Electrochemistry | IB - Wyatt's Notes

1. Redox Review (SL Topic 9)

Oxidation and Reduction

Definition. A redox reaction is a reaction in which electrons are transferred between Species. Oxidation is the loss of electrons; reduction is the gain of electrons.

Mnemonic: OIL RIG — Oxidation Is Loss, Reduction Is Gain.

Definition. An oxidizing agent is a species that causes oxidation in another species by Accepting electrons (it is itself reduced). A reducing agent is a species that causes reduction In another species by donating electrons (it is itself oxidized).

Oxidation States (Oxidation Numbers)

Rules for assigning oxidation states:

The oxidation state of an element in its standard state is zero (e.g., $\mathrm{O}_2$ $\mathrm{Fe}$ , $\mathrm{Cl}_2$ ).
For a monatomic ion, the oxidation state equals the charge (e.g., $\mathrm{Na}^+$ is $+1$ $\mathrm{O}^{2-}$ is $-2$ ).
Oxygen is $-2$ Except in peroxides ( $-1$ ) and with fluorine ( $+2$ ).
Hydrogen is $+1$ Except in metal hydrides where it is $-1$ .
The sum of oxidation states in a neutral compound is zero; in a polyatomic ion it equals the ion charge.
Fluorine is always $-1$ in compounds.
Group 1 metals are $+1$ ; Group 2 metals are $+2$ in compounds.

Identifying Redox Reactions

A reaction is redox if there is a change in oxidation state of any element. To identify:

Assign oxidation states to all elements in reactants and products.
Identify which elements change oxidation state.
The element that increases in oxidation state is oxidized.
The element that decreases in oxidation state is reduced.

Writing Half-Equations

Half-equations separate the oxidation and reduction processes. Steps:

Write the skeletal equation for the species involved.
Balance all atoms except $\mathrm{O}$ and $\mathrm{H}$ .
Balance $\mathrm{O}$ by adding $\mathrm{H}_2\mathrm{O}$ .
Balance $\mathrm{H}$ by adding $\mathrm{H}^+$ .
Balance charge by adding electrons ( $e^-$ ).

Balancing Redox in Acidic Conditions

Example: Balance $\mathrm{MnO}_4^- + \mathrm{Fe}^{2+} \to \mathrm{Mn}^{2+} + \mathrm{Fe}^{3+}$ In acidic solution.

Reduction half-reaction:

\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}

Oxidation half-reaction:

\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+} + e^-

Multiply the oxidation half-reaction by 5 to balance electrons:

5\mathrm{Fe}^{2+} \to 5\mathrm{Fe}^{3+} + 5e^-

Add both half-reactions:

\mathrm{MnO}_4^- + 5\mathrm{Fe}^{2+} + 8\mathrm{H}^+ \to \mathrm{Mn}^{2+} + 5\mathrm{Fe}^{3+} + 4\mathrm{H}_2\mathrm{O}

Balancing Redox in Basic Conditions

Follow the same steps as acidic conditions, then add $\mathrm{OH}^-$ to both sides to neutralize $\mathrm{H}^+$ :

Balance as if in acidic conditions.
Add $\mathrm{OH}^-$ to both sides equal to the number of $\mathrm{H}^+$ .
Combine $\mathrm{H}^+$ and $\mathrm{OH}^-$ to form $\mathrm{H}_2\mathrm{O}$ .
Cancel any $\mathrm{H}_2\mathrm{O}$ that appears on both sides.

Example: Balance $\mathrm{CrO}_4^{2-} + \mathrm{SO}_3^{2-} \to \mathrm{Cr(OH)}_3 + \mathrm{SO}_4^{2-}$ in basic Solution.

Step 1 — Balance in acidic conditions:

\mathrm{CrO}_4^{2-} + 4\mathrm{H}^+ + 3e^- \to \mathrm{Cr(OH)}_3 + \mathrm{H}_2\mathrm{O}

\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \to \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2e^-

Step 2 — Multiply to balance electrons (LCM of 3 and 2 is 6):

2\mathrm{CrO}_4^{2-} + 8\mathrm{H}^+ + 6e^- \to 2\mathrm{Cr(OH)}_3 + 2\mathrm{H}_2\mathrm{O}

3\mathrm{SO}_3^{2-} + 3\mathrm{H}_2\mathrm{O} \to 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+ + 6e^-

Step 3 — Add:

2\mathrm{CrO}_4^{2-} + 3\mathrm{SO}_3^{2-} + 5\mathrm{H}_2\mathrm{O} \to 2\mathrm{Cr(OH)}_3 + 3\mathrm{SO}_4^{2-} + 4\mathrm{H}_2\mathrm{O}

Simplify:

2\mathrm{CrO}_4^{2-} + 3\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \to 2\mathrm{Cr(OH)}_3 + 3\mathrm{SO}_4^{2-}

IB Exam Tip When the question specifies acidic or basic conditions, you MUST show the balanced half-equations Separately before combining them. Showing working is essential for full marks.

2. Standard Electrode Potentials

The Standard Hydrogen Electrode (SHE)

Definition. The standard hydrogen electrode is the reference electrode against which all Other electrode potentials are measured. It consists of:

A platinum electrode coated with platinum black (fine Pt particles for catalytic surface).
Hydrogen gas at $100\mathrm{ kPa}$ bubbling over the electrode.
A solution of $\mathrm{H}^+$ (aq) at $1.0\mathrm{ mol/dm}^3$ ( $298\mathrm{ K}$ ).
Defined potential: $E^\circ = 0.00\mathrm{ V}$ .

The half-reaction is:

2\mathrm{H}^+(aq, 1\mathrm{ M}) + 2e^- \rightleftharpoons \mathrm{H}_2(g, 100\mathrm{ kPa}) \quad E^\circ = 0.00\mathrm{ V}

Standard Conditions

All standard electrode potentials ( $E^\circ$ ) are measured under:

Temperature: $298\mathrm{ K}$ ( $25\degree\mathrm{C}$ )
Pressure: $100\mathrm{ kPa}$ for all gases
Concentration: $1.0\mathrm{ mol/dm}^3$ for all aqueous species
Solid elements in their standard states

Standard Reduction Potentials Table

The electrochemical series ranks species by their tendency to be reduced. Some key values:

Half-Reaction	$E^\circ$ (V)
$\mathrm{F}_2 + 2e^- \rightleftharpoons 2\mathrm{F}^-$	$+2.87$
$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightleftharpoons \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$	$+1.51$
$\mathrm{Cl}_2 + 2e^- \rightleftharpoons 2\mathrm{Cl}^-$	$+1.36$
$\mathrm{Ag}^+ + e^- \rightleftharpoons \mathrm{Ag}$	$+0.80$
$\mathrm{Cu}^{2+} + 2e^- \rightleftharpoons \mathrm{Cu}$	$+0.34$
$2\mathrm{H}^+ + 2e^- \rightleftharpoons \mathrm{H}_2$	$0.00$
$\mathrm{Fe}^{2+} + 2e^- \rightleftharpoons \mathrm{Fe}$	$-0.44$
$\mathrm{Zn}^{2+} + 2e^- \rightleftharpoons \mathrm{Zn}$	$-0.76$
$\mathrm{Na}^+ + e^- \rightleftharpoons \mathrm{Na}$	$-2.71$
$\mathrm{Li}^+ + e^- \rightleftharpoons \mathrm{Li}$	$-3.04$

Definition. A more positive $E^\circ$ value indicates a greater tendency for the species to be Reduced (stronger oxidizing agent). A more negative $E^\circ$ indicates a greater tendency for the Species to be oxidized (stronger reducing agent).

Measuring Standard Electrode Potential

To measure the standard electrode potential of a half-cell, connect it to the SHE and measure the Cell potential using a high-resistance voltmeter (so negligible current flows).

If the half-cell is the cathode (reduction occurs): $E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} = E^\circ_{\mathrm{half-cell}} - 0.00$
If the half-cell is the anode (oxidation occurs): $E^\circ_{\mathrm{cell}} = 0.00 - E^\circ_{\mathrm{half-cell}}$

Predicting Spontaneity

For a redox reaction to be spontaneous under standard conditions:

E^\circ_{\mathrm{cell}} \gt 0

This means the reduction half-reaction (cathode) must have a more positive $E^\circ$ than the Oxidation half-reaction (anode).

