Topic 14: Redox II and electrode potentials

Chemistry (9CH0) · Pearson Edexcel A Level · 13 min read · Updated: 7 Sept 2026

Topic 14: Redox II — Advanced Electrochemistry and Redox Titrations

Welcome to Redox II! In Year 1 (Redox I), you learned the basics of electron transfer and oxidation numbers. In this chapter, we take those fundamentals into the world of electrochemistry, thermodynamic feasibility, fuel cells, and high-precision quantitative titrations. These notes are structured specifically for Pearson Edexcel A Level Chemistry (9CH0) Paper 1. Take your time, walk through the step-by-step examples, and master the key concepts.

1. Advanced Half-Equations and Redox Balancing

In advanced inorganic chemistry, redox processes often happen in acidic aqueous solutions. To balance complex multi-electron reactions, we use a systematic method involving water (\(\text{H}_2\text{O}\)), hydrogen ions (\(\text{H}^+\)), and electrons (\(\text{e}^-\)).

Step-by-Step Balancing Method (Acidic Conditions)

1. Balance the main element: Balance all atoms other than oxygen and hydrogen first.
2. Balance oxygen atoms: Add \(\text{H}_2\text{O}(\text{l})\) molecules to the side deficient in oxygen.
3. Balance hydrogen atoms: Add \(\text{H}^+(\text{aq})\) ions to the side deficient in hydrogen.
4. Balance overall charge: Add electrons (\(\text{e}^-\)) to the more positive side so that total charges on both sides match.

Key Specification Half-Equations You Must Know

• Reduction of manganate(VII) ions:
\(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightleftharpoons \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)
(Oxidation state of manganese changes from \(+7\) to \(+2\))

• Reduction of dichromate(VI) ions:
\(\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{e}^- \rightleftharpoons 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})\)
(Oxidation state of chromium changes from \(+6\) to \(+3\))

• Reduction of iodine:
\(\text{I}_2(\text{aq}) + 2\text{e}^- \rightleftharpoons 2\text{I}^-(\text{aq})\)

• Oxidation of thiosulfate ions to tetrathionate ions:
\(2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightleftharpoons \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{e}^-\)
(Average oxidation state of sulfur increases from \(+2\) in \(\text{S}_2\text{O}_3^{2-}\) to \(+2.5\) in \(\text{S}_4\text{O}_6^{2-}\))

Combining Half-Equations to Form Overall Equations

To combine half-equations, multiply one or both half-equations by integers so that the number of electrons lost equals the number of electrons gained. When added together, the electrons must completely cancel out.

Worked Example: Combine the oxidation of \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\) with the reduction of \(\text{MnO}_4^-\).
Oxidation half-equation: \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^-\)
Reduction half-equation: \(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)
Multiply the iron half-equation by 5 so both have \(5\text{e}^-\):
\(5\text{Fe}^{2+}(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + 5\text{e}^-\)
Add the equations together:
\(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)

Key Takeaway

Always balance atoms first (\(\text{O}\) using \(\text{H}_2\text{O}\), \(\text{H}\) using \(\text{H}^+\)), then balance charge using \(\text{e}^-\). Always ensure electrons cancel out fully in the final overall equation.

2. Standard Electrode Potentials and Electrochemical Cells

What is Standard Electrode Potential (\(E^\ominus\))?

The Standard Electrode Potential (\(E^\ominus\)) of a half-cell is the electromotive force (EMF) generated when the half-cell is connected to a Standard Hydrogen Electrode (SHE) under standard conditions.

Standard Conditions Required

• Temperature: \(298\text{ K}\) (\(25^\circ\text{C}\))
• Pressure: \(100\text{ kPa}\) (\(1\text{ bar}\)) for all gases
• Concentration: \(1.00\text{ mol dm}^{-3}\) of all aqueous ions participating in the reaction

The Standard Hydrogen Electrode (SHE) — The Universal Reference

Since we cannot measure the absolute potential of a single isolated half-cell, we compare everything to a universal reference: the Standard Hydrogen Electrode. By international agreement, its potential is defined as:

\(E^\ominus = 0.00\text{ V}\)

