Galvanic Cells and Standard Cell Potentials

⚡ Galvanic Cells — Redox Review

Part 1 of 7 — Half-Reactions and Electron Transfer

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

🔄 Redox Review

Oxidation and Reduction

🔑 Key Concept: Remember OIL RIG — Oxidation Is Loss, Reduction Is Gain.

Oxidation Numbers

Oxidation numbers (states) help track electron transfer:

• Elements in standard state: 0
• Monatomic ions: charge = oxidation number
• O is usually −2 (except peroxides: −1)
• H is usually +1 (except metal hydrides: −1)
• Sum of oxidation numbers = charge of species

Identifying Redox

• The species that is oxidized is the reducing agent (it reduces something else)
• The species that is reduced is the oxidizing agent (it oxidizes something else)

✍️ Writing Half-Reactions

Every redox reaction can be split into two half-reactions — one for oxidation and one for reduction.

Worked Example

Overall Reaction:

$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

Oxidation half-reaction (anode):

$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$

Zinc loses 2 electrons — oxidation number changes from $0$ to $+2$.

Reduction half-reaction (cathode):

$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

Copper gains 2 electrons — oxidation number changes from $+2$ to $0$.

Key Points

• Electrons must balance — the number lost in oxidation equals the number gained in reduction
• The electrode where oxidation occurs is the anode
• The electrode where reduction occurs is the cathode

🔑 Key Concept — AN OX, RED CAT:

ANode = OXidation  |  REDuction = CAThode

Redox Fundamentals Quiz 🎯

Oxidation State Practice 🧮

Determine the oxidation state of the underlined element:

1) The oxidation state of Mn in $\text{MnO}_4^-$ is:

2) The oxidation state of Cr in $\text{Cr}_2\text{O}_7^{2-}$ is:

3) The oxidation state of N in $\text{NO}_3^-$ is:

Redox Terminology 🔽

Exit Quiz — Redox Review ✅

🔋 Galvanic Cell Structure

Part 2 of 7 — Salt Bridges, Electron Flow, and Ion Flow

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

🏗️ Anatomy of a Galvanic Cell

The Two Half-Cells

A galvanic cell consists of two half-cells, each containing:

• An electrode (solid conductor, often a metal)
• An electrolyte solution (containing the relevant ions)

Key Components

The Zn-Cu Cell (Daniell Cell)

Anode (oxidation): $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$

Cathode (reduction): $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$

Overall: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

🔀 Flow Directions

⚡ Electron Flow (through the wire)

$\text{Anode} \xrightarrow{e^-} \text{Cathode}$

Electrons flow from anode to cathode through the external circuit.

🧂 Ion Flow (through the salt bridge)

🤔 Why Is the Salt Bridge Necessary?

Without a salt bridge, the cell would stop working almost immediately.

Here's why:

1. The anode solution would become too positive (excess $\text{Zn}^{2+}$ produced)
2. The cathode solution would become too negative ($\text{Cu}^{2+}$ consumed)
3. This charge imbalance would halt the reaction

The salt bridge maintains electrical neutrality by allowing ion migration between the two half-cells.

± Anode Sign Convention

In a galvanic cell:

• Anode = negative terminal (−)
• Cathode = positive terminal (+)

⚠️ This is opposite to electrolytic cells!

Cell Structure Quiz 🎯

Cell Component Identification 🔽

Cell Analysis 🧮

For a galvanic cell with the overall reaction: $\text{Mg}(s) + \text{Fe}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Fe}(s)$

1) Which metal is the anode? (type the element symbol)

2) Which metal is the cathode? (type the element symbol)

3) How many electrons are transferred in the balanced reaction?

Exit Quiz — Cell Structure ✅

⚡ Standard Reduction Potentials

Part 3 of 7 — E° and Calculating Cell Voltage

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

⚡ Standard Reduction Potential Table

All half-reactions are written as reductions (gaining electrons):

🔑 Key Concept — Reading the Table:

• More positive $E°$: stronger tendency to be reduced (stronger oxidizing agent)
• More negative $E°$: stronger tendency to be oxidized (stronger reducing agent)
• The Standard Hydrogen Electrode (SHE) is the reference: $E° = 0.00$ V

🔢 Calculating Standard Cell Potential

🔑 The Master Equation:

$\boxed{E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}}$

⚠️ Important Rules

⚠️ Warning: Never multiply $E°$ values by stoichiometric coefficients. $E°$ is an intensive property — it does not change when you scale the equation.

