FAD1018 Electrochemistry — Reading Note

Purpose: Build deep conceptual understanding of Electrochemistry by connecting ideas across three lectures and the tutorial. Read this before attempting the drill pack.

Big Picture: What is Electrochemistry Really About?

Electrochemistry is the study of electron transfer between species. Strip away the jargon, and it's about one thing: which species wants electrons more badly.

Think of it as a tug-of-war. Every species has a certain "electron hunger" (its reduction potential $E°$). The species with the bigger hunger pulls electrons away from the weaker one. When you put them in a circuit, electrons flow from the weaker to the stronger — that's your current. The difference in hunger is your voltage.

The Central Concept: Reduction Potential

Every half-cell has a standard reduction potential ($E°$) — a number that tells you how strongly that species wants to be reduced (gain electrons).

If $E°$ is... The species is a... It wants to...
More positive Stronger oxidising agent Gain electrons (be reduced)
More negative Stronger reducing agent Lose electrons (be oxidised)

[!important] The Counter-Intuitive Insight The half-reaction with the more negative $E°$ is the one that gets oxidised (the anode). This seems backwards — why does the "weaker" species oxidise? Because it has less desire to hold onto its electrons, so it gives them up more easily. The "stronger" species ($E°$ more positive) is more hungry and wins the tug-of-war — it gets reduced at the cathode.

Your mental shortcut:

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

  1. Find the two half-reactions and their $E°$ values
  2. The more positive $E°$ → cathode (reduction)
  3. The more negative $E°$ → anode (oxidation)
  4. Subtract: $E°_{cell} = \text{more positive} - \text{more negative}$

If the result is positive ($E°_{cell} > 0$), the reaction is spontaneous — it's a galvanic cell.

Understanding Galvanic vs Electrolytic

Heres the single most important distinction:

Galvanic Electrolytic
Who supplies energy? The chemicals themselves An external power source
$E°_{cell}$ Positive (spontaneous) Negative (non-spontaneous)
Anode polarity Negative (−) Positive (+)
Real-world example Battery powering a device Charging a battery

[!tip] Why does anode polarity flip? In both cells, oxidation happens at the anode. But in a galvanic cell, the anode produces electrons (so it's negative). In an electrolytic cell, the anode pulls electrons out (so it's positive — the power source is forcing the oxidation).

Mnemonic: GAL-AN — GALvanic → ANode is Negative. Electrolytic is the opposite.

The Driving Force: Why Do Electrons Flow?

Electrons flow because of a potential difference — exactly like water flows downhill. The bigger the difference in $E°$ between the two half-cells, the bigger the driving force.

At equilibrium ($E_{cell} = 0$), both forward and reverse reactions happen at equal rates. No net flow. This is what happens when a battery "dies" — it's not out of chemicals, it's at equilibrium. The chemical reaction has run until the concentrations make $E_{cell} = 0$.

The Nernst equation quantifies exactly this:

$$E_{cell} = E°_{cell} - \frac{0.0592}{n} \log Q$$

The $\log Q$ term is the "reaction progress penalty". As the reaction progresses:

  • Reactants decrease → $Q$ increases → $\log Q$ increases → $E_{cell}$ drops
  • Eventually $Q = K$ → $\log Q = \log K$ → $E_{cell} = 0$ → dead battery

Why $\Delta G = -nFE$ Matters

This equation connects thermodynamics (Gibbs free energy) to electrochemistry (cell potential):

$$\Delta G° = -nFE°_{cell}$$

The negative sign is the key:

  • $E°_{cell} > 0$ → $\Delta G° < 0$ → spontaneous
  • $E°_{cell} < 0$ → $\Delta G° > 0$ → non-spontaneous (needs an electrolytic cell)

You can go further: $$E°_{cell} = \frac{0.0592}{n} \log K \quad \text{(at 25°C)}$$

So the cell potential tells you the equilibrium constant directly. A $1\ \text{V}$ cell with $n = 2$ has $\log K \approx 34$ → $K \approx 10^{34}$ — that reaction is practically irreversible.

Electrolysis: Forcing the Unnatural

When you electrolyse something, you're running a reaction backwards — against its natural direction. You're supplying external electrical energy to force a non-spontaneous reaction.

Selective Discharge: Who Reacts First?

In an aqueous solution, you don't just have the dissolved ions — you also have $\text{H}^+$ and $\text{OH}^-$ from water autoprotolysis. When multiple cations and anions are present, only one of each gets discharged. Priority is determined by three factors:

1. Electrochemical Series (E° values) — The primary factor. Species with more positive $E°$ are reduced first (cations); species with more negative $E°$ are oxidised first (anions — but remember oxidation reverses the sign logic).

2. Concentration — High concentration can override the electrochemical series. If you have a lot of $\text{Cl}^-$ in solution, it can be oxidised before $\text{OH}^-$ even though $\text{OH}^-$ is "easier" based on $E°$ alone. (This is a common exam trap.)

3. Electrode type — Active electrodes (Cu, Ag) participate in the reaction and may dissolve. Inert electrodes (Pt, graphite) just provide a surface.

[!question] Self-Check: What products form when you electrolyse dilute aqueous NaCl? Ions present: $\text{Na}^+$, $\text{Cl}^-$, $\text{H}^+$, $\text{OH}^-$

At the cathode: $\text{Na}^+$ has $E° = -2.71\ \text{V}$, $\text{H}^+$ has $E° = 0.00\ \text{V}$. $\text{H}^+$ is more positive → $\text{H}_2$ gas is produced.