Components of the SHE:
• Hydrogen gas (\(\text{H}_2\)) bubbled at \(100\text{ kPa}\).
• An aqueous solution of hydrogen ions with a concentration of \(1.00\text{ mol dm}^{-3}\) \(\text{H}^+(\text{aq})\) (e.g., \(1.00\text{ mol dm}^{-3}\ \text{HCl}(\text{aq})\) or \(0.50\text{ mol dm}^{-3}\ \text{H}_2\text{SO}_4(\text{aq})\)).
• A platinised platinum electrode (\(\text{Pt}\)), which provides an inert conducting surface and acts as a catalyst for the equilibrium.
• Maintained at \(298\text{ K}\).
• Equilibrium half-reaction: \(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2(\text{g})\)

Types of Half-Cells

1. Metal / Metal Ion Half-Cell:
Consists of a solid metal strip dipping into a \(1.00\text{ mol dm}^{-3}\) solution of its own ions (e.g., a strip of \(\text{Zn}(\text{s})\) in \(1.00\text{ mol dm}^{-3}\ \text{Zn}^{2+}(\text{aq})\)).
Equilibrium: \(\text{Zn}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Zn}(\text{s})\)

2. Gas / Non-Metal Ion Half-Cell:
A gas is bubbled over an inert platinum electrode in contact with a \(1.00\text{ mol dm}^{-3}\) solution of its ions (e.g., \(\text{Cl}_2(\text{g})\) gas bubbled over \(\text{Pt}(\text{s})\) in \(1.00\text{ mol dm}^{-3}\ \text{Cl}^-(\text{aq})\)).
Equilibrium: \(\text{Cl}_2(\text{g}) + 2\text{e}^- \rightleftharpoons 2\text{Cl}^-(\text{aq})\)

3. Ion / Ion Half-Cell (Two Oxidation States in Solution):
An inert platinum electrode dipping into a solution containing equimolar concentrations (\(1.00\text{ mol dm}^{-3}\) of each) of two different oxidation states of the same element (e.g., \(\text{Pt}(\text{s})\) in a solution containing both \(1.00\text{ mol dm}^{-3}\ \text{Fe}^{2+}(\text{aq})\) and \(1.00\text{ mol dm}^{-3}\ \text{Fe}^{3+}(\text{aq})\)).
Equilibrium: \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})\)

Apparatus Functions: High-Resistance Voltmeter and Salt Bridge

• High-Resistance Voltmeter: Connects the two electrodes. It draws virtually zero current, preventing the redox reactions from running to equilibrium and allowing the maximum potential difference (electromotive force / EMF) to be measured.
• Salt Bridge: Completes the electrical circuit by allowing ions to move between half-cells to balance charge build-up, without allowing the bulk solutions to mix. It is typically a piece of filter paper soaked in an unreactive electrolyte such as \(\text{KNO}_3(\text{aq})\) or \(\text{KCl}(\text{aq})\) (provided it will not form a precipitate with either half-cell solution).

IUPAC Standard Cell Notation

Chemists use a standard shorthand notation to represent electrochemical cells without having to draw elaborate diagrams:

• Single vertical line (\(\mid\)): Represents a phase boundary (e.g., between solid metal and aqueous solution).
• Double vertical line (\(\parallel\)): Represents the salt bridge.
• Comma (\(,\)): Separates species that are in the same physical phase (e.g., \(\text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq})\)).
• Arrangement: By convention, the half-cell undergoing oxidation (anode) is placed on the left, and the half-cell undergoing reduction (cathode) is placed on the right. The most reduced species are written on the extreme outer edges, and the most oxidised species are adjacent to the salt bridge.

