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Galvanic Cells and Standard Cell Potentials | Study Mondo
Topics / Electrochemistry / Galvanic Cells and Standard Cell Potentials Galvanic Cells and Standard Cell Potentials Understand voltaic cells, half-reactions, standard reduction potentials, and calculating cell voltage.
BC Written and reviewed by Brendan Cusack , Study Mondo Education Team • Last updated April 7, 2026 🎯 ⭐ INTERACTIVE LESSON
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Galvanic Cells and Standard Cell Potentials
Redox Review
Oxidation: Loss of electrons (increase in oxidation number)
Reduction: Gain of electrons (decrease in oxidation number)
OIL RIG: Oxidation Is Loss, Reduction Is Gain
Redox reaction: Both oxidation and reduction occur simultaneously
Galvanic (Voltaic) Cells Galvanic cell: Spontaneous redox reaction produces electric current
Anode: Oxidation occurs (electrons leave)
Negative electrode
Loses mass over time
Cathode: Reduction occurs (electrons enter)
Positive electrode
Gains mass over time
Salt bridge: Allows ion flow to maintain neutrality
Contains inert electrolyte (KNO₃, Na₂SO₄)
Completes circuit
External circuit: Electrons flow from anode to cathode
Mnemonic: "AN OX and a RED CAT"
ANode = OXidation
REDuction = CAThode
Cell Notation (Line Notation) Anode | Anode solution || Cathode solution | Cathode \text{Anode | Anode solution || Cathode solution | Cathode} Anode | Anode solution || Cathode solution | Cathode
Example: Zn/Zn²⁺ || Cu²⁺/Cu
Single line (|): Phase boundary
Double line (||): Salt bridge
Anode always on left
Cathode always on right
Zn(s) | Zn²⁺(aq, 1 M) || Cu²⁺(aq, 1 M) | Cu(s)
Standard Reduction Potentials (E°)
25°C (298 K)
1 M concentrations
1 atm pressure for gases
Standard reduction potential (E°): Voltage when half-reaction occurs under standard conditions
Measured relative to standard hydrogen electrode (SHE):
2H + ( a q ) + 2 e − → H 2 ( g ) E ° = 0.00 V \text{2H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \quad E° = 0.00 \text{ V} 2H + ( a q ) + 2 e − → H 2 ( g ) E ° = 0.00 V
All potentials compared to this reference
Using Standard Reduction Potential Table More positive E°: Stronger oxidizing agent (easier to reduce)
More negative E°: Stronger reducing agent (easier to oxidize)
Half-Reaction E° (V) F₂ + 2e⁻ → 2F⁻ +2.87 Au³⁺ + 3e⁻ → Au +1.50 Ag⁺ + e⁻ → Ag +0.80 Cu²⁺ + 2e⁻ → Cu +0.34 2H⁺ + 2e⁻ → H₂ 0.00 Zn²⁺ + 2e⁻ → Zn -0.76 Li⁺ + e⁻ → Li -3.05
Top of table: Best oxidizing agents (F₂, Au³⁺)
Bottom of table: Best reducing agents (Li, Zn)
Calculating Standard Cell Potential E ° cell = E ° cathode − E ° anode E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} E ° cell = E ° cathode − E ° anode
E ° cell = E ° red − E ° ox E°_{\text{cell}} = E°_{\text{red}} - E°_{\text{ox}} E ° cell = E ° red − E ° ox
Identify oxidation and reduction half-reactions
Look up E° for each (as reduction)
E°_cell = E°_cathode - E°_anode
If E°_cell > 0: spontaneous
If E°_cell < 0: non-spontaneous
Predicting Spontaneity E ° cell E°_{\text{cell}} E ° cell Δ G ° \Delta G° Δ G ° Spontaneous? Positive Negative Yes (galvanic) Zero Zero At equilibrium Negative Positive No (needs energy)
Δ G ° = − n F E ° cell \Delta G° = -nFE°_{\text{cell}} Δ G ° = − n FE ° cell
n = moles of electrons transferred
F = Faraday constant = 96,485 C/mol e⁻
E°_cell in volts
Example Cell: Zn-Cu Cell Zn ( s ) + Cu 2 + ( a q ) → Zn 2 + ( a q ) + Cu ( s ) \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) Zn ( s ) + Cu 2 + ( a q ) → Zn 2 + ( a q ) + Cu ( s )
Anode (oxidation): Zn → Zn²⁺ + 2e⁻
E° = -0.76 V (reverse of reduction)
Cathode (reduction): Cu²⁺ + 2e⁻ → Cu
E ° cell = 0.34 − ( − 0.76 ) = 1.10 V E°_{\text{cell}} = 0.34 - (-0.76) = 1.10 \text{ V} E ° cell = 0.34 − ( − 0.76 ) = 1.10 V
Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
Balancing Redox Equations
Separate into oxidation and reductionBalance atoms (except O and H)Balance O with H₂OBalance H with H⁺Balance charge with e⁻Multiply to equalize electronsAdd half-reactionsCancel common terms In base: Add OH⁻ to neutralize H⁺
Intensive vs Extensive Properties
Does NOT depend on amount
Don't multiply E° when balancing
E° same whether 1 electron or 1000 electrons
Depends on amount (moles)
Must multiply by n (electrons transferred)
Predicting Reactions Will reaction occur spontaneously?
