📉 Non-Standard Conditions — The Nernst Equation

Part 1 of 7 — E = E° − (RT/nF) ln Q

Standard cell potentials ($E°$) apply only when all concentrations are 1 M and all pressures are 1 atm. Real cells rarely operate at standard conditions. The Nernst equation tells us the cell potential at any set of conditions.

Deriving the Nernst Equation

We know: $\Delta G = \Delta G° + RT\ln Q$

And: $\Delta G = -nFE$ and $\Delta G° = -nFE°$

Substituting: $-nFE = -nFE° + RT\ln Q$

Dividing by $-nF$:

$\boxed{E = E° - \frac{RT}{nF}\ln Q}$

Variables

Interpreting the Nernst Equation

$E = E° - \frac{RT}{nF}\ln Q$

How Q Affects E

Key Insight

As a galvanic cell operates:

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

A "dead" battery is simply a cell that has reached equilibrium!

Worked Example

For the Daniell cell: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

$E° = +1.10$ V, $n = 2$

Find $E$ when $[\text{Zn}^{2+}] = 0.10$ M, $[\text{Cu}^{2+}] = 2.0$ M at 298 K.

Step 1: Write Q (solids are excluded)

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

Step 2: Apply Nernst Equation

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

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

$E = 1.10 - (0.01284)(-3.00) = 1.10 + 0.039 = 1.14 \text{ V}$

$E > E°$ because $Q < 1$ — there are excess reactants (Cu²⁺), driving a higher voltage.

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

At 25°C (298 K), the Nernst equation simplifies to a convenient form using base-10 logarithms. This is the version most commonly used on the AP exam.

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

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

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

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

Common Q Expressions

Remember: solids and pure liquids are excluded from Q!

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

A concentration cell is a special galvanic cell where both electrodes are the same metal and the same half-reaction occurs in both compartments — but at different concentrations. The voltage comes entirely from the concentration difference.

How Concentration Cells Work

Setup

Both half-cells contain the same electrode and same ion, but at different concentrations:

• Dilute side: $[\text{M}^{n+}]_{\text{dilute}}$
• Concentrated side: $[\text{M}^{n+}]_{\text{conc}}$

E° = 0!

Since both half-reactions are identical, $E° = 0$:

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

The Nernst Equation Gives the Voltage

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

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

Which Side Is Which?

• Anode (oxidation): the dilute side — metal dissolves to increase $[\text{M}^{n+}]$
• Cathode (reduction): the concentrated side — $\text{M}^{n+}$ deposits to decrease concentration
• The cell drives toward equal concentrations (equilibrium)

Worked Example

A Cu concentration cell at 25°C:

• Left compartment: $[\text{Cu}^{2+}] = 0.010$ M
• Right compartment: $[\text{Cu}^{2+}] = 1.0$ M

$n = 2$ (Cu²⁺ + 2e⁻ → Cu)

Step 1: Identify Anode and Cathode

• Dilute side (0.010 M) = anode (oxidation)
• Concentrated side (1.0 M) = cathode (reduction)

Step 2: Calculate Q

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

Step 3: Apply Nernst

$E = 0 - \frac{0.0592}{2}\log(0.010)$ $= -0.0296 \times (-2) = +0.0592 \text{ V}$

The cell produces 59.2 mV. Small but measurable!

What Happens Over Time?

• Dilute side: $[\text{Cu}^{2+}]$ increases (Cu dissolves)
• Concentrated side: $[\text{Cu}^{2+}]$ decreases (Cu²⁺ plates out)
• Eventually: both sides equal → $E = 0$ (equilibrium)

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 — E° = (0.0592/n) log K

At equilibrium, $E = 0$ and $Q = K$. This gives us a direct relationship between the standard cell potential and the equilibrium constant — one of the most tested connections on the AP exam.

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°

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

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

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

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

Example 2: Find E° from K

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

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

Example 3: The Complete Thermodynamic Triangle

$\Delta G° = -nFE° = -RT\ln K$

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

Batteries are galvanic cells engineered for practical use. Understanding the chemistry behind common batteries connects electrochemistry to everyday life — and appears frequently on the AP exam.

Primary Batteries (Non-Rechargeable)

Primary batteries involve irreversible reactions — once the reactants are consumed, the battery is dead.

Alkaline Battery (Zinc-Manganese Dioxide)

Anode: $\text{Zn}(s) + 2\text{OH}^-(aq) \rightarrow \text{ZnO}(s) + \text{H}_2\text{O}(l) + 2e^-$

Cathode: $2\text{MnO}_2(s) + \text{H}_2\text{O}(l) + 2e^- \rightarrow \text{Mn}_2\text{O}_3(s) + 2\text{OH}^-(aq)$

$E \approx 1.5$ V per cell

• Most common household battery (AA, AAA, C, D)
• Uses alkaline (KOH) electrolyte
• Cannot be recharged (structural changes are irreversible)

Zinc-Air Battery

• Uses oxygen from air as the cathode reactant
• Very high energy density (used in hearing aids)
• $E \approx 1.4$ V

Secondary Batteries (Rechargeable)

Secondary batteries involve reversible reactions — applying external voltage reverses the cell chemistry.

Lead-Acid Battery (Car Battery)

Anode: $\text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^-$

Cathode: $\text{PbO}_2(s) + 4\text{H}^+(aq) + \text{SO}_4^{2-}(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$

$E \approx 2.0$ V per cell (6 cells in series → 12 V car battery)

Lithium-Ion Battery (Li-ion)

• Used in phones, laptops, electric vehicles
• $E \approx 3.7$ V per cell
• Li⁺ ions shuttle between graphite anode and metal oxide cathode
• Lightweight, high energy density
• Intercalation: Li⁺ inserts into layered structures without destroying them

Nickel-Metal Hydride (NiMH)

• Used in hybrid cars, rechargeable AA batteries
• $E \approx 1.2$ V per cell
• More environmentally friendly than older Ni-Cd batteries

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

This workshop provides practice with the Nernst equation, concentration cells, the E°-K relationship, and battery applications. These are the types of calculations you will encounter on the AP exam.

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

This final review integrates the Nernst equation, concentration cells, the E°-K relationship, and battery applications. Master these connections and you will be fully prepared for the electrochemistry portion of the AP exam.

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

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 ✅