Definition. A spontaneous redox reaction will occur when a species with a more positive Reduction potential is paired with a species with a more negative reduction potential.

Common Mistake $E^\circ$ values are intensive properties. They are NOT multiplied by stoichiometric coefficients When calculating $E^\circ_{\mathrm{cell}}$ . Do not multiply $E^\circ$ values by the number of Electrons transferred.

3. Electrochemical Cells (Galvanic/Voltaic Cells)

Overview

Definition. A galvanic (voltaic) cell converts chemical energy from spontaneous redox Reactions into electrical energy. Electrons flow from the anode to the cathode through an external Circuit.

Cell Components

Component	Function
Anode	Electrode where oxidation occurs. Electrons are released. Negative electrode in galvanic cells.
Cathode	Electrode where reduction occurs. Electrons are consumed. Positive electrode in galvanic cells.
Salt bridge	Completes the internal circuit by allowing ion migration. Maintains electrical neutrality. Contains an inert electrolyte (e.g., $\mathrm{KNO}_3$ (aq), $\mathrm{KCl}$ (aq)).
External circuit	Wire connecting the two electrodes through which electrons flow.
Voltmeter	Measures the potential difference between the two half-cells.

Salt Bridge

The salt bridge prevents charge buildup in each half-cell. Without it, the cell would quickly stop Functioning as excess charge accumulates. The ions in the salt bridge migrate to neutralize charge:

Anions ( $\mathrm{NO}_3^-$ , $\mathrm{Cl}^-$ ) migrate toward the anode (to balance $\mathrm{M}^{n+}$ produced).
Cations ( $\mathrm{K}^+$ ) migrate toward the cathode (to balance the loss of $\mathrm{M}^{n+}$ consumed).

Definition. The salt bridge must contain ions that do not react with the cell components. $\mathrm{KNO}_3$ is commonly used because $\mathrm{K}^+$ and $\mathrm{NO}_3^-$ are spectator ions For most systems.

Cell Diagram Notation (SL)

The IUPAC convention for representing galvanic cells:

\mathrm{Anode} \mid \mathrm{oxidized species} \mid \mathrm{reduced species} \parallel \mathrm{oxidized species} \mid \mathrm{reduced species} \mid \mathrm{Cathode}

Conventions:

The anode (oxidation) is written on the left.
The cathode (reduction) is written on the right.
A single vertical line $\mid$ represents a phase boundary.
A double vertical line $\parallel$ represents the salt bridge.
If there are multiple species in the same phase, they are separated by commas.
Inert electrodes (e.g., $\mathrm{Pt}$ , $\mathrm{C}$ ) are included when the half-cell does not involve a solid metal.

Example: $\mathrm{Zn}\mathrm{-Cu}$ Daniell cell:

\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Cu}^{2+}(aq) \mid \mathrm{Cu}(s)

Example: $\mathrm{Fe}^{2+}$ / $\mathrm{Fe}^{3+}$ with $\mathrm{MnO}_4^-$ / $\mathrm{Mn}^{2+}$ :

\mathrm{Pt}(s) \mid \mathrm{Fe}^{2+}(aq),\mathrm{Fe}^{3+}(aq) \parallel \mathrm{MnO}_4^-(aq),\mathrm{H}^+(aq),\mathrm{Mn}^{2+}(aq) \mid \mathrm{Pt}(s)

Calculating EMF (SL)

The standard cell potential (electromotive force) is:

E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}}

Or equivalently:

E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{reduction}} - E^\circ_{\mathrm{oxidation}}

Example: Calculate $E^\circ_{\mathrm{cell}}$ for $\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Cu}^{2+}(aq) \mid \mathrm{Cu}(s)$ :

Cathode (reduction): $\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$ , $E^\circ = +0.34\mathrm{ V}$
Anode (oxidation): $\mathrm{Zn} \to \mathrm{Zn}^{2+} + 2e^-$ , $E^\circ = -0.76\mathrm{ V}$

E^\circ_{\mathrm{cell}} = 0.34 - (-0.76) = +1.10\mathrm{ V}

Since $E^\circ_{\mathrm{cell}} \gt 0$ The reaction is spontaneous.

Anode/Cathode Sign Convention

In a galvanic cell:

Feature	Anode	Cathode
Process	Oxidation	Reduction
Electron flow	Electrons leave	Electrons arrive
Sign in galvanic cell	Negative ( $-$ )	Positive ( $+$ )
Ion migration	Anions migrate toward	Cations migrate toward

IB Exam Tip When asked to predict whether a reaction is spontaneous, always:

Identify the two relevant half-reactions from the data booklet.
Assign cathode (more positive $E^\circ$ ) and anode (more negative $E^\circ$ ).
Calculate $E^\circ_{\mathrm{cell}}$ .
If $E^\circ_{\mathrm{cell}} \gt 0$ The reaction is spontaneous.

4. Electrolytic Cells

Overview

Definition. An electrolytic cell uses electrical energy from an external power supply to Drive a non-spontaneous redox reaction. The external source forces electrons to flow in the opposite Direction to what would occur spontaneously.

Key Differences: Galvanic vs. Electrolytic

Feature	Galvanic Cell	Electrolytic Cell
Energy conversion	Chemical to electrical	Electrical to chemical
Spontaneity	Spontaneous ( $E^\circ_{\mathrm{cell}} \gt 0$ )	Non-spontaneous ( $E^\circ_{\mathrm{cell}} \lt 0$ )
Anode sign	Negative	Positive
Cathode sign	Positive	Negative
External power supply	Not required	Required
Salt bridge	Yes	Not required (single cell)

Electrolysis of Molten Compounds

When a molten ionic compound is electrolyzed, the cation is reduced at the cathode and the anion is Oxidized at the anode.

Example: Electrolysis of molten $\mathrm{NaCl}$ :

Cathode: $\mathrm{Na}^+ + e^- \to \mathrm{Na}(l)$
Anode: $2\mathrm{Cl}^- \to \mathrm{Cl}_2(g) + 2e^-$
Overall: $2\mathrm{NaCl}(l) \to 2\mathrm{Na}(l) + \mathrm{Cl}_2(g)$

Example: Electrolysis of molten $\mathrm{Al}_2\mathrm{O}_3$ :

Cathode: $\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}(l)$
Anode: $2\mathrm{O}^{2-} \to \mathrm{O}_2(g) + 4e^-$

Electrolysis of Aqueous Solutions

In aqueous solutions, both the cation and $\mathrm{H}^+$ can be reduced at the cathode. Both the Anion and $\mathrm{H}_2\mathrm{O}$ can be oxidized at the anode. The species that is discharged Depends on the relative ease of discharge (electrode potential and overpotential).

Discharge at the Cathode (Reduction)

Cation present	Species discharged	Reason
Group 1 ( $\mathrm{Li}^+$$\mathrm{Na}^+$$\mathrm{K}^+$ )	$\mathrm{H}_2\mathrm{O}$ (producing $\mathrm{H}_2$ )	These cations are very hard to reduce ( $E^\circ$ very negative)
Group 2 ( $\mathrm{Mg}^{2+}$$\mathrm{Ca}^{2+}$ )	$\mathrm{H}_2\mathrm{O}$ (producing $\mathrm{H}_2$ )	Still too hard to reduce in aqueous solution
Transition metals and below ( $\mathrm{Cu}^{2+}$$\mathrm{Ag}^+$$\mathrm{Au}^+$ )	Metal cation	These are easier to reduce than water

The reduction of water at the cathode:

2\mathrm{H}_2\mathrm{O}(l) + 2e^- \to \mathrm{H}_2(g) + 2\mathrm{OH}^-(aq)

Discharge at the Anode (Oxidation)

Anion present	Species discharged	Reason
Halides ( $\mathrm{Cl}^-$$\mathrm{Br}^-$$\mathrm{I}^-$ )	Halide ion (except with concentrated $\mathrm{Cl}^-$ )	Easier to oxidize than water (except $\mathrm{F}^-$ )
$\mathrm{SO}_4^{2-}$$\mathrm{NO}_3^-$$\mathrm{CO}_3^{2-}$	$\mathrm{H}_2\mathrm{O}$ (producing $\mathrm{O}_2$ )	These anions are not discharged in aqueous solution

The oxidation of water at the anode:

4\mathrm{H}_2\mathrm{O}(l) \to \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^-

Note on $\mathrm{Cl}^-$ discharge: In dilute $\mathrm{NaCl}$ solution, water is oxidized Preferentially. In concentrated $\mathrm{NaCl}$ solution, $\mathrm{Cl}^-$ is discharged due to high Concentration and overpotential effects on $\mathrm{O}_2$ evolution.