Species with higher E° gets reduced (cathode)
Species with lower E° gets oxidized (anode)
Calculate E°_cell
If positive → yes!
📚 Practice Problems
1 Problem 1easy ❓ Question:Calculate the standard cell potential for: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s). E°(Ag⁺/Ag) = +0.80 V, E°(Zn²⁺/Zn) = -0.76 V. Is the reaction spontaneous?
💡 Show Solution Given:
Reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
E°(Ag⁺/Ag) = +0.80 V
E°(Zn²⁺/Zn) = -0.76 V
Identify half-reactions:
Oxidation (anode): Zn → Zn²⁺ + 2e⁻
Zn loses electrons
E°_anode = -0.76 V (as reduction)
Reduction (cathode): Ag⁺ + e⁻ → Ag (×2 for balance)
Ag⁺ gains electrons
E°_cathode = +0.80 V
Calculate E°_cell:
E ° cell = E ° cathode − E ° anode E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} E ° cell = E ° cathode − E °
E ° cell = 0.80 − ( − 0.76 ) E°_{\text{cell}} = 0.80 - (-0.76) E ° cell = 0.80 − ( − 0.76 )
E ° cell = 0.80 + 0.76 E°_{\text{cell}} = 0.80 + 0.76 E ° cell = 0.80 + 0.76
E ° cell = 1.56 V E°_{\text{cell}} = 1.56 \text{ V} E ° cell = 1.56 V
Answer: E°_cell = 1.56 V
Is reaction spontaneous?
E°_cell = +1.56 V > 0
Yes, reaction is spontaneous!
Cell notation:
Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)
Explanation:
Positive cell potential indicates spontaneous redox
Electrons flow from Zn (anode) to Ag⁺ (cathode)
Zn is stronger reducing agent (lower E°)
Ag⁺ is stronger oxidizing agent (higher E°)
Note: We did NOT multiply E° by 2 even though 2Ag⁺ are reduced. E° is intensive property!
2 Problem 2medium ❓ Question:A galvanic cell is constructed with a zinc electrode in 1.0 M Zn²⁺ solution and a copper electrode in 1.0 M Cu²⁺ solution. Given: E°(Zn²⁺/Zn) = -0.76 V and E°(Cu²⁺/Cu) = +0.34 V. (a) Write the half-reactions and overall cell reaction. (b) Calculate E°_cell. (c) Which electrode is the anode?
💡 Show Solution Solution:
(a) Half-reactions:
Oxidation (anode): Zn(s) → Zn²⁺(aq) + 2e⁻ E° = -0.76 V
Reduction (cathode): Cu²⁺(aq) + 2e⁻ → Cu(s) E° = +0.34 V
Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
(b) Calculate E°_cell:
E°_cell = E°_cathode - E°_anode
E°_cell = 0.34 V - (-0.76 V)
E°_cell = +1.10 V
(Positive E°_cell confirms spontaneous reaction)
(c) Anode identification:
Zinc is the anode (oxidation occurs here)
Copper is the cathode (reduction occurs here)
3 Problem 3medium ❓ Question:Calculate ΔG° for the reaction: 3Mg(s) + 2Al³⁺(aq) → 3Mg²⁺(aq) + 2Al(s). E°(Al³⁺/Al) = -1.66 V, E°(Mg²⁺/Mg) = -2.37 V. F = 96,485 C/mol e⁻.