Example (Daniell Cell):
\(\text{Zn}(\text{s}) \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu}(\text{s})\)

Example with Ion/Ion System:
\(\text{Pt}(\text{s}) \mid \text{Fe}^{2+}(\text{aq}), \text{Fe}^{3+}(\text{aq}) \parallel \text{Ag}^+(\text{aq}) \mid \text{Ag}(\text{s})\)

Calculating Standard Cell Potential (\(E^\ominus_{\text{cell}}\))

To calculate the standard potential of a cell:

\(E^\ominus_{\text{cell}} = E^\ominus_{\text{right}} - E^\ominus_{\text{left}}\) (or \(E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\))

Worked Calculation:
Given the standard reduction potentials:
\(\text{Zn}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Zn}(\text{s}) \quad E^\ominus = -0.76\text{ V}\)
\(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s}) \quad E^\ominus = +0.34\text{ V}\)
If \(\text{Zn}\) is on the left and \(\text{Cu}\) is on the right:
\(E^\ominus_{\text{cell}} = (+0.34) - (-0.76) = +1.10\text{ V}\)

Key Takeaway

Electrons flow through the external wire from the more negative electrode (oxidation) to the more positive electrode (reduction). Ions migrate through the salt bridge to maintain electrical neutrality.

3. Thermodynamic Feasibility vs. Kinetic Inhibition

Predicting Feasibility from \(E^\ominus\) Values

• A redox reaction is thermodynamically feasible under standard conditions if the overall standard cell potential is positive: \(E^\ominus_{\text{cell}} > 0\text{ V}\).
• The more positive the \(E^\ominus\) value of a half-cell, the greater its tendency to gain electrons (act as an oxidising agent / undergo reduction).
• The more negative (or less positive) the \(E^\ominus\) value, the greater its tendency to lose electrons (act as a reducing agent / undergo oxidation).

Connection to Entropy (\(\Delta S_{\text{total}}\)) and Equilibrium Constant (\(K\))

Standard electrode potential is directly linked to thermodynamic stability and equilibrium positions:

• \(E^\ominus_{\text{cell}}\) is directly proportional to the total entropy change (\(\Delta S_{\text{total}}\)):
\(\Delta S_{\text{total}} = n F E^\ominus_{\text{cell}}\)
(where \(n\) is the number of moles of electrons transferred and \(F\) is the Faraday constant)

• \(E^\ominus_{\text{cell}}\) is directly proportional to \(\ln K\):
\(\ln K \propto E^\ominus_{\text{cell}}\)
Because \(\Delta G^\ominus = -n F E^\ominus_{\text{cell}} = -R T \ln K\), rearranging gives:
\(\ln K = \frac{n F E^\ominus_{\text{cell}}}{R T}\)

Edexcel focuses on the proportional relationships: A large positive \(E^\ominus_{\text{cell}}\) corresponds to a large positive \(\Delta S_{\text{total}}\) and a very large equilibrium constant \(K\), meaning the reaction goes virtually to completion.

Why Might a Reaction with \(E^\ominus_{\text{cell}} > 0\text{ V}\) Not Occur in Practice?

Don't fall into the trap of thinking a feasible reaction always happens instantly! There are two major reasons why a reaction might not be observed:

1. Kinetic Inhibition (High Activation Energy):
\(E^\ominus_{\text{cell}}\) only tells us about the thermodynamics (the energy difference between reactants and products). It says nothing about the rate of the reaction. If the reaction has a very high activation energy (\(E_a\)), the reaction rate will be imperceptibly slow at room temperature.

2. Non-Standard Conditions:
Standard electrode potentials apply strictly at \(298\text{ K}\), \(100\text{ kPa}\), and \(1.00\text{ mol dm}^{-3}\). If you alter concentrations, pressure, or temperature, the position of equilibrium shifts according to Le Chatelier's principle, changing the actual electrode potential (\(E\)) away from \(E^\ominus\). This can make a previously unfeasible reaction become feasible, or vice versa.

Disproportionation Reactions

A disproportionation reaction occurs when a single chemical species is simultaneously oxidised and reduced to form two different products. We can evaluate whether disproportionation is feasible by calculating \(E^\ominus_{\text{cell}}\) for the coupled reduction and oxidation half-reactions.