Preferential Discharge Summary

Definition. Preferential discharge is the rule that determines which ion is discharged at an Electrode when multiple species are present. On:

The position in the electrochemical series (standard reduction potential).
The concentration of ions (higher concentration favours discharge).
The nature of the electrode (e.g., carbon vs. Mercury).

IB Exam Tip For IB exams, use the simplified rules:

At the cathode: if the metal is below aluminium in the reactivity series, the metal is deposited; otherwise, hydrogen gas is produced.
At the anode: if halide ions are present (except fluoride), the halogen is produced; otherwise, oxygen gas from water oxidation is produced.

5. Electroplating and Industrial Applications

Electroplating

Definition. Electroplating is the process of depositing a thin layer of metal onto the Surface of another object (the cathode) by electrolysis.

The object to be plated is made the cathode. The anode is made of the plating metal. The electrolyte Contains ions of the plating metal.

Example: Electroplating silver onto a spoon:

Cathode (spoon): $\mathrm{Ag}^+(aq) + e^- \to \mathrm{Ag}(s)$
Anode (silver bar): $\mathrm{Ag}(s) \to \mathrm{Ag}^+(aq) + e^-$
Electrolyte: $\mathrm{AgNO}_3$ (aq) or $\mathrm{K}[\mathrm{Ag(CN)}_2]$ (aq)

As silver deposits on the spoon, the silver anode dissolves to maintain the silver ion concentration In solution.

Factors Affecting Electroplating Quality

Factor	Effect
Current density	Too high causes rough, brittle deposits; too low causes slow, uneven plating
Temperature	Higher temperature increases ion mobility but may cause uneven deposits
Ion concentration	Higher concentration generally improves deposit quality
Electrolyte composition	Adding complexing agents (e.g., cyanide) improves smoothness
Cathode preparation	Surface must be clean and free of oxides

Electrorefining of Copper

Copper is purified by electrolysis using impure copper as the anode and pure copper as the cathode:

Anode (impure Cu): $\mathrm{Cu}(s) \to \mathrm{Cu}^{2+}(aq) + 2e^-$
Cathode (pure Cu): $\mathrm{Cu}^{2+}(aq) + 2e^- \to \mathrm{Cu}(s)$
Electrolyte: $\mathrm{CuSO}_4$ (aq) with dilute $\mathrm{H}_2\mathrm{SO}_4$

Impurities settle as anode sludge beneath the anode. This sludge is economically significant Because it contains precious metals (Ag, Au, Pt).

Aluminum Extraction — Hall-Heroult Process (HL)

Definition. The Hall-Heroult process is the industrial method for extracting aluminum from Alumina ( $\mathrm{Al}_2\mathrm{O}_3$ ) by electrolysis. Alumina is dissolved in molten cryolite ( $\mathrm{Na}_3\mathrm{AlF}_6$ ) at approximately $950\degree\mathrm{C}$ .

Cryolite is used to:

Lower the melting point of $\mathrm{Al}_2\mathrm{O}_3$ from $2050\degree\mathrm{C}$ to approximately $950\degree\mathrm{C}$ .
Increase the conductivity of the molten mixture.
Reduce energy costs significantly.

Reactions:

Cathode (carbon lining): $\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}(l)$
Anode (carbon): $2\mathrm{O}^{2-} + \mathrm{C}(s) \to \mathrm{CO}_2(g) + 4e^-$
Overall: $2\mathrm{Al}_2\mathrm{O}_3(l) + 3\mathrm{C}(s) \to 4\mathrm{Al}(l) + 3\mathrm{CO}_2(g)$

The carbon anodes are consumed and must be replaced periodically.

Chlor-Alkali Process (HL)

Definition. The chlor-alkali process is the electrolysis of concentrated aqueous sodium Chloride to produce chlorine gas, hydrogen gas, and sodium hydroxide.

Using a membrane cell:

Cathode: $2\mathrm{H}_2\mathrm{O}(l) + 2e^- \to \mathrm{H}_2(g) + 2\mathrm{OH}^-(aq)$
Anode: $2\mathrm{Cl}^-(aq) \to \mathrm{Cl}_2(g) + 2e^-$
Overall: $2\mathrm{NaCl}(aq) + 2\mathrm{H}_2\mathrm{O}(l) \to \mathrm{Cl}_2(g) + \mathrm{H}_2(g) + 2\mathrm{NaOH}(aq)$

The ion-exchange membrane allows $\mathrm{Na}^+$ to pass but prevents $\mathrm{OH}^-$ from reaching The anode, which would react with $\mathrm{Cl}_2$ to form hypochlorite.

Comparison of chlor-alkali cell types:

Feature	Membrane Cell	Diaphragm Cell	Mercury Cell
$\mathrm{NaOH}$ purity	High (up to 50%)	Lower (~11%)	Very high (50%)
Energy consumption	Moderate	Higher	Highest
Environmental impact	Low	Moderate	High (Hg pollution)
Current usage	Primary method	Declining	Phased out

IB Exam Tip The membrane cell is the preferred method for the chlor-alkali process. Know the half-equations at Each electrode and the purpose of the membrane.

6. Nernst Equation (HL)

Derivation from Thermodynamics

The standard Gibbs free energy change relates to the cell potential by:

\Delta G = -nFE_{\mathrm{cell}}

Under non-standard conditions, the Gibbs free energy is given by:

\Delta G = \Delta G^\circ + RT\ln Q

Substituting $\Delta G = -nFE_{\mathrm{cell}}$ and $\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}$ :

-nFE_{\mathrm{cell}} = -nFE^\circ_{\mathrm{cell}} + RT\ln Q

Dividing both sides by $-nF$ :

\boxed{E_{\mathrm{cell}} = E^\circ_{\mathrm{cell}} - \frac`\{RT}``\{nF}`\ln Q}

This is the Nernst equation. At $298\mathrm{ K}$ Substituting $R = 8.314\mathrm{ J/(mol K)}$ $F = 96485\mathrm{ C/mol}$ And converting to $\log_{10}$ :

E_{\mathrm{cell}} = E^\circ_{\mathrm{cell}} - \frac{0.0592}{n}\log_{10} Q

Where:

$E_{\mathrm{cell}}$ = cell potential under non-standard conditions (V)
$E^\circ_{\mathrm{cell}}$ = standard cell potential (V)
$R$ = universal gas constant ( $8.314\mathrm{ J/(mol K)}$ )
$T$ = temperature (K)
$n$ = number of moles of electrons transferred
$F$ = Faraday constant ( $96485\mathrm{ C/mol}$ )
$Q$ = reaction quotient

The Reaction Quotient $Q$

For the general reaction:

A\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}

Q = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

Solids and pure liquids are omitted from $Q$ .

Effect of Concentration on Cell Potential

From the Nernst equation, increasing the concentration of reactants (or decreasing products) Increases $E_{\mathrm{cell}}$ While decreasing reactant concentration (or increasing products) Decreases $E_{\mathrm{cell}}$ .

Example: For the cell $\mathrm{Zn} \mid \mathrm{Zn}^{2+} \parallel \mathrm{Cu}^{2+} \mid \mathrm{Cu}$ :

Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]}

If $[\mathrm{Cu}^{2+}]$ increases, $Q$ decreases, and $E_{\mathrm{cell}}$ increases (the reaction is More spontaneous).