🧪 Worked Example: Zn-Cu Cell

Problem: Calculate $E^\circ_{\text{cell}}$ for a Zn-Cu galvanic cell.

• Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$    ($E° = +0.34$ V)
• Anode: $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$    ($E° = -0.76$ V)

Solution:

$\boxed{E^\circ_{\text{cell}} = (+0.34) - (-0.76) = +1.10 \text{ V}}$

✅ The positive $E^\circ_{\text{cell}}$ confirms the reaction is spontaneous.

Reduction Potential Quiz 🎯

Cell Potential Calculations 🧮

Standard Reduction Potentials:

Half-Reaction	$E^\circ$ (V)
$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$	$+0.80$
$\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$	$+0.34$
$\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s)$	$-0.26$
$\text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s)$	$-0.44$

1) $E^\circ_{\text{cell}}$ for a cell with Ag cathode and Fe anode: (in V, to 3 significant figures)

2) $E^\circ_{\text{cell}}$ for a cell with Cu cathode and Ni anode: (in V, to 3 significant figures)

3) $E^\circ_{\text{cell}}$ for a cell with Ni cathode and Fe anode: (in V, to 3 significant figures)

Reduction Potential Concepts 🔽

Exit Quiz — Standard Reduction Potentials ✅

📝 Cell Notation (Line Notation)

Part 4 of 7 — Shorthand for Electrochemical Cells

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 4

• Understanding the core concepts covered in Part 4
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

📝 Cell Notation Rules

The Format

$\boxed{\text{Anode} \mid \text{Anode ion} \| \text{Cathode ion} \mid \text{Cathode}}$

Conventions

🔑 Key Concept: Read cell notation left to right — anode (oxidation) → salt bridge → cathode (reduction).

Example: Daniell Cell

$\boxed{\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \| \text{Cu}^{2+}(aq) \mid \text{Cu}(s)}$

Read left to right:

1. Zn solid electrode (anode)
2. Phase boundary
3. $Zn^{2+}$ ions in solution
4. Salt bridge
5. $Cu^{2+}$ ions in solution
6. Phase boundary
7. Cu solid electrode (cathode)

⭐ Special Cases in Cell Notation

🔩 Inert Electrodes

When a half-reaction involves only aqueous species (no solid metal), we use an inert electrode — typically Pt (platinum) or C (graphite):

Example:

$\text{Pt} \mid \text{Fe}^{2+}(aq), \text{Fe}^{3+}(aq) \| \text{Ag}^+(aq) \mid \text{Ag}(s)$

The comma separates species in the same phase.

💨 Gas Electrodes

For reactions involving gases, the gas contacts the Pt electrode and is separated by a phase boundary:

Example:

$\text{Pt} \mid \text{H}_2(g) \mid \text{H}^+(aq) \| \text{Ag}^+(aq) \mid \text{Ag}(s)$

The $\text{H}_2$ gas bubbles over the Pt surface.

🎯 Key Points for AP

Cell Notation Quiz 🎯

Reading Cell Notation 🧮

For the cell: $\text{Al}(s) \mid \text{Al}^{3+}(aq) \| \text{Ni}^{2+}(aq) \mid \text{Ni}(s)$

1) Which metal is the anode? (element symbol)

2) Which metal is the cathode? (element symbol)

3) How many electrons are transferred in the balanced reaction? ($Al^{3+}$ needs $3e^{-}$, $Ni^{2+}$ needs $2e^{-}$)

Cell Notation Elements 🔽

Exit Quiz — Cell Notation ✅

🔗 Connecting Free Energy and Cell Potential

Part 5 of 7 — $\Delta G^\circ = -nFE^\circ$

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

🔑 The Key Equation

The bridge between thermodynamics and electrochemistry:

$\Delta G^\circ = -nFE^\circ$

🤔 Why the Negative Sign?

The negative sign ensures these are always opposite in sign — positive $E^\circ$ gives negative $\Delta G^\circ$. ✓

📏 Unit Check

$\Delta G^\circ = -(\text{mol})(\text{C/mol})(\text{J/C}) = \text{J}$

💡 The units work out to joules. Divide by 1000 to convert to kJ.

🔗 The Thermodynamic Triangle

Three key relationships connect $\Delta G^\circ$, $E^\circ$, and $K$:

Master Equations:

$\Delta G^\circ = -nFE^\circ = -RT\ln K$

$E^\circ = \frac{RT}{nF}\ln K$

At 25°C (298 K):

$E^\circ = \frac{0.0257}{n}\ln K = \frac{0.0592}{n}\log K$

🗺️ The Web of Connections

✅ All Three Must Be Consistent

💡 If you know any one of these three values, you can determine the other two!