Example: Disproportionation of Copper(I) ions
Given the half-equations:
1. Reduction: \(\text{Cu}^+(\text{aq}) + \text{e}^- \rightleftharpoons \text{Cu}(\text{s}) \quad E^\ominus = +0.52\text{ V}\)
2. Oxidation: \(\text{Cu}^{2+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Cu}^+(\text{aq}) \quad E^\ominus = +0.15\text{ V}\)
Overall reaction: \(2\text{Cu}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + \text{Cu}(\text{s})\)
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = (+0.52\text{ V}) - (+0.15\text{ V}) = +0.37\text{ V}\)
Since \(E^\ominus_{\text{cell}} > 0\text{ V}\), the disproportionation of aqueous \(\text{Cu}^+\) into \(\text{Cu}^{2+}\) and solid \(\text{Cu}\) is thermodynamically feasible under standard conditions.

Key Takeaway

Positive \(E^\ominus_{\text{cell}}\) means thermodynamically feasible (\(\Delta S_{\text{total}} > 0\) and \(K > 1\)), but a high activation energy can kinetically stop it from happening.

4. Commercial Cells and Fuel Cells

Categories of Commercial Cells

• Primary Cells: Non-rechargeable. The electrochemical reactions inside the cell are irreversible. Once the chemicals are consumed, the cell is spent (e.g., zinc-carbon or standard alkaline batteries).
• Secondary Cells: Rechargeable. The electrochemical reactions are reversible by applying an external electrical current in the opposite direction, regenerating the original reactants (e.g., lithium-ion batteries, lead-acid car batteries).
• Fuel Cells: Use a continuous external supply of fuel (such as hydrogen) and an oxidant (such as oxygen). The fuel is oxidised electrochemically at an electrode rather than burned in combustion, producing electricity with high efficiency and lower greenhouse gas emissions.

The Hydrogen-Oxygen Fuel Cell

The hydrogen-oxygen fuel cell can operate under either acidic or alkaline conditions. You must know the electrode reactions and overall equations for both electrolytes.

Acidic Electrolyte Fuel Cell

• Anode (Oxidation):
\(\text{H}_2(\text{g}) \rightarrow 2\text{H}^+(\text{aq}) + 2\text{e}^- \quad (E^\ominus = 0.00\text{ V})\)

• Cathode (Reduction):
\(\text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}(\text{l}) \quad (E^\ominus = +1.23\text{ V})\)

• Overall Cell Reaction:
\(2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l}) \quad (E^\ominus_{\text{cell}} = +1.23\text{ V})\)

Alkaline Electrolyte Fuel Cell

• Anode (Oxidation):
\(\text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \quad (E^\ominus = -0.83\text{ V})\)

• Cathode (Reduction):
\(\text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^- \rightarrow 4\text{OH}^-(\text{aq}) \quad (E^\ominus = +0.40\text{ V})\)

• Overall Cell Reaction:
\(2\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{H}_2\text{O}(\text{l}) \quad (E^\ominus_{\text{cell}} = (+0.40) - (-0.83) = +1.23\text{ V})\)

Notice that the overall equation and the overall standard cell potential (\(+1.23\text{ V}\)) are identical regardless of whether acidic or alkaline conditions are used! The only product formed is water (\(\text{H}_2\text{O}\)).

Key Takeaway

Fuel cells operate continuously as long as fuel and oxygen are supplied. The overall equation in both acidic and alkaline hydrogen fuel cells is always \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\) with an overall \(E^\ominus_{\text{cell}} = +1.23\text{ V}\).

5. Quantitative Redox Titrations and Core Practical 11

Core Practical 11: Determination of Iron Content in Iron Tablets Using Potassium Manganate(VII)

Potassium manganate(VII), \(\text{KMnO}_4\), is a powerful oxidising agent used in quantitative volumetric analysis.

• The Reaction:
\(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)
• Stoichiometry: \(1\text{ mol }\text{MnO}_4^- \equiv 5\text{ mol }\text{Fe}^{2+}\)

Crucial Practical Details & Acid Choice (Common Exam Questions)

• Why must the solution be acidified with dilute sulfuric acid (\(\text{H}_2\text{SO}_4\))?
Dilute sulfuric acid provides the necessary \(\text{H}^+\) ions without taking part in any competing redox reaction.

• Why can we NOT use hydrochloric acid (\(\text{HCl}\))?
The \(\text{MnO}_4^-\) ions are strong enough to oxidise chloride ions (\(\text{Cl}^-\)) to chlorine gas (\(\text{Cl}_2\)). This would consume extra manganate(VII) solution, resulting in an artificially high titre value and overestimating the iron content.