💡 Show Solution Given:
Reaction: 3Mg(s) + 2Al³⁺(aq) → 3Mg²⁺(aq) + 2Al(s)
E°(Al³⁺/Al) = -1.66 V
E°(Mg²⁺/Mg) = -2.37 V
F = 96,485 C/mol e⁻
Identify half-reactions:
Oxidation: Mg → Mg²⁺ + 2e⁻ (×3)
3Mg → 3Mg²⁺ + 6e⁻
E°_anode = -2.37 V
Reduction: Al³⁺ + 3e⁻ → Al (×2)
2Al³⁺ + 6e⁻ → 2Al
E°_cathode = -1.66 V
4 Problem 4hard ❓ Question:For the reaction: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s) with E°_cell = +2.00 V at 25°C. Calculate ΔG° for this reaction.
💡 Show Solution Solution:
Relationship between ΔG° and E°_cell:
ΔG° = -nFE°_cell
where:
n = moles of electrons transferred
F = Faraday constant = 96,485 C/mol
E°_cell = standard cell potential
Determine n:
From balanced equation:
Al → Al³⁺ + 3e⁻ (×2) = 6 electrons
Cu²⁺ + 2e⁻ → Cu (×3) = 6 electrons
n = 6
Calculate ΔG°:
ΔG° = -(6 mol)(96,485 C/mol)(2.00 V)
ΔG° = -1,157,820 J
ΔG° = -1158 kJ or -1.16 × 10³ kJ
5 Problem 5hard ❓ Question:Will the reaction Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g) occur spontaneously under standard conditions? E°(Cu²⁺/Cu) = +0.34 V, E°(H⁺/H₂) = 0.00 V.
💡 Show Solution Given:
Proposed reaction: Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g)
E°(Cu²⁺/Cu) = +0.34 V
E°(H⁺/H₂) = 0.00 V (SHE reference)
Analyze proposed reaction:
Oxidation half: Cu → Cu²⁺ + 2e⁻
Copper is oxidized (loses electrons)
Would be anode
E°_anode = +0.34 V (as reduction)
Reduction half: 2H⁺ + 2e⁻ → H₂
Protons are reduced (gain electrons)
Would be cathode
Explain using: 📝 Simple words 🔗 Analogy 🎨 Visual desc. 📐 Example 💡 Explain
📋 AP Chemistry — Exam Format Guide⏱ 3 hours 15 minutes 📝 67 questions 📊 3 sections
Section Format Questions Time Weight Calculator Multiple Choice MCQ 60 90 min 50% ✅ Free Response (Long) FRQ 3 69 min 30% ✅ Free Response (Short) FRQ 4 36 min 20% ✅
💡 Key Test-Day Tips✓ Memorize common polyatomic ions✓ Practice dimensional analysis✓ Know your gas laws⚠️ Common Mistakes: Galvanic Cells and Standard Cell PotentialsAvoid these 3 frequent errors
1 Not balancing equations before doing stoichiometry
▾ 2 Confusing molarity (M) with molality (m)
▾ 3 Forgetting to convert temperature to Kelvin for gas law problems
▾ 🌍 Real-World Applications: Galvanic Cells and Standard Cell PotentialsSee how this math is used in the real world
🌍 Water Purification
Environment
▾ 💻 Battery Technology
Technology
▾
📝 Worked Example: Stoichiometry — Limiting ReagentProblem: 2 2 2 H 2 H_2 H 2 1 1 1 O 2 O_2 O 2 2 H 2 + O 2 → 2 H 2 O 2H_2 + O_2 \to 2H_2O 2 H 2 + O 2 → 2 H 2 O
1 Write the balanced equation Click to reveal →
2 Determine the limiting reagent
3 Calculate moles of product
🧪 Practice Lab Interactive practice problems for Galvanic Cells and Standard Cell Potentials
▾ 📌 Related Topics in Electrochemistry❓ Frequently Asked QuestionsWhat is Galvanic Cells and Standard Cell Potentials?▾ Understand voltaic cells, half-reactions, standard reduction potentials, and calculating cell voltage.