Equilibrium Connection

At equilibrium, $\Delta G = 0$ So $E_{\mathrm{cell}} = 0$ and $Q = K$ (the equilibrium constant). Substituting into the Nernst equation:

0 = E^\circ_{\mathrm{cell}} - \frac`\{RT}``\{nF}`\ln K

E^\circ_{\mathrm{cell}} = \frac`\{RT}``\{nF}`\ln K = \frac{0.0592}{n}\log_{10} K \quad \mathrm{at } 298\mathrm{ K}

Definition. At equilibrium, the cell potential is zero. The standard cell potential directly Determines the equilibrium constant.

IB Exam Tip A useful relationship: a larger positive $E^\circ_{\mathrm{cell}}$ means a larger equilibrium Constant $K$ Meaning the reaction proceeds further to completion. For $E^\circ_{\mathrm{cell}} \gt 0.3\mathrm{ V}$ (approximately), $K \gt 1$ And the reaction can be Considered to go essentially to completion.

Nernst Equation for Half-Cells

The Nernst equation can be applied to individual half-cells:

E = E^\circ - \frac{0.0592}{n}\log_{10}\frac{[\mathrm{reduced form}]}{[\mathrm{oxidized form}]}

Example: For $\mathrm{Fe}^{3+} + e^- \rightleftharpoons \mathrm{Fe}^{2+}$ :

E = E^\circ - \frac{0.0592}{1}\log_{10}\frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}

If $[\mathrm{Fe}^{2+}] \gt [\mathrm{Fe}^{3+}]$ Then $E \lt E^\circ$ (less tendency to be reduced).

Common Mistake When applying the Nernst equation to a half-cell, ensure the reduced form is in the numerator and The oxidized form is in the denominator. Reversing this will give the wrong answer.

7. Standard Free Energy and Electrochemistry (HL)

$\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}$

Definition. The standard Gibbs free energy change of a redox reaction is related to the standard Cell potential by:

\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}

Where:

$\Delta G^\circ$ = standard Gibbs free energy change (J)
$n$ = number of moles of electrons transferred in the balanced equation
$F$ = Faraday constant ( $96485\mathrm{ C/mol}$ )
$E^\circ_{\mathrm{cell}}$ = standard cell potential (V)

Units check: $n$ is dimensionless (mol of $e^-$ ), $F$ has units of $\mathrm{C/mol}$ $E^\circ_{\mathrm{cell}}$ has units of $\mathrm{J/C}$ (since $\mathrm{V} = \mathrm{J/C}$ ). So:

\Delta G^\circ = (\mathrm{mol})(\mathrm{C/mol})(\mathrm{J/C}) = \mathrm{J}

Example: For $\mathrm{Zn} \mid \mathrm{Zn}^{2+} \parallel \mathrm{Cu}^{2+} \mid \mathrm{Cu}$ Where $n = 2$ and $E^\circ_{\mathrm{cell}} = 1.10\mathrm{ V}$ :

\Delta G^\circ = -2 \times 96485 \times 1.10 = -212\,267\mathrm{ J} = -212\mathrm{ kJ/mol}

The negative value confirms the reaction is spontaneous.

Relationship to Equilibrium Constant

Combining $\Delta G^\circ = -RT\ln K$ with $\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}$ :

-nFE^\circ_{\mathrm{cell}} = -RT\ln K

\boxed{E^\circ_{\mathrm{cell}} = \frac`\{RT}``\{nF}`\ln K}

At $298\mathrm{ K}$ :

E^\circ_{\mathrm{cell}} = \frac{0.0592}{n}\log_{10} K

Rearranging:

\log_{10} K = \frac{nE^\circ_{\mathrm{cell}}}{0.0592}

Example: For the reaction $\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s)$ :

\log_{10} K = \frac{2 \times 1.10}{0.0592} = 37.2

K = 10^{37.2} = 1.6 \times 10^{37}

This extremely large $K$ confirms the reaction goes essentially to completion.

Summary of Relationships

Equation	What it Gives
$\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}$	Spontaneity ( $\Delta G^\circ \lt 0$ when $E^\circ_{\mathrm{cell}} \gt 0$ )
$E^\circ_{\mathrm{cell}} = \frac{RT}{nF}\ln K$	Equilibrium constant from cell potential
$E_{\mathrm{cell}} = E^\circ_{\mathrm{cell}} - \frac{RT}{nF}\ln Q$	Cell potential under non-standard conditions
$\Delta G = -nFE_{\mathrm{cell}}$	Gibbs free energy under non-standard conditions

Common Mistake When using $\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}$ Ensure $n$ is the number of moles of Electrons for the balanced overall equation, not per half-reaction. If you write the equation as $2\mathrm{Ag}^+ + \mathrm{Cu} \to 2\mathrm{Ag} + \mathrm{Cu}^{2+}$ Then $n = 2$ Not $n = 1$ .

8. Electrolysis Quantitative Problems (HL)

Faraday”s Laws

Faraday’s First Law. The mass of substance deposited or liberated at an electrode during Electrolysis is directly proportional to the quantity of electricity passed.

M = \frac{Q \cdot M}{n \cdot F}

Or equivalently:

M = \frac{It \cdot M}{n \cdot F}

Where:

$m$ = mass of substance deposited or liberated (g)
$Q$ = total charge passed (C)
$I$ = current (A)
$t$ = time (s)
$M$ = molar mass of the substance (g/mol)
$n$ = number of electrons transferred per ion
$F$ = Faraday constant ( $96485\mathrm{ C/mol}$ )

Faraday’s Second Law. When the same quantity of electricity is passed through different Electrolytes, the masses of substances deposited are proportional to their equivalent masses ( $M/n$ ).

Key Constants

F = 96485\mathrm{ C/mol} \approx 96500\mathrm{ C/mol}

One mole of electrons carries a charge of one Faraday.

Quantitative Examples

Example 1: Calculate the mass of copper deposited when a current of $2.50\mathrm{ A}$ is passed Through $\mathrm{CuSO}_4$ (aq) for $45.0$ minutes.

Q = It = 2.50 \times 45.0 \times 60 = 6750\mathrm{ C}

N_e = \frac{Q}{F} = \frac{6750}{96500} = 0.06995\mathrm{ mol of } e^-

For $\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$$n = 2$ :

N(\mathrm{Cu}) = \frac{n_e}{2} = \frac{0.06995}{2} = 0.03498\mathrm{ mol}

M(\mathrm{Cu}) = 0.03498 \times 63.55 = 2.22\mathrm{ g}

Example 2: What current is needed to deposit $5.00\mathrm{ g}$ of aluminum from $\mathrm{Al}_2\mathrm{O}_3$ in $2.00$ hours?

N(\mathrm{Al}) = \frac{5.00}{26.98} = 0.1853\mathrm{ mol}

For $\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}$$n = 3$ :

N_e = 3 \times 0.1853 = 0.5559\mathrm{ mol of } e^-

Q = n_e \times F = 0.5559 \times 96500 = 53\,644\mathrm{ C}

I = \frac{Q}{t} = \frac{53\,644}{2.00 \times 3600} = 7.45\mathrm{ A}

Example 3: What volume of $\mathrm{Cl}_2$ gas (at STP) is produced when a current of $5.00\mathrm{ A}$ is passed through concentrated $\mathrm{NaCl}$ solution for $30.0$ minutes?

Q = It = 5.00 \times 30.0 \times 60 = 9000\mathrm{ C}

N_e = \frac{9000}{96500} = 0.09326\mathrm{ mol of } e^-

For $2\mathrm{Cl}^- \to \mathrm{Cl}_2 + 2e^-$$n = 2$ :

N(\mathrm{Cl}_2) = \frac{0.09326}{2} = 0.04663\mathrm{ mol}

V(\mathrm{Cl}_2) = nRT/P = 0.04663 \times 22.7\mathrm{ dm}^3 = 1.06\mathrm{ dm}^3 \mathrm{ at STP}

IB Exam Tip Always convert time to seconds before calculating charge. Always check whether the question gives Volume conditions (STP, RTP, or specific T and P). The IB data booklet uses $22.7\mathrm{ dm}^3\mathrm{/mol}$ at STP ( $273\mathrm{ K}$$100\mathrm{ kPa}$ ) and $24.0\mathrm{ dm}^3\mathrm{/mol}$ at RTP ( $298\mathrm{ K}$$100\mathrm{ kPa}$ ).