• Why can we NOT use nitric acid (\(\text{HNO}_3\))?
Nitric acid is itself an oxidising agent and would oxidise some of the \(\text{Fe}^{2+}\) ions to \(\text{Fe}^{3+}\), resulting in a smaller volume of \(\text{MnO}_4^-\) being used (artificially low titre).

• Why can we NOT use concentrated \(\text{H}_2\text{SO}_4\) or ethanoic acid (\(\text{CH}_3\text{COOH}\))?
Concentrated \(\text{H}_2\text{SO}_4\) is an oxidising agent. Ethanoic acid is a weak acid and cannot supply the high concentration of \(\text{H}^+\) ions required for the reaction.

• Self-Indicating End Point:
No external indicator is needed! The \(\text{MnO}_4^-\) ion is deep purple, while \(\text{Mn}^{2+}\) is virtually colourless. As \(\text{MnO}_4^-\) is added from the burette into the conical flask, it is immediately decolourised by \(\text{Fe}^{2+}\). The end point is reached when all \(\text{Fe}^{2+}\) has reacted and one drop of excess \(\text{MnO}_4^-\) turns the solution from colourless to the first permanent faint pink.

• Burette Reading Tip:
Because potassium manganate(VII) is an intensely dark purple liquid, it is difficult to see the bottom of the meniscus. You must read from the top of the meniscus consistently for both initial and final readings.

Iodine-Thiosulfate Titrations

Iodine-thiosulfate titrations are used to determine the concentration of various oxidising agents (such as \(\text{Cu}^{2+}\), \(\text{ClO}^-\), or \(\text{IO}_3^-\)) in a two-step procedure.

Step 1: Generation of Iodine
An excess of potassium iodide (\(\text{I}^-\)) is added to the oxidising agent. The oxidising agent oxidises \(\text{I}^-\) to liberate iodine (\(\text{I}_2\)).
Example (with copper(II) ions):
\(2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})\)
(Note: \(\text{CuI}\) is a white precipitate, and \(\text{I}_2\) makes the solution brown)

Step 2: Titration of Liberated Iodine
The liberated iodine is titrated against a standard solution of sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)):
\(\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})\)
Stoichiometry: \(1\text{ mol }\text{I}_2 \equiv 2\text{ mol }\text{S}_2\text{O}_3^{2-}\)
Therefore: \(2\text{ mol }\text{Cu}^{2+} \equiv 1\text{ mol }\text{I}_2 \equiv 2\text{ mol }\text{S}_2\text{O}_3^{2-}\), giving a \(1:1\) molar ratio between \(\text{Cu}^{2+}\) and \(\text{S}_2\text{O}_3^{2-}\).

The Indicator and Timing:
• Indicator: Starch solution.
• When to add: Titrate the brown iodine solution until it becomes a pale straw-yellow colour, then add a few drops of starch indicator. The solution turns blue-black.
• Why not add starch at the beginning? If starch is added when iodine concentration is high, iodine binds irreversibly to the starch, forming an insoluble complex and making the end point inaccurate.
• End Point: The sharp disappearance of the blue-black colour, turning completely colourless (with a white precipitate of \(\text{CuI}\) present if analyzing copper).

Key Takeaway

In manganate titrations, use dilute \(\text{H}_2\text{SO}_4\) and look for the first permanent pale pink colour. In thiosulfate titrations, add starch only when pale straw-yellow, watching for the blue-black to colourless transition.

6. Summary of Key Pitfalls and Examiner Warnings

• Multiplying \(E^\ominus\) values: Never multiply standard electrode potential values by integers when balancing electrons! \(E^\ominus\) is an intensive property and does not depend on the quantity of material reacting.
• Omitting inert platinum electrodes: In IUPAC cell notations involving gases (e.g., \(\text{Cl}_2/\text{Cl}^-\)) or ion/ion mixtures (e.g., \(\text{Fe}^{3+}/\text{Fe}^{2+}\)), always remember to write \(\text{Pt}(\text{s})\) at the outer edge.
• Vague explanations for acid choice: Stating that "hydrochloric acid reacts" will not earn full marks. You must explicitly state that \(\text{MnO}_4^-\) oxidises \(\text{Cl}^-\) to \(\text{Cl}_2\).
• Confusing thermodynamics with kinetics: A positive \(E^\ominus_{\text{cell}}\) means a reaction is thermodynamically feasible, but does not guarantee it will occur at a measurable rate due to high activation energy.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the systematic 4-step method for balancing redox half-equations in acidic aqueous conditions?