🧪 Worked Example

Daniell Cell:

$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

$E^\circ = +1.10$ V,   $n = 2$ mol $e^-$

📐 Calculate $\Delta G^\circ$

$\Delta G^\circ = -nFE^\circ = -(2)(96{,}485)(1.10)$

$= -212{,}267 \text{ J} = \boxed{-212.3 \text{ kJ}}$

📐 Calculate $K$ at 298 K

$\ln K = \frac{nFE^\circ}{RT} = \frac{(2)(96{,}485)(1.10)}{(8.314)(298)} = \frac{212{,}267}{2478} = 85.66$

$K = e^{85.66} = \boxed{1.6 \times 10^{37}}$

🔥 This enormous $K$ confirms the reaction is virtually complete at equilibrium — products are overwhelmingly favored.

$\Delta G^\circ$ and $E^\circ$ Quiz 🎯

Thermodynamic Triangle Calculations 🧮

1) $E^\circ = +0.80$ V, $n = 1$. Calculate $\Delta G^\circ$ in kJ. (to 1 decimal)

2) $\Delta G^\circ = -579$ kJ, $n = 6$. Calculate $E^\circ$ in V. (to 3 significant figures)

3) If $E^\circ > 0$ for a cell, is $K$ greater than or less than 1? (type "greater" or "less")

Connecting the Three Quantities 🔽

Exit Quiz — $\Delta G^\circ$ and $E^\circ$ ✅

🛠️ Problem-Solving Workshop — Galvanic Cells

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 Strategy

Step-by-Step Approach

🔑 Key Concept: Follow this systematic approach for every galvanic cell problem:

1. Identify the two half-reactions
2. Determine which is oxidized (anode) and which is reduced (cathode) using $E^\circ$ values
3. Calculate $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
4. Balance electrons (find $n$)
5. Calculate $\Delta G^\circ = -nFE^\circ$ if needed
6. Write cell notation if asked

Common Mistakes to Avoid

⚠️ Warning: These are the most frequent errors on the AP exam:

Mixed Galvanic Cell Problems 🎯

Given: $\text{Ag}^+/\text{Ag} = +0.80$ V, $\text{Zn}^{2+}/\text{Zn} = -0.76$ V, $\text{Fe}^{2+}/\text{Fe} = -0.44$ V, $\text{Cu}^{2+}/\text{Cu} = +0.34$ V

Calculation Workshop 🧮

Given: $\text{Cu}^{2+}/\text{Cu} = +0.34$ V, $\text{Fe}^{2+}/\text{Fe} = -0.44$ V

For the cell: Fe(s) | $Fe^{2+}(aq)$ || $Cu^{2+}(aq)$ | Cu(s)

1) $E^\circ_{\text{cell}} = ?$ (in V, to 3 significant figures)

2) $n = ?$ (electrons transferred)

3) $\Delta G^\circ = ?$ (in kJ, to nearest whole number)

Cell Analysis 🔽

For the cell: Al(s) | $Al^{3+}(aq)$ || $Ag^{+}(aq)$ | Ag(s)

Given: $Al^{3+}/Al$ = −1.66 V, $Ag^{+}/Ag$ = +0.80 V

Exit Quiz — Problem-Solving Workshop ✅

🎯 Synthesis & AP Review — Galvanic Cells

Part 7 of 7 — Mastery Check

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 Summary

Essential Equations

🔑 Key Equations — Memorize These:

$\boxed{E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}}$

$\boxed{\Delta G^{\circ} = -nFE^{\circ}}$

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

Cell Components

Spontaneity Criteria

🔑 Key Concept: These three quantities are always linked — know one, know them all.

Comprehensive AP Review 🎯

Given: $Zn^{2+}/Zn$ = −0.76 V, $Cu^{2+}/Cu$ = +0.34 V, $Ag^{+}/Ag$ = +0.80 V, $Fe^{2+}/Fe$ = −0.44 V

Integration Problems 🧮

1) A cell has $E^\circ = +2.00$ V and $n = 3$. What is $\Delta G^\circ$ in kJ? (to nearest whole number)

2) A cell has $\Delta G^\circ = -386$ kJ and $n = 4$. What is $E^\circ$ in V? (to 3 significant figures)

3) If $E^\circ = +0.50$ V and $n = 2$ at 298 K, is $K$ greater or less than 1? (type "greater" or "less")

Final Concept Review 🔽

Final Exit Quiz — Galvanic Cells Mastery ✅