Current Efficiency

In practice, not all the current is used for the desired reaction. Current efficiency is defined As:

\mathrm{Current efficiency} = \frac{\mathrm{actual mass deposited}}{\mathrm{theoretical mass deposited}} \times 100\%

Side reactions (e.g., water electrolysis) and impurities reduce current efficiency.

9. Fuel Cells (HL)

Overview

Definition. A fuel cell is an electrochemical cell that converts the chemical energy of a Fuel ( hydrogen) and an oxidant ( oxygen) directly into electrical energy. Unlike Batteries, fuel cells consume reactants that must be continuously supplied.

Hydrogen Fuel Cells (AFC)

In an alkaline fuel cell, the electrolyte is hot concentrated $\mathrm{KOH}$ :

Anode: $2\mathrm{H}_2(g) + 4\mathrm{OH}^-(aq) \to 4\mathrm{H}_2\mathrm{O}(l) + 4e^-$
Cathode: $\mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^- \to 4\mathrm{OH}^-(aq)$
Overall: $2\mathrm{H}_2(g) + \mathrm{O}_2(g) \to 2\mathrm{H}_2\mathrm{O}(l)$

PEM Fuel Cells (Proton Exchange Membrane)

Definition. A PEM fuel cell uses a solid polymer membrane ( Nafion) as the Electrolyte. Protons ( $\mathrm{H}^+$ ) pass through the membrane while electrons flow through the External circuit.

Anode: $\mathrm{H}_2(g) \to 2\mathrm{H}^+(aq) + 2e^-$
Cathode: $\frac{1}{2}\mathrm{O}_2(g) + 2\mathrm{H}^+(aq) + 2e^- \to \mathrm{H}_2\mathrm{O}(l)$
Overall: $\mathrm{H}_2(g) + \frac{1}{2}\mathrm{O}_2(g) \to \mathrm{H}_2\mathrm{O}(l)$

The PEM allows only $\mathrm{H}^+$ ions to pass. Both electrodes contain a platinum catalyst.

Comparison of Fuel Cell Types

Feature	Alkaline Fuel Cell (AFC)	PEM Fuel Cell
Electrolyte	Concentrated $\mathrm{KOH}$	Solid polymer membrane
Operating temp	$100\mathrm{-}250\degree\mathrm{C}$	$50\mathrm{-}100\degree\mathrm{C}$
Mobile ion	$\mathrm{OH}^-$	$\mathrm{H}^+$
Catalyst	Platinum or nickel	Platinum
Efficiency	~60%	~40-50%
Applications	Space missions (Apollo, Shuttle)	Vehicles, portable power

Thermodynamic Efficiency

The theoretical maximum efficiency of a fuel cell is:

\mathrm{Efficiency} = \frac{\Delta G^\circ}{\Delta H^\circ} \times 100\%

For the hydrogen fuel cell:

\Delta G^\circ = -237\mathrm{ kJ/mol}, \quad \Delta H^\circ = -286\mathrm{ kJ/mol}

\mathrm{Maximum efficiency} = \frac{237}{286} \times 100\% = 83\%

Actual operating efficiencies are lower due to activation overpotential, ohmic losses, and mass Transport limitations, $40\mathrm{-}60\%$ .

Advantages and Disadvantages of Fuel Cells

Advantages	Disadvantages
High efficiency compared to combustion engines	Hydrogen storage and transport challenges
No direct greenhouse gas emissions (only $\mathrm{H}_2\mathrm{O}$ )	Hydrogen production often relies on fossil fuels
Quiet operation	Platinum catalysts are expensive
Continuous operation with fuel supply	Limited infrastructure for hydrogen refuelling
Scalable design	Water management in PEM cells

IB Exam Tip When comparing fuel cells to combustion engines, emphasize that fuel cells are more efficient Because they are not limited by the Carnot cycle. Also, note that the overall reaction is the same As combustion of hydrogen, but the energy conversion pathway is different.

10. Corrosion

Rusting Mechanism

Definition. Corrosion is the deterioration of a metal by an electrochemical reaction with Its environment. The rusting of iron is the most common form of corrosion.

Rusting requires:

Iron (or steel) in contact with both water and oxygen.
An electrolyte (water with dissolved ions accelerates the process).

Mechanism

Rusting is an electrochemical process involving two half-reactions:

Anode region (oxidation of iron):

\mathrm{Fe}(s) \to \mathrm{Fe}^{2+}(aq) + 2e^-

Cathode region (reduction of oxygen):

\mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^- \to 4\mathrm{OH}^-(aq)

The $\mathrm{Fe}^{2+}$ ions then react with $\mathrm{OH}^-$ and oxygen to form rust:

4\mathrm{Fe}^{2+}(aq) + \mathrm{O}_2(g) + 4\mathrm{OH}^-(aq) + 2\mathrm{H}_2\mathrm{O}(l) \to 4\mathrm{Fe(OH)}_3(s)

$\mathrm{Fe(OH)}_3$ dehydrates to form $\mathrm{Fe}_2\mathrm{O}_3 \cdot n\mathrm{H}_2\mathrm{O}$ (rust), which is a flaky, porous solid that does not protect the underlying metal.

Factors Affecting Corrosion Rate

Factor	Effect on Corrosion
Presence of water and oxygen	Both required for rusting
Dissolved salts (electrolytes)	Accelerate corrosion by improving conductivity
Acids (low pH)	Accelerate corrosion (more $\mathrm{H}^+$ available for reduction)
Temperature	Higher temperature generally accelerates corrosion
Contact with less active metals	Iron acts as anode and corrodes faster
Contact with more active metals	Iron acts as cathode and is protected (galvanic protection)
Stress in metal	Stressed regions are more anodic and corrode faster

Prevention Methods

1. Barrier Coatings

Paint, oil, grease, plastic, or enamel coatings prevent contact between the iron surface and Water/oxygen. This is effective only as long as the coating remains intact. If scratched, rusting Occurs at the defect.

2. Galvanizing

Definition. Galvanizing is the process of coating iron or steel with a layer of zinc. Zinc Serves two purposes:

Barrier protection: Zinc coating prevents water and oxygen from reaching the iron.
Sacrificial protection: If the coating is scratched, zinc acts as the anode (more reactive than iron, $E^\circ = -0.76\mathrm{ V}$ vs. $E^\circ = -0.44\mathrm{ V}$ ) and is preferentially oxidized, protecting the iron.

\mathrm{Zn}(s) \to \mathrm{Zn}^{2+}(aq) + 2e^- \quad (\mathrm{zinc oxidized, iron protected})

3. Sacrificial Anodes (Cathodic Protection)

Definition. Sacrificial anodes are blocks of a more reactive metal ( zinc, magnesium, Or aluminum) attached to the iron structure. The more reactive metal acts as the anode and corrodes Instead of the iron.

Applications: ship hulls, underground pipelines, water heaters, offshore oil platforms.

4. Alloying

Stainless steel contains chromium (minimum 10.5%) which forms a thin, adherent layer of Chromium(III) oxide ( $\mathrm{Cr}_2\mathrm{O}_3$ ) on the surface. This oxide layer is self-healing And prevents further corrosion.

5. Cathodic Protection (Impressed Current)

An external DC power supply forces the iron to be the cathode. Electrons are pumped onto the iron Surface, preventing its oxidation. Used for large structures like pipelines and storage tanks.

Comparison of Protection Methods

Method	Mechanism	Advantages	Limitations
Barrier coating	Physical barrier	Simple, inexpensive	Fails if scratched
Galvanizing	Barrier + sacrificial	Self-healing, long-lasting	Limited to zinc-coated items
Sacrificial anodes	Galvanic protection	Replaceable, effective	Anodes must be periodically replaced
Alloying (stainless steel)	Passive oxide layer	Very durable	Expensive, not suitable for all applications
Impressed current	Forced cathode	Effective for large structures	Requires continuous power supply

IB Exam Tip When explaining why zinc protects iron in galvanizing, reference the electrochemical series: zinc Has a more negative $E^\circ$ than iron, so zinc is preferentially oxidized. This is the same Principle as sacrificial anodes.

11. HL-Only Extensions

Concentration Cells

Definition. A concentration cell is an electrochemical cell where both half-cells contain The same species but at different concentrations. The cell potential arises solely from the Concentration difference.

