Galvanic Cells and Standard Cell Potentials

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

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
- 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)

Standard format:

$\text{Anode | Anode solution || Cathode solution | Cathode}$

Example: Zn/Zn²⁺ || Cu²⁺/Cu

Single line (|): Phase boundary
Double line (||): Salt bridge
Anode always on left
Cathode always on right

Full example:

Zn(s) | Zn²⁺(aq, 1 M) || Cu²⁺(aq, 1 M) | Cu(s)

Standard Reduction Potentials (E°)

Standard conditions:

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):

$\ce{2H^+(aq) + 2e^- -> H2(g)} \quad E° = 0.00 \text{ V}$

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)

Common half-reactions:

| 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

Formula:

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

Or:

$E°_{\text{cell}} = E°_{\text{red}} - E°_{\text{ox}}$

Steps:

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

From cell potential:

| E°_cell | ΔG° | Spontaneous? | |---------|-----|--------------| | Positive | Negative | Yes (galvanic) | | Zero | Zero | At equilibrium | | Negative | Positive | No (needs energy) |

Relationship to ΔG°:

$\Delta G° = -nFE°_{\text{cell}}$

Where:

n = moles of electrons transferred
F = Faraday constant = 96,485 C/mol e⁻
E°_cell in volts

Example Cell: Zn-Cu Cell

Overall reaction:

$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$

Half-reactions:

Anode (oxidation): Zn → Zn²⁺ + 2e⁻

E° = -0.76 V (reverse of reduction)

Cathode (reduction): Cu²⁺ + 2e⁻ → Cu

E° = +0.34 V

Cell potential:

$E°_{\text{cell}} = 0.34 - (-0.76) = 1.10 \text{ V}$

Positive → spontaneous!

Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Balancing Redox Equations

Half-reaction method:

Separate into oxidation and reduction
Balance atoms (except O and H)
Balance O with H₂O
Balance H with H⁺
Balance charge with e⁻
Multiply to equalize electrons
Add half-reactions
Cancel common terms

In base: Add OH⁻ to neutralize H⁺

Intensive vs Extensive Properties

E° is intensive:

Does NOT depend on amount
Don't multiply E° when balancing
E° same whether 1 electron or 1000 electrons

ΔG° is extensive:

Depends on amount (moles)
Must multiply by n (electrons transferred)

Predicting Reactions

Will reaction occur spontaneously?

Compare E° values:

Species with higher E° gets reduced (cathode)
Species with lower E° gets oxidized (anode)
Calculate E°_cell
If positive → yes!

📚 Practice Problems

1Problem 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:

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

$E°_{\text{cell}} = 0.80 - (-0.76)$

$E°_{\text{cell}} = 0.80 + 0.76$

$E°_{\text{cell}} = 1.56 \text{ V}$

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!

2Problem 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)

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

3Problem 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

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

4Problem 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

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

5Problem 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
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

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