Electrochemistry

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:

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:

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:

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.

IB Exam Tip

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

1. Identify the two relevant half-reactions from the data booklet.
2. Assign cathode (more positive $E^\circ$) and anode (more negative $E^\circ$).
3. Calculate $E^\circ_{\mathrm{cell}}$.
4. 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

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)

The reduction of water at the cathode:

Discharge at the Anode (Oxidation)

The oxidation of water at the anode:

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:

1. The position in the electrochemical series (standard reduction potential).
2. The concentration of ions (higher concentration favours discharge).
3. 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

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:

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:

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

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

Dividing both sides by $-nF$:

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 $\log_{10}$:

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:

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 $\mathrm{Zn} \mid \mathrm{Zn}^{2+} \parallel \mathrm{Cu}^{2+} \mid \mathrm{Cu}$:

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:

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:

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

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

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:

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

Thermodynamic Efficiency

The theoretical maximum efficiency of a fuel cell is:

For the hydrogen fuel cell:

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

Advantages and Disadvantages of Fuel 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:

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

Mechanism

Rusting is an electrochemical process involving two half-reactions:

Anode region (oxidation of iron):

Cathode region (reduction of oxygen):

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

$\mathrm{Fe(OH)}_3$ 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

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:

1. Barrier protection: Zinc coating prevents water and oxygen from reaching the iron.
2. 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.

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

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:

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):

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.

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.

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:

(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:

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

(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:

Markscheme

Reduction half-reaction:

Oxidation half-reaction:

Multiply oxidation by 3:

Add both half-reactions:

Question 5 (HL, 5 marks)

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

\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.

Summary of Key Equations

Practice Problems