Example: $\mathrm{Cu}(s) \mid \mathrm{Cu}^{2+}(0.10\mathrm{ M}) \parallel \mathrm{Cu}^{2+}(1.0\mathrm{ M}) \mid \mathrm{Cu}(s)$

Since $E^\circ_{\mathrm{cell}} = 0$ (same half-reaction), the potential comes from the Nernst Equation:

E_{\mathrm{cell}} = E^\circ_{\mathrm{cell}} - \frac{0.0592}{n}\log_{10}\frac{[\mathrm{Cu}^{2+}]_{\mathrm{anode}}}{[\mathrm{Cu}^{2+}]_{\mathrm{cathode}}}

E_{\mathrm{cell}} = 0 - \frac{0.0592}{2}\log_{10}\frac{0.10}{1.0} = -\frac{0.0592}{2} \times (-1) = +0.0296\mathrm{ V}

The half-cell with the lower concentration undergoes oxidation (anode), and the half-cell with the Higher concentration undergoes reduction (cathode). The cell operates until the concentrations Equalize.

Lead-Acid Batteries

Definition. A lead-acid battery is a rechargeable battery commonly used in automobiles. It Consists of lead dioxide ( $\mathrm{PbO}_2$ ) as the cathode, lead ( $\mathrm{Pb}$ ) as the anode, and Sulfuric acid ( $\mathrm{H}_2\mathrm{SO}_4$ ) as the electrolyte.

Discharging:

Anode (oxidation): $\mathrm{Pb}(s) + \mathrm{SO}_4^{2-}(aq) \to \mathrm{PbSO}_4(s) + 2e^-$
Cathode (reduction): $\mathrm{PbO}_2(s) + \mathrm{SO}_4^{2-}(aq) + 4\mathrm{H}^+(aq) + 2e^- \to \mathrm{PbSO}_4(s) + 2\mathrm{H}_2\mathrm{O}(l)$
Overall: $\mathrm{Pb}(s) + \mathrm{PbO}_2(s) + 2\mathrm{H}_2\mathrm{SO}_4(aq) \to 2\mathrm{PbSO}_4(s) + 2\mathrm{H}_2\mathrm{O}(l)$

Charging (reverse reaction):

2\mathrm{PbSO}_4(s) + 2\mathrm{H}_2\mathrm{O}(l) \to \mathrm{Pb}(s) + \mathrm{PbO}_2(s) + 2\mathrm{H}_2\mathrm{SO}_4(aq)

The state of charge can be determined by measuring the density of the electrolyte. As the battery Discharges, $\mathrm{H}_2\mathrm{SO}_4$ is consumed and $\mathrm{H}_2\mathrm{O}$ is produced, Decreasing the electrolyte density.

Parameter	Fully Charged	Discharged
Electrolyte density	~1.28 g/cm $^3$	~1.10 g/cm $^3$
Cell voltage	~2.1 V	~1.8 V
$\mathrm{H}_2\mathrm{SO}_4$ concentration	High	Low

Lithium-Ion Batteries

Definition. A lithium-ion battery is a rechargeable battery where lithium ions move between The anode and cathode during charging and discharging. The electrodes are intercalation materials — Lithium ions are inserted into and extracted from layered crystal structures.

Discharging:

Anode (oxidation): $\mathrm{LiC}_6(s) \to \mathrm{C}_6(s) + \mathrm{Li}^+ + e^-$
Cathode (reduction): $\mathrm{CoO}_2(s) + \mathrm{Li}^+ + e^- \to \mathrm{LiCoO}_2(s)$
Overall: $\mathrm{LiC}_6(s) + \mathrm{CoO}_2(s) \to \mathrm{C}_6(s) + \mathrm{LiCoO}_2(s)$

Charging: The reverse of the above reactions.

Key features:

The electrolyte is a non-aqueous lithium salt (e.g., $\mathrm{LiPF}_6$ in organic solvent) — water cannot be used because lithium reacts violently with it.
Typical cell voltage: ~3.7 V (much higher than lead-acid).
High energy density: suitable for portable electronics and electric vehicles.
Memory effect is negligible.
Degradation occurs over time due to electrolyte decomposition and electrode material degradation.

Feature	Lead-Acid	Lithium-Ion
Cell voltage	2.0 V	3.7 V
Energy density	30-40 Wh/kg	150-250 Wh/kg
Cycle life	200-300 cycles	500-1000+ cycles
Self-discharge rate	~5% per month	~1-3% per month
Toxicity	Lead is toxic	Less toxic materials
Cost	Lower	Higher
Weight	Heavy	Lightweight

IB Exam Tip Know the half-reactions for the lead-acid battery. The key insight is that during discharge, both Electrodes are converted to $\mathrm{PbSO}_4$ And during charging, the reaction is reversed. This Reversibility is what makes the battery rechargeable.

12. Exam Practice

Question 1 (SL, 4 marks)

Consider the following standard reduction potentials:

Half-Reaction	$E^\circ$ (V)
$\mathrm{Mg}^{2+} + 2e^- \rightleftharpoons \mathrm{Mg}$	$-2.37$
$\mathrm{Ag}^+ + e^- \rightleftharpoons \mathrm{Ag}$	$+0.80$

(a) Write the cell diagram notation for a galvanic cell constructed from these two half-cells. (1 mark)

(b) Calculate the standard cell potential. (1 mark)

(c) Identify the oxidizing agent and the reducing agent. (2 marks)

Markscheme

(a) $\mathrm{Mg}(s) \mid \mathrm{Mg}^{2+}(aq) \parallel \mathrm{Ag}^+(aq) \mid \mathrm{Ag}(s)$

The anode (oxidation) is $\mathrm{Mg}$ (more negative $E^\circ$ ), placed on the left. The cathode (reduction) is $\mathrm{Ag}$ (more positive $E^\circ$ ), placed on the right. Accept any reasonable State symbols.

(b) $E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} = 0.80 - (-2.37) = +3.17\mathrm{ V}$

(c) The oxidizing agent is $\mathrm{Ag}^+$ (it is reduced, gaining electrons). The reducing Agent is $\mathrm{Mg}$ (it is oxidized, losing electrons).

Question 2 (SL, 5 marks)

An aqueous solution of copper(II) sulfate is electrolyzed using inert graphite electrodes.

(a) Write the half-equation for the reaction at the cathode. (1 mark)

(b) Write the half-equation for the reaction at the anode. (1 mark)

(c) State the observation at each electrode. (2 marks)

(d) Calculate the volume of gas produced at the anode when a current of $0.500\mathrm{ A}$ is Passed for $30.0$ minutes at RTP. (3 marks)

Markscheme

(a) $\mathrm{Cu}^{2+}(aq) + 2e^- \to \mathrm{Cu}(s)$

Copper is below aluminium in the reactivity series, so $\mathrm{Cu}^{2+}$ is preferentially Discharged over $\mathrm{H}_2\mathrm{O}$ .

(b) $2\mathrm{H}_2\mathrm{O}(l) \to \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^-$

The sulfate ion is not discharged; water is oxidized instead.

(c) Cathode: orange/brown solid (copper metal) deposits on the electrode. Anode: colourless gas Bubbles (oxygen) are evolved.