1. Balance atoms other than \(\text{O}\) and \(\text{H}\).
2. Balance \(\text{O}\) by adding \(\text{H}_2\text{O}\).
3. Balance \(\text{H}\) by adding \(\text{H}^+\).
4. Balance overall charge by adding electrons (\(\text{e}^-\)).

What is the reduction half-equation and oxidation state change for manganate(VII) ions in acidic solution?

\(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightleftharpoons \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\), where manganese changes from oxidation state \(+7\) to \(+2\).

What is the definition of standard electrode potential (\(E^\ominus\))?

The electromotive force (EMF) generated when a half-cell is connected to a Standard Hydrogen Electrode (SHE) under standard conditions ( \(298\text{ K}\), \(100\text{ kPa}\), and \(1.00\text{ mol dm}^{-3}\) ion concentrations).

What are the essential components and operating conditions of the Standard Hydrogen Electrode (SHE)?

\(\text{H}_2(\text{g})\) at \(100\text{ kPa}\), \(1.00\text{ mol dm}^{-3}\ \text{H}^+(\text{aq})\), a platinised platinum electrode, at \(298\text{ K}\) with \(E^\ominus = 0.00\text{ V}\).

What roles do the high-resistance voltmeter and salt bridge play in an electrochemical cell?

The voltmeter measures maximum potential difference (EMF) with virtually zero current drawn; the salt bridge completes the circuit and maintains electrical neutrality by allowing ion migration without mixing solutions.

How is IUPAC standard cell notation structured for electrochemical cells?

Oxidation (anode) is on the left, reduction (cathode) on the right; phase boundaries are shown as \(\mid\), salt bridges as \(\parallel\), and species in the same phase separated by commas (\(,\)).

How are standard cell potential (\(E^\ominus_{\text{cell}}\)), total entropy change (\(\Delta S_{\text{total}}\)), and the equilibrium constant (\(K\)) related?

\(E^\ominus_{\text{cell}}\) is directly proportional to \(\Delta S_{\text{total}}\) (\(\Delta S_{\text{total}} = nFE^\ominus_{\text{cell}}\)) and proportional to \(\ln K\) (\(\ln K \propto E^\ominus_{\text{cell}}\)).

Why might a reaction with a positive \(E^\ominus_{\text{cell}}\) fail to occur under laboratory conditions?

Due to kinetic inhibition (a high activation energy, \(E_a\)) or non-standard conditions that shift electrode potentials away from standard values.

What is a disproportionation reaction, and how can \(E^\ominus\) values be used to determine its thermodynamic feasibility?

A reaction where a single species is simultaneously oxidised and reduced; it is feasible if \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} > 0\text{ V}\).

What are the electrode half-reactions and overall equation for an alkaline hydrogen-oxygen fuel cell?

Anode: \(\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2\text{e}^-\); Cathode: \(\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightarrow 4\text{OH}^-\); Overall: \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\) (\(E^\ominus_{\text{cell}} = +1.23\text{ V}\)).

Why must dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)) be used instead of \(\text{HCl}\) or \(\text{HNO}_3\) in manganate(VII) titrations?

\(\text{HCl}\) is oxidised by \(\text{MnO}_4^-\) to \(\text{Cl}_2\) (giving an artificially high titre); \(\text{HNO}_3\) oxidises \(\text{Fe}^{2+}\) itself (giving an artificially low titre); dilute \(\text{H}_2\text{SO}_4\) does not undergo competing redox.

In an iodine-thiosulfate titration, when should the starch indicator be added and what is the end-point colour change?

Add starch when the iodine solution reaches a pale straw-yellow colour; the end point is the sharp change from blue-black to completely colourless.

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