• Contains inert electrolyte (KNO₃, Na₂SO₄)
• Completes circuit

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):

$\text{2H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \quad E° = 0.00 \text{ V}$

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:

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:

1. Identify oxidation and reduction half-reactions
2. Look up E° for each (as reduction)
3. E°_cell = E°_cathode - E°_anode
4. If E°_cell > 0: spontaneous
5. If E°_cell < 0: non-spontaneous

Predicting Spontaneity

From cell potential:

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:

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

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:

1. Separate into oxidation and reduction
2. Balance atoms (except O and H)
3. Balance O with H₂O
4. Balance H with H⁺
5. Balance charge with e⁻
6. Multiply to equalize electrons
7. Add half-reactions
8. 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!

• 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!