculate $E$. (to 3 significant figures)

2) $E° = 1.10$ V, $n = 2$, $Q = 10^4$. Calculate $E$. (to 3 significant figures)

3) $E° = 0.80$ V, $n = 1$, $Q = 10^{-3}$. Calculate $E$. (to 3 significant figures)

Nernst at 25°C Concepts 🔽

Exit Quiz — Simplified Nernst ✅

🔄 Concentration Cells

Part 3 of 7 — Same Electrodes, Different Concentrations

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

• Understanding the core concepts covered in Part 3
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

🔧 How Concentration Cells Work

The Setup — Same Metal, Different Concentrations

Both half-cells contain the same electrode and the same ion — the only difference is concentration:

⚡ E° = 0 — But the Cell Still Works!

Since both half-reactions are identical:

$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = E° - E° = 0$

🔑 So where does the voltage come from? Entirely from the concentration difference!

📐 The Nernst Equation for Concentration Cells

Starting from the Nernst equation with $E° = 0$:

$E = 0 - \frac{0.0592}{n}\log Q = -\frac{0.0592}{n}\log\frac{[\text{dilute}]}{[\text{conc}]}$

Flipping the fraction removes the negative sign:

$\boxed{E = \frac{0.0592}{n}\log\frac{[\text{conc}]}{[\text{dilute}]}}$

💡 The bigger the concentration ratio, the higher the voltage. A 10× ratio gives $0.0592/n$ V per factor of 10.

🧭 Why Does Dilute = Anode?

The system wants to reach equilibrium (equal concentrations). It does this by:

1. Dissolving metal on the dilute side → increases $[\text{M}^{n+}]$ there (oxidation = anode)
2. Plating out ions on the concentrated side → decreases $[\text{M}^{n+}]$ there (reduction = cathode)
3. Equilibrium is reached when both sides are equal → $Q = 1$ → $E = 0$

🧪 Worked Example — Copper Concentration Cell

Problem: A $Cu/Cu^{2+}$ concentration cell at 25°C has $[\text{Cu}^{2+}]_{\text{left}} = 0.010$ M (dilute, anode) and $[\text{Cu}^{2+}]_{\text{right}} = 1.0$ M (concentrated, cathode). Given $n = 2$ (from $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$) and $E° = 0$, calculate the cell potential.

Solution:

Step 1 — Identify Anode and Cathode

Dilute side = anode (metal dissolves to increase concentration) Concentrated side = cathode (ions plate out to decrease concentration)

Step 2 — Calculate the Reaction Quotient

$Q = \frac{[\text{Cu}^{2+}]_{\text{anode}}}{[\text{Cu}^{2+}]_{\text{cathode}}} = \frac{0.010}{1.0} = 0.010$

Step 3 — Apply the Nernst Equation

$E = 0 - \frac{0.0592}{2}\log(0.010)$

$E = -0.0296 \times (-2.00)$

$\boxed{E = +0.0592 \text{ V} = 59.2 \text{ mV}}$

📏 Small but measurable! This is exactly the principle behind pH meters and ion-selective electrodes.

⏱️ What Happens Over Time?

The cell spontaneously equalizes the concentrations — just like entropy demands!

Concentration Cell Quiz 🎯

Concentration Cell Calculations 🧮

At 25°C:

1) An Ag concentration cell has $[\text{Ag}^+]_{\text{dilute}} = 0.0010$ M and $[\text{Ag}^+]_{\text{conc}} = 1.0$ M. $n = 1$. Calculate $E$. (in V, to 3 significant figures)

2) A Zn concentration cell has $[\text{Zn}^{2+}] = 0.10$ M and $1.0$ M. $n = 2$. Calculate $E$. (in V, to 3 significant figures)

3) If both compartments have the same concentration, $E = ?$ (in V)

Concentration Cell Concepts 🔽

Exit Quiz — Concentration Cells ✅

🔗 Relationship Between E° and K

Part 4 of 7

At equilibrium, the cell potential drops to zero ($E = 0$) and the reaction quotient equals the equilibrium constant ($Q = K$). Substituting into the Nernst equation gives one of the most powerful connections in electrochemistry:

$\boxed{E° = \frac{0.0592}{n}\log K \quad \text{(at 25°C)}}$

🔑 Know any one of $\Delta G°$, $E°$, or $K$ — and you can calculate the other two. This is the "Thermodynamic Triangle" and it's one of the most frequently tested relationships on the AP exam.

🗺️ What You'll Learn in This Section

🔗 Deriving the E°-K Relationship

Starting from the Nernst equation at equilibrium ($E = 0$, $Q = K$):

$0 = E° - \frac{0.0592}{n}\log K$

Rearranging:

$\boxed{E° = \frac{0.0592}{n}\log K} \quad \text{(at 25°C)}$

Or equivalently:

$\log K = \frac{nE°}{0.0592}$

$K = 10^{nE°/0.0592}$

What This Tells Us

How Sensitive Is K to E°?

For a 2-electron process:

• $E° = +0.10$ V → $K = 10^{2(0.10)/0.0592} = 10^{3.38} \approx 2400$
• $E° = +0.50$ V → $K = 10^{16.9} \approx 10^{17}$
• $E° = +1.00$ V → $K = 10^{33.8}$

Even small $E°$ values correspond to enormous equilibrium constants!

🧪 Worked Examples

Example 1: Find K from E°

Problem: For the Daniell cell: $E° = 1.10$ V, $n = 2$

Solution:

$\log K = \frac{nE°}{0.0592} = \frac{(2)(1.10)}{0.0592} = 37.2$

$K = 10^{37.2} = 1.6 \times 10^{37}$

This enormous $K$ means the reaction goes essentially to completion.