How can I study Galvanic Cells and Standard Cell Potentials effectively?▾ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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💡 Study Tips✓ Work through examples step-by-step ✓ Practice with flashcards daily ✓ Review common mistakes anode
Mnemonic: "An Ox" (Anode = Oxidation), "Red Cat" (Reduction = Cathode)
Electrons transferred: n = 6
E ° cell = E ° cathode − E ° anode E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} E ° cell = E ° cathode − E ° anode
E ° cell = − 1.66 − ( − 2.37 ) E°_{\text{cell}} = -1.66 - (-2.37) E ° cell = − 1.66 − ( − 2.37 )
E ° cell = − 1.66 + 2.37 E°_{\text{cell}} = -1.66 + 2.37 E ° cell = − 1.66 + 2.37
E ° cell = 0.71 V E°_{\text{cell}} = 0.71 \text{ V} E ° cell = 0.71 V
Δ G ° = − n F E ° cell \Delta G° = -nFE°_{\text{cell}} Δ G ° = − n FE ° cell
Δ G ° = − ( 6 ) ( 96 , 485 ) ( 0.71 ) \Delta G° = -(6)(96,485)(0.71) Δ G ° = − ( 6 ) ( 96 , 485 ) ( 0.71 )
Δ G ° = − 6 × 96 , 485 × 0.71 \Delta G° = -6 \times 96,485 \times 0.71 Δ G ° = − 6 × 96 , 485 × 0.71
Δ G ° = − 410 , 858 J \Delta G° = -410,858 \text{ J} Δ G ° = − 410 , 858 J
Δ G ° = − 411 kJ \Delta G° = -411 \text{ kJ} Δ G ° = − 411 kJ
E°_cell = +0.71 V → spontaneous
ΔG° = -411 kJ → spontaneous (negative ΔG°)
Consistent! Both indicate reaction is spontaneous.
Energy released: 411 kJ per 3 mol Mg reacted
Δ G ° = − n F E ° cell \Delta G° = -nFE°_{\text{cell}} Δ G ° = − n FE ° cell
Positive E° → Negative ΔG° → Spontaneous
Negative E° → Positive ΔG° → Non-spontaneous
E° = 0 → ΔG° = 0 → Equilibrium
Interpretation: Large negative ΔG° confirms the reaction is highly spontaneous and product-favored.
E°_cathode = 0.00 V
E ° cell = E ° cathode − E ° anode E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} E ° cell = E ° cathode − E ° anode
E ° cell = 0.00 − 0.34 E°_{\text{cell}} = 0.00 - 0.34 E ° cell = 0.00 − 0.34
E ° cell = − 0.34 V E°_{\text{cell}} = -0.34 \text{ V} E ° cell = − 0.34 V
Answer: NO, reaction is NOT spontaneous
Reason: E°_cell = -0.34 V < 0
Cu has higher E° (+0.34) than H⁺ (0.00):
Cu²⁺ is better oxidizing agent than H⁺
Cu²⁺ prefers to be reduced, not oxidized
Copper is "noble" - resists oxidation by acids
Spontaneous reaction is REVERSE:
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)
E°_cell = 0.34 - 0.00 = +0.34 V ✓
Spontaneous!
Activity series application:
Metals below H in activity series:
Cu, Ag, Au (E° > 0)
Do NOT react with non-oxidizing acids (HCl, H₂SO₄)
Need oxidizing acid (HNO₃) to dissolve
Metals above H in activity series:
Zn, Fe, Mg (E° < 0)
DO react with acids
Produce H₂ gas
Higher E° species gets reduced (cathode)
Lower E° species gets oxidized (anode)
If proposed reaction matches this: E°_cell > 0, spontaneous
If proposed reaction reverses this: E°_cell < 0, non-spontaneous