(d) $Q = It = 0.500 \times 30.0 \times 60 = 900\mathrm{ C}$

$n_e = 900 / 96500 = 0.00933\mathrm{ mol of } e^-$

$n(\mathrm{O}_2) = 0.00933 / 4 = 0.00233\mathrm{ mol}$

$V(\mathrm{O}_2) = 0.00233 \times 24.0 = 0.0560\mathrm{ dm}^3$

Question 3 (HL, 6 marks)

A galvanic cell is constructed as follows:

\mathrm{Ni}(s) \mid \mathrm{Ni}^{2+}(0.010\mathrm{ M}) \parallel \mathrm{Ag}^+(1.0\mathrm{ M}) \mid \mathrm{Ag}(s)

Given: $E^\circ(\mathrm{Ni}^{2+}/\mathrm{Ni}) = -0.25\mathrm{ V}$ $E^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\mathrm{ V}$

(a) Calculate $E^\circ_{\mathrm{cell}}$ . (1 mark)

(b) Calculate the cell potential under the given non-standard conditions at $298\mathrm{ K}$ . (3 Marks)

(c) Calculate $\Delta G^\circ$ for the cell reaction. (2 marks)

Markscheme

(a) $E^\circ_{\mathrm{cell}} = 0.80 - (-0.25) = +1.05\mathrm{ V}$

(b) Overall reaction: $\mathrm{Ni}(s) + 2\mathrm{Ag}^+(aq) \to \mathrm{Ni}^{2+}(aq) + 2\mathrm{Ag}(s)$$n = 2$

Q = \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Ag}^+]^2} = \frac{0.010}{(1.0)^2} = 0.010

E_{\mathrm{cell}} = 1.05 - \frac{0.0592}{2}\log_{10}(0.010) = 1.05 - \frac{0.0592}{2} \times (-2) = 1.05 + 0.0592 = 1.109\mathrm{ V}

(c) $\Delta G^\circ = -nFE^\circ_{\mathrm{cell}} = -2 \times 96500 \times 1.05 = -202\,650\mathrm{ J} = -203\mathrm{ kJ/mol}$

Question 4 (HL, 4 marks)

Balance the following redox equation in acidic solution:

\mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{H}^+ + \mathrm{I}^- \to \mathrm{Cr}^{3+} + \mathrm{I}_2 + \mathrm{H}_2\mathrm{O}

Markscheme

Reduction half-reaction:

\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \to 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}

Oxidation half-reaction:

2\mathrm{I}^- \to \mathrm{I}_2 + 2e^-

Multiply oxidation by 3:

6\mathrm{I}^- \to 3\mathrm{I}_2 + 6e^-

Add both half-reactions:

\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{I}^- \to 2\mathrm{Cr}^{3+} + 3\mathrm{I}_2 + 7\mathrm{H}_2\mathrm{O}

Question 5 (HL, 5 marks)

A hydrogen-oxygen fuel cell operates at $298\mathrm{ K}$ . The overall reaction is:

\mathrm{H}_2(g) + \frac{1}{2}\mathrm{O}_2(g) \to \mathrm{H}_2\mathrm{O}(l)

$\Delta H^\circ = -286\mathrm{ kJ/mol}$$\Delta G^\circ = -237\mathrm{ kJ/mol}$

(a) Calculate the standard cell potential. (2 marks)

(b) Calculate the maximum theoretical efficiency of the fuel cell. (1 mark)

(c) Explain why the actual efficiency is lower than the theoretical maximum. (2 marks)

Markscheme

(a) $\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}$

From the half-reactions, $n = 2$ (for the equation $\mathrm{H}_2 + \mathrm{O}_2 \to 2\mathrm{H}_2\mathrm{O}$ with $n = 4$ But per mole of $\mathrm{H}_2$ as written, $n = 2$ ).

$E^\circ_{\mathrm{cell}} = \frac{-\Delta G^\circ}{nF} = \frac{237\,000}{2 \times 96500} = +1.23\mathrm{ V}$

(b) $\mathrm{Efficiency} = \frac{\Delta G^\circ}{\Delta H^\circ} \times 100\% = \frac{237}{286} \times 100\% = 82.9\%$

(c) Actual efficiency is lower due to:

Activation overpotential (energy required to initiate reactions at the electrode surface).
Ohmic losses (resistance of the electrolyte and electrodes).
Mass transport limitations (slow diffusion of reactants to the electrodes).
Heat losses to the surroundings.

Question 6 (SL/HL, 4 marks)

A piece of iron piping is connected to a block of magnesium using a conducting wire. Both are buried In moist soil.

(a) Identify which metal acts as the anode and which acts as the cathode. (1 mark)

(b) Write the half-equation for the reaction at the anode. (1 mark)

(c) Explain why this arrangement protects the iron from corrosion. (2 marks)

Markscheme

(a) Magnesium is the anode; iron is the cathode. Magnesium has a more negative $E^\circ$ ( $-2.37\mathrm{ V}$ ) compared to iron ( $-0.44\mathrm{ V}$ ).

(b) $\mathrm{Mg}(s) \to \mathrm{Mg}^{2+}(aq) + 2e^-$

(c) Since magnesium is more reactive than iron, it is preferentially oxidized (corrodes instead Of iron). Electrons flow from magnesium to iron, making the iron surface electron-rich and Preventing the oxidation of iron. This is called sacrificial (cathodic) protection. The magnesium Block must be replaced periodically as it is consumed.

Question 7 (HL, 4 marks)

The standard cell potential for the reaction $\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s)$ is $+1.10\mathrm{ V}$ .

(a) Calculate the equilibrium constant $K$ at $298\mathrm{ K}$ . (3 marks)

(b) Comment on the feasibility of the reverse reaction under standard conditions. (1 mark)

Markscheme

(a) $\log_{10} K = \frac{nE^\circ_{\mathrm{cell}}}{0.0592} = \frac{2 \times 1.10}{0.0592} = 37.16$

$K = 10^{37.16} = 1.4 \times 10^{37}$

(b) Since $K$ is extremely large, the forward reaction is essentially irreversible under Standard conditions. The reverse reaction is not feasible ( $K_{\mathrm{reverse}} = 1/K = 7.1 \times 10^{-38}$ ), meaning the equilibrium lies overwhelmingly Toward the products.

flowchart TD
    A[3_Electrochemistry] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary of Key Equations

Equation	Application
$E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}}$	Calculating standard cell potential
$E_{\mathrm{cell}} = E^\circ_{\mathrm{cell}} - \frac{0.0592}{n}\log_{10} Q$	Nernst equation at $298\mathrm{ K}$
$\Delta G^\circ = -nFE^\circ_{\mathrm{cell}}$	Free energy from cell potential
$E^\circ_{\mathrm{cell}} = \frac{0.0592}{n}\log_{10} K$	Equilibrium constant from cell potential
$m = \frac{ItM}{nF}$	Faraday’s law (mass deposited)
$\mathrm{Efficiency} = \frac{\Delta G^\circ}{\Delta H^\circ} \times 100\%$	Fuel cell theoretical efficiency
$E^\circ_{\mathrm{cell}} = \frac{\Delta G^\circ}{-nF} = \frac{RT}{nF}\ln K$	Thermodynamic relationships

Practice Problems

Question 1: Predicting Spontaneous Redox Reactions

Given the following standard reduction potentials:

Half-Reaction	$E^\circ$ (V)
$\mathrm{Fe}^{3+} + e^- \rightleftharpoons \mathrm{Fe}^{2+}$	$+0.77$
$\mathrm{I}_2 + 2e^- \rightleftharpoons 2\mathrm{I}^-$	$+0.54$
$\mathrm{Br}_2 + 2e^- \rightleftharpoons 2\mathrm{Br}^-$	$+1.07$
$\mathrm{Zn}^{2+} + 2e^- \rightleftharpoons \mathrm{Zn}$	$-0.76$

(a) Will $\mathrm{Fe}^{3+}$ oxidise $\mathrm{I}^-$ to $\mathrm{I}_2$ ? Calculate $E^\circ_{\mathrm{cell}}$ .

(b) Will $\mathrm{Br}_2$ oxidise $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ ? Calculate $E^\circ_{\mathrm{cell}}$ .

Answer

(a) Cathode (reduction): $\mathrm{Fe}^{3+} + e^- \to \mathrm{Fe}^{2+}$$E^\circ = +0.77\mathrm{ V}$

Anode (oxidation): $2\mathrm{I}^- \to \mathrm{I}_2 + 2e^-$$E^\circ = +0.54\mathrm{ V}$

$E^\circ_{\mathrm{cell}} = 0.77 - 0.54 = +0.23\mathrm{ V}$

Since $E^\circ_{\mathrm{cell}} \gt 0$ Yes, $\mathrm{Fe}^{3+}$ will spontaneously oxidise $\mathrm{I}^-$ .