Example 2: Find E° from K

Problem: A reaction has $K = 1.0 \times 10^{10}$ and $n = 2$.

Solution:

$E° = \frac{0.0592}{2}\log(10^{10}) = 0.0296 \times 10 = 0.296 \text{ V}$

Example 3: The Complete Thermodynamic Triangle

$\boxed{\Delta G° = -nFE° = -RT\ln K}$

All three quantities are interconnected:

• Know any one → calculate the other two

E° and K Quiz 🎯

E° and K Calculations 🧮

At 25°C:

1) $E° = 0.46$ V, $n = 2$. Calculate $\log K$. (to 3 significant figures)

2) $K = 10^{20}$, $n = 4$. Calculate $E°$. (in V, to 3 significant figures)

3) $E° = -0.10$ V, $n = 1$. Is $K$ greater or less than 1? (type "greater" or "less")

E° and K Connections 🔽

Exit Quiz — E° and K ✅

🔋 Batteries — Primary, Secondary, and Fuel Cells

Part 5 of 7 — Real-World Applications

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

• Understanding the core concepts covered in Part 5
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

� Primary Batteries (Non-Rechargeable)

One-way trip! Primary batteries involve irreversible reactions — once the reactants are consumed, the battery is dead. You use it, then recycle it.

⚡ Alkaline Battery — The Household Workhorse

💡 Why can't you recharge it? The solid products ($\text{ZnO}$, $\text{Mn}_2\text{O}_3$) undergo structural changes that can't be cleanly reversed.

🌬️ Zinc-Air Battery — Breathing Electricity

🔑 AP Tip: Zinc-air is a favorite exam topic because it blurs the line between a battery and a fuel cell — the oxidant ($\text{O}_2$) comes from outside!

� Secondary Batteries (Rechargeable)

Round trip! Secondary batteries involve reversible reactions — applying external voltage reverses the cell chemistry, restoring the original reactants.

🚗 Lead-Acid Battery — Under Every Hood

💡 Exam alert: Both electrodes produce $\text{PbSO}_4$ — so as the battery discharges, $[\text{H}_2\text{SO}_4]$ decreases. That's why mechanics test battery health with a hydrometer!

📱 Lithium-Ion Battery — The Modern Standard

🔑 Key term — Intercalation: $Li^{+}$ ions slip between layers of the electrode material without breaking the crystal structure. This is what makes Li-ion reversible and long-lasting.

🔋 Nickel-Metal Hydride (NiMH)

⚖️ Quick Comparison

⚠️ Warning: Don't confuse primary and secondary! Primary = irreversible = non-rechargeable. Secondary = reversible = rechargeable. A fuel cell is neither — reactants are continuously supplied from outside.

📌 Fuel Cells

A fuel cell is a galvanic cell where the reactants are continuously supplied from outside. Unlike batteries, fuel cells don't run down — they operate as long as fuel and oxidant are fed in.

Hydrogen Fuel Cell

Anode: $2\text{H}_2(g) \rightarrow 4\text{H}^+(aq) + 4e^-$

Cathode: $\text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \rightarrow 2\text{H}_2\text{O}(l)$

Overall: $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$

$E \approx 1.23$ V

Why Fuel Cells Are Important

Battery Chemistry Quiz 🎯

Battery Types 🔽

Exit Quiz — Batteries ✅

🛠️ Problem-Solving Workshop — Nernst Equation

Part 6 of 7 — Practice and Integration

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

• Working through complete multi-step problems from start to finish
• Building problem-solving strategies you can apply on the AP exam
• Identifying which concepts to apply and in what order

🛠️ Problem-Solving Toolkit

Which Equation to Use?

⚠️ Common Pitfalls:

1. Forgetting to exclude solids/liquids from Q
2. Using 0.0592 at temperatures other than 25°C
3. Confusing log and ln ($\ln K = 2.303\log K$)
4. Getting Q upside down (products over reactants!)

Mixed Nernst Problems 🎯

Calculation Workshop 🧮

1) $E° = 0.80$ V, $n = 1$, $Q = 0.001$ at 25°C. Calculate $E$. (to 3 significant figures)

2) A concentration cell: $[\text{Cu}^{2+}] = 0.01$ M and $1.0$ M, $n = 2$. Calculate $E$. (to 3 significant figures)

3) $E° = 1.50$ V, $n = 3$. Calculate $\log K$. (to 3 significant figures)

Problem Strategy 🔽

Exit Quiz — Problem-Solving Workshop ✅

🎯 Synthesis & AP Review — Nernst Equation

Part 7 of 7 — Complete Mastery

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

• Solving AP-style questions that integrate multiple concepts from this unit
• Writing clear, concise explanations using proper chemistry terminology
• Identifying and avoiding common AP exam traps and mistakes

📋 Master Equation Summary

The Core Equations

The Thermodynamic Triangle (at 25°C)

$\Delta G° \xleftrightarrow{-nF} E° \xleftrightarrow{0.0592/n} \log K \xleftrightarrow{-RT\ln} \Delta G°$

Battery Classification

🔑 Key Concept: Know any one of $\Delta G°$, $E°$, or $K$ — and you can calculate the other two. This "thermodynamic triangle" unifies equilibrium, electrochemistry, and thermodynamics.

Comprehensive AP Review 🎯

Integration Problems 🧮

1) E° = 0.80 V, n = 2, T = 298 K. What is $\Delta G°$ in kJ? (to 1 decimal)

2) E° = 0.40 V, n = 2. What is $\log K$? (to 1 decimal)

3) A dead battery has E = ___ V and Q = ___ (type "0" and "K" separated by a comma)

Round all answers to 3 significant figures.

Final Concept Review 🔽

Final Exit Quiz — Nernst Equation Mastery ✅