What is Galvanic Cells and Standard Cell Potentials?

Understand voltaic cells, half-reactions, standard reduction potentials, and calculating cell voltage.

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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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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

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

Components:

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

❓ 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?

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)

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%	✅

$2$ mol of $H_2$ reacts with $1$ mol of $O_2$ . How many grams of water are produced? Which is the limiting reagent? ( $2H_2 + O_2 \to 2H_2O$ )

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

$E°_{\text{cell}}$	$\Delta G°$	Spontaneous?
Positive	Negative	Yes (galvanic)
Zero	Zero	At equilibrium
Negative	Positive	No (needs energy)

❓ 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?

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)

Mnemonic: "An Ox" (Anode = Oxidation), "Red Cat" (Reduction = Cathode)

❓ 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⁻.

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

Electrons transferred: n = 6

Calculate E°_cell:

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

$E°_{\text{cell}} = -1.66 - (-2.37)$

$E°_{\text{cell}} = -1.66 + 2.37$

$E°_{\text{cell}} = 0.71 \text{ V}$

Calculate ΔG°:

$\Delta G° = -nFE°_{\text{cell}}$

Substitute values:

$\Delta G° = -(6)(96,485)(0.71)$

$\Delta G° = -6 \times 96,485 \times 0.71$

$\Delta G° = -410,858 \text{ J}$

$\Delta G° = -411 \text{ kJ}$

Answer: ΔG° = -411 kJ

Interpretation:

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

Key relationship:

$\Delta G° = -nFE°_{\text{cell}}$

Positive E° → Negative ΔG° → Spontaneous
Negative E° → Positive ΔG° → Non-spontaneous
E° = 0 → ΔG° = 0 → Equilibrium

❓ 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.

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

Interpretation: Large negative ΔG° confirms the reaction is highly spontaneous and product-favored.

❓ 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.

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
E°_cathode = 0.00 V

Calculate E°_cell:

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

$E°_{\text{cell}} = 0.00 - 0.34$

$E°_{\text{cell}} = -0.34 \text{ V}$

E°_cell is negative!

Answer: NO, reaction is NOT spontaneous

Reason: E°_cell = -0.34 V < 0

Explanation:

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)

For this reverse:

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

General rule:

To predict spontaneity:

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

Galvanic Cells and Standard Cell Potentials

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Galvanic Cells and Standard Cell Potentials

Redox Review

Galvanic (Voltaic) Cells

📚 Practice Problems

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⚠️ Common Mistakes: Galvanic Cells and Standard Cell Potentials

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❓ Frequently Asked Questions

Cell Notation (Line Notation)

Standard Reduction Potentials (E°)

Using Standard Reduction Potential Table

Calculating Standard Cell Potential

Predicting Spontaneity

Example Cell: Zn-Cu Cell

Balancing Redox Equations

Intensive vs Extensive Properties

Predicting Reactions

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3Problem 3medium

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