At the anode: Without overpotential, $\text{OH}^-$ ($E° = -0.40\ \text{V}$) would oxidise before $\text{Cl}^-$ ($E° = -1.36\ \text{V}$), producing $\text{O}_2$. But oxygen overvoltage adds ~0.5 V, making $\text{Cl}^-$ oxidation more favourable. Result: $\text{Cl}_2$ gas is produced.

Products: $\text{H}_2$ at cathode, $\text{Cl}_2$ at anode, $\text{NaOH}$ remains in solution.

If you answered $\text{Na}$ metal at the cathode — that only happens in molten NaCl (Downs Cell).

Overpotential: The Hidden Factor

Overpotential is the single most commonly missed concept on electrolysis exams. Here's what it means:

The standard electrode potential tells you the minimum voltage needed. But in reality, some reactions (especially those producing gases) need extra voltage. The extra is the overpotential.

Why does oxygen have high overpotential? Forming $\text{O}_2$ gas at an electrode surface involves multiple electron-transfer steps and the formation of $\text{O}$—$\text{O}$ bonds. It's kinetically sluggish. Bubbles of $\text{O}_2$ also physically block the electrode surface, reducing the active area.

The practical consequence: $\text{Cl}^-$, $\text{Br}^-$, and $\text{I}^-$ can be oxidised at the anode before water, even though their standard potentials suggest water should go first. The overpotential effectively "shifts" water's discharge potential.

[!tip] Exam Pattern Recognition If a question says "aqueous NaCl" and asks for anode product → the answer is $\text{Cl}_2$ (not $\text{O}_2$) because of overpotential. If it says "dilute aqueous $\text{Na}_2\text{SO}_4$" → $\text{SO}_4^{2-}$ has very high overpotential and won't discharge; $\text{OH}^-$ is oxidised to $\text{O}_2$ instead.

Faraday's Laws: The Quantitative Side

Faraday's laws connect mass to electrical charge. This is the calculation-heavy part of electrochemistry.

The unified formula:

$$m = \frac{ItM}{nF}$$

Every variable has a clear physical meaning:

  • $I$ (current) × $t$ (time) = total charge that passed through the cell
  • $M$ (molar mass) = how heavy one mole of the substance is
  • $n$ (electrons per ion) = how many electrons each ion needs; for $\text{Cu}^{2+}$ it's 2, for $\text{Ag}^+$ it's 1
  • $F$ = Faraday constant = charge of one mole of electrons

Think of it as: total electrons passed → moles of electrons → moles of substance → mass of substance.

[!question] Self-Check: Copper Electroplating A current of 2.0 A is passed through $\text{CuSO}_4(aq)$ for 30 minutes. How much Cu is deposited?

$Q = It = 2.0\ \text{A} \times 30 \times 60\ \text{s} = 3600\ \text{C}$ $m = \dfrac{3600 \times 63.5}{2 \times 96500} = 1.18\ \text{g}$

Notice: $n=2$ because $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$. Getting $n$ wrong is the single biggest mistake students make.

Industrial Applications: Why We Care

Downs Cell (Na from NaCl)

Electrolysis of molten NaCl (not aqueous!) with CaCl₂ added to lower melting point. The key insight: you must use molten salt because in aqueous solution, $\text{H}^+$ (not $\text{Na}^+$) gets reduced at the cathode.

Hall Process (Al from Al₂O₃)

Aluminium oxide dissolved in molten cryolite ($\text{Na}_3\text{AlF}_6$) at ~950°C. The carbon anodes are consumed: $\text{C} + 2\text{O}^{2-} \rightarrow \text{CO}_2 + 4e^-$. This makes aluminium extraction expensive — you're burning carbon and electricity.

Electroplating

Metal cations are reduced at the cathode, depositing as a thin layer. The anode is usually the plating metal (active electrode). Example: silver plating uses an Ag anode that dissolves to replenish $\text{Ag}^+$ in solution.

Connecting the Dots: One Unified Framework

Here is how everything connects:

Redox fundamentals (OIL RIG, half-equations)
    ↓
Cell types: Galvanic (spontaneous) vs Electrolytic (forced)
    ↓
Standard potentials (E°, SHE) → calculate E°cell
    ↓
Nernst equation → concentration effects → when does cell die?
    ↓
ΔG = -nFE → thermodynamics → spontaneity & equilibrium
    ↓
Electrolysis → selective discharge (E°, conc, electrode type) → overpotential
    ↓
Faraday's laws → quantitative (mass = f(current, time))
    ↓
Industrial applications (Downs, Hall, electroplating)

Common Exam Traps

Trap What students do wrong Correct thinking
Anode polarity Think anode is always negative Galvanic: anode (−), Electrolytic: anode (+)
Aqueous vs molten Apply same rules to both Molten = only salt ions present; Aqueous = H⁺/OH⁻ compete
Overpotential Ignore it, predict O₂ for all aqueous Cl⁻, Br⁻, I⁻ can oxidise before water due to high O₂ overvoltage
Faraday's $n$ Use wrong number of electrons Count e⁻ in balanced half-equation: Ag⁺ needs 1, Cu²⁺ needs 2, Al³⁺ needs 3
Nernst sign Forget the minus, add instead of subtract $E = E° - \frac{0.0592}{n}\log Q$ — always subtract
Dead battery Think chemicals are used up It's at equilibrium: $E_{cell} = 0$, forward = reverse rate

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