(b) Cathode (reduction): $\mathrm{Br}_2 + 2e^- \to 2\mathrm{Br}^-$ , $E^\circ = +1.07\mathrm{ V}$

Anode (oxidation): $\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+} + e^-$ , $E^\circ = +0.77\mathrm{ V}$

$E^\circ_{\mathrm{cell}} = 1.07 - 0.77 = +0.30\mathrm{ V}$

Since $E^\circ_{\mathrm{cell}} \gt 0$ Yes, $\mathrm{Br}_2$ will spontaneously oxidise $\mathrm{Fe}^{2+}$ .

Question 2: Electrolysis of Aqueous Solutions

An aqueous solution of $\mathrm{CuSO}_4$ is electrolysed using inert graphite electrodes.

(a) Write the half-equation at the cathode and identify the product.

(b) Write the half-equation at the anode and identify the product.

Answer

(a) Copper(II) ions are below aluminium in the reactivity series, so $\mathrm{Cu}^{2+}$ is Preferentially discharged over $\mathrm{H}_2\mathrm{O}$ :

$\mathrm{Cu}^{2+}(aq) + 2e^- \to \mathrm{Cu}(s)$

Product: orange-brown solid (copper metal) deposits on the cathode.

(b) Sulfate ions are not discharged; water is oxidised instead:

$2\mathrm{H}_2\mathrm{O}(l) \to \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^-$

Product: colourless oxygen gas bubbles at the anode.

(c) Cathode: orange-brown coating of copper forms on the electrode. The blue colour of the solution Fades as $\mathrm{Cu}^{2+}$ is removed. Anode: colourless gas bubbles (oxygen) are evolved.

Question 3: Nernst Equation Application

A galvanic cell is constructed as:

$\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(0.0010\mathrm{ M}) \parallel \mathrm{Cu}^{2+}(0.10\mathrm{ M}) \mid \mathrm{Cu}(s)$

Given $E^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\mathrm{ V}$ and $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\mathrm{ V}$ Calculate the cell potential at $298\mathrm{ K}$ .

Answer

$E^\circ_{\mathrm{cell}} = 0.34 - (-0.76) = +1.10\mathrm{ V}$

Overall reaction: $\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s)$ $n = 2$

$Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]} = \frac{0.0010}{0.10} = 0.010$

$E_{\mathrm{cell}} = E^\circ_{\mathrm{cell}} - \frac{0.0592}{n}\log_{10} Q = 1.10 - \frac{0.0592}{2}\log_{10}(0.010)$

$= 1.10 - 0.0296 \times (-2) = 1.10 + 0.0592 = 1.16\mathrm{ V}$

Question 4: Faraday's Law Calculation

What mass of aluminium is deposited when a current of $5.00\mathrm{ A}$ is passed through molten $\mathrm{Al}_2\mathrm{O}_3$ for $2.00$ hours?

Answer

$Q = It = 5.00 \times 2.00 \times 3600 = 36000\mathrm{ C}$

$n_e = \frac{Q}{F} = \frac{36000}{96500} = 0.373\mathrm{ mol of } e^-$

For $\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}$ , $n = 3$ :

$n(\mathrm{Al}) = \frac{n_e}{3} = \frac{0.373}{3} = 0.124\mathrm{ mol}$

$m(\mathrm{Al}) = 0.124 \times 26.98 = 3.35\mathrm{ g}$

Question 5: Balancing Redox in Basic Solution

Balance the following equation in basic solution:

$\mathrm{MnO}_4^- + \mathrm{SO}_3^{2-} \to \mathrm{MnO}_2 + \mathrm{SO}_4^{2-}$

Answer

Reduction half-reaction (in acidic conditions):

$\mathrm{MnO}_4^- + 4\mathrm{H}^+ + 3e^- \to \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$

Oxidation half-reaction (in acidic conditions):

$\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \to \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2e^-$

Multiply reduction by 2 and oxidation by 3 to balance electrons:

$2\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 6e^- \to 2\mathrm{MnO}_2 + 4\mathrm{H}_2\mathrm{O}$

$3\mathrm{SO}_3^{2-} + 3\mathrm{H}_2\mathrm{O} \to 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+ + 6e^-$

Add both half-reactions:

$2\mathrm{MnO}_4^- + 3\mathrm{SO}_3^{2-} + 2\mathrm{H}^+ \to 2\mathrm{MnO}_2 + 3\mathrm{SO}_4^{2-} + \mathrm{H}_2\mathrm{O}$

Now convert to basic conditions by adding $2\mathrm{OH}^-$ to both sides:

$2\mathrm{MnO}_4^- + 3\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \to 2\mathrm{MnO}_2 + 3\mathrm{SO}_4^{2-} + 2\mathrm{OH}^-$

Common Pitfalls

Forgetting to balance equations before performing calculations. Always check that atoms and charges balance on both sides.
Writing half-equations without balancing charges or atoms. Always check electrons, hydrogen ions, and water molecules.
Confusing the terms ‘molar’ and ‘molecular’. Molar refers to per mole ( $\text{mol}^{-1}$ ), while molecular refers to individual molecules.
Drawing structural formulae incorrectly. Check the number of bonds each atom can form and the overall charge.

Summary

Galvanic: chemical → electrical; electrolytic: electrical → chemical
Anode: oxidation; cathode: reduction (“an ox, red cat”)
$E^\circ$ is intensive (does not change sign when equation reversed)
Faraday’s laws: $Q = nF$ ; mass deposited $= \frac{MIt}{nF}$

Cross-References

Topic	Site	Link
[Electrochemistry]	A-Level	View
[Electrochemistry]	IB	View
[Electrochemistry]	DSE	View

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Intuition

Electrochemistry connects chemistry and electricity. A battery (galvanic cell) converts chemical energy into electrical energy by spontaneous redox reactions. An electrolytic cell does the reverse — it uses electricity to force non-spontaneous reactions. In both cases, oxidation happens at the anode and reduction at the cathode (“an ox, red cat”).

Think of standard electrode potentials as a “tendency to gain electrons.” A more positive $E^\circ$ means a greater tendency to be reduced (gain electrons). In a galvanic cell, the half-cell with the more positive $E^\circ$ undergoes reduction (cathode), and the one with the less positive (or more negative) $E^\circ$ undergoes oxidation (anode). The cell potential is $E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ , and it must be positive for the reaction to be spontaneous. Faraday’s laws of electrolysis link the amount of substance deposited to the charge passed — more charge means more electrons transferred, which means more product formed.

Common Mistakes

Confusing the sign of $E^\circ$ when reversing half-equations. $E^\circ$ values are intensive properties and do not change sign when you reverse a half-equation. The cell potential is always $E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ . If you reverse both half-equations, the cell potential magnitude stays the same but the sign of the overall reaction changes — not the individual $E^\circ$ values.
Mixing up anode and cathode in galvanic vs electrolytic cells. In both cell types, oxidation occurs at the anode and reduction at the cathode (“an ox, red cat”). The difference is direction: galvanic cells produce electricity from spontaneous reactions, while electrolytic cells use electricity to drive non-spontaneous reactions. Students often assume the anode is always negative — it is negative in galvanic cells but positive in electrolytic cells.
Applying Faraday’s laws without converting to moles. Faraday’s first law states $m = \frac{MIt}{nF}$ , where $M$ is molar mass and $n$ is the number of electrons transferred. A common error is using mass directly instead of molar mass, or forgetting to divide by $n$ (the electrons per ion). For example, depositing aluminium requires 3 electrons per atom ( $n = 3$ ), not 1.
Assuming a positive $E^\circ_{\text{cell}}$ means the reaction is fast. Standard electrode potentials predict thermodynamic spontaneity, not kinetic rate. A reaction can have a large positive $E^\circ_{\text{cell}}$ but be extremely slow due to high activation energy (e.g., the rusting of iron is spontaneous but slow without a catalyst).
Forgetting to balance charge when combining half-equations. When adding two half-equations to get the overall cell reaction, the number of electrons lost must equal the number gained. Multiply half-equations by integers to equalise electrons before adding, and cancel electrons from both sides.

Electrochemistry | IB - Wyatt's Notes

1. Redox Review (SL Topic 9)

Oxidation and Reduction

Oxidation States (Oxidation Numbers)

Identifying Redox Reactions

Writing Half-Equations

Balancing Redox in Acidic Conditions

Balancing Redox in Basic Conditions

Intuition

Common Mistakes

See Also