Electrolysis

Overview

Electrolysis uses electrical energy to drive non-spontaneous redox reactions. It's the opposite of a galvanic cell—electrical energy is converted to chemical energy.

Comparison: Galvanic vs Electrolytic Cells

Feature	Galvanic Cell	Electrolytic Cell
Energy	Chemical → Electrical	Electrical → Chemical
Spontaneity	Spontaneous ($\Delta G < 0$)	Non-spontaneous ($\Delta G > 0$)
$E°_{\text{cell}}$	Positive	Negative
Anode	Negative (-)	Positive (+)
Cathode	Positive (+)	Negative (-)

In Both Types

• Oxidation at anode
• Reduction at cathode
• Electrons flow anode → cathode (external circuit)

Electrolysis of Molten Salts

Example: Molten NaCl

Cathode: $\text{Na}^+ + e^- \rightarrow \text{Na}(l)$ (reduction) Anode: $2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^-$ (oxidation) Overall: $2\text{NaCl}(l) \rightarrow 2\text{Na}(l) + \text{Cl}_2(g)$

Simple case—only ions present are Na⁺ and Cl⁻.

Electrolysis of Aqueous Solutions

More complex—water can also be oxidized or reduced.

Possible Cathode Reactions

$$\text{Metal ion} + e^- \rightarrow \text{Metal} \quad (\text{if } E° > -0.83 \text{ V})$$ $$2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \quad E° = -0.83 \text{ V}$$

Possible Anode Reactions

$$\text{Anion} \rightarrow \text{Product} + e^- \quad (\text{depends on anion})$$ $$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \quad E° = +1.23 \text{ V}$$

Prediction Rules

At Cathode (reduction):

• Active metals (Li, Na, K, Ca, Mg, Al): H₂O reduced to H₂
• Less active metals (Zn and below): Metal deposited

At Anode (oxidation):

• Halides (Cl⁻, Br⁻, I⁻): Halogen produced
• Other anions (NO₃⁻, SO₄²⁻): O₂ from water
• Inert electrode required for O₂

Example: Aqueous NaCl

Cathode: $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ (not Na⁺) Anode: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ (not H₂O) Overall: $2\text{NaCl}(aq) + 2\text{H}_2\text{O} \rightarrow \text{H}_2 + \text{Cl}_2 + 2\text{NaOH}$

Example: Aqueous CuSO₄

Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (Cu deposits) Anode: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ (O₂ evolved) Overall: $2\text{CuSO}_4 + 2\text{H}_2\text{O} \rightarrow 2\text{Cu} + \text{O}_2 + 2\text{H}_2\text{SO}_4$

Faraday's Laws of Electrolysis

First Law

Mass deposited is proportional to charge passed:

$$m \propto Q$$

Second Law

Mass deposited is proportional to molar mass and inversely proportional to number of electrons:

$$m = \frac{Q \times M}{n \times F}$$

Or:

$$m = \frac{I \times t \times M}{n \times F}$$

Where:

• $m$ = mass deposited (g)
• $Q$ = charge (C) = $I \times t$
• $I$ = current (A)
• $t$ = time (s)
• $M$ = molar mass (g/mol)
• $n$ = electrons transferred
• $F$ = Faraday constant = 96,485 C/mol

Calculations

Example 1: Mass from Current and Time

Calculate mass of Cu deposited by 5.0 A for 2.0 hours.

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ ($n = 2$) $M(\text{Cu}) = 63.5$ g/mol

$$t = 2.0 \text{ h} \times 3600 \text{ s/h} = 7200 \text{ s}$$ $$Q = It = 5.0 \times 7200 = 36,000 \text{ C}$$ $$m = \frac{Q \times M}{n \times F} = \frac{36,000 \times 63.5}{2 \times 96,485} = 11.8 \text{ g}$$

Example 2: Time for Given Mass

How long to deposit 10.0 g of silver at 3.0 A?

$\text{Ag}^+ + e^- \rightarrow \text{Ag}$ ($n = 1$) $M(\text{Ag}) = 108$ g/mol

$$\text{mol Ag} = \frac{10.0}{108} = 0.0926 \text{ mol}$$ $$\text{mol } e^- = 0.0926 \text{ mol} \quad (n = 1)$$ $$Q = 0.0926 \times 96,485 = 8935 \text{ C}$$ $$t = \frac{Q}{I} = \frac{8935}{3.0} = 2978 \text{ s} \approx 50 \text{ min}$$

Example 3: Current for Given Rate

What current deposits 5.0 g of aluminum per hour?

$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ ($n = 3$) $M(\text{Al}) = 27.0$ g/mol

$$\text{mol Al} = \frac{5.0}{27.0} = 0.185 \text{ mol}$$ $$\text{mol } e^- = 3 \times 0.185 = 0.556 \text{ mol}$$ $$Q = 0.556 \times 96,485 = 53,600 \text{ C}$$ $$I = \frac{Q}{t} = \frac{53,600}{3600} = 14.9 \text{ A}$$

Industrial Applications

Aluminum Production (Hall-Héroult Process)

Electrolyte: Al₂O₃ dissolved in molten cryolite (Na₃AlF₆) Cathode: $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(l)$ Anode: $\text{C} + \text{O}^{2-} \rightarrow \text{CO}_2 + 4e^-$

• Temperature: ~950°C
• Carbon anodes consumed and replaced

Copper Refining

$$\text{Impure Cu anode} \rightarrow \text{Cu}^{2+} \rightarrow \text{Pure Cu cathode}$$

• Impure copper is oxidized at anode
• Pure copper deposits at cathode
• Impurities fall to bottom as "anode mud"

Chlor-Alkali Process

$$2\text{NaCl}(aq) + 2\text{H}_2\text{O} \rightarrow \text{Cl}_2(g) + \text{H}_2(g) + 2\text{NaOH}(aq)$$

Products: Chlorine, hydrogen, sodium hydroxide

Electroplating

Object as cathode, plating metal as anode Solution contains metal ions

• Chrome plating: $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}$
• Gold plating: $\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$

Summary of Key Equations

Find	Formula
Charge	$Q = It$
Moles of electrons	$\text{mol } e^- = Q/F$
Moles of substance	$\text{mol} = Q/(nF)$
Mass	$m = \frac{QM}{nF} = \frac{ItM}{nF}$
Current	$I = \frac{nFm}{Mt}$
Time	$t = \frac{nFm}{MI}$