كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

5.1Unit 5. Transition Metals and Organic Nitrogen Chemistry

Redox Equilibria — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

10 min read 12 parts Lesson 1 of 43 1 quiz

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision

Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution

pH = -log[H⁺]  |  pOH = -log[OH⁻]  |  pH + pOH = 14  |  [H⁺][OH⁻] = 10⁻¹⁴

Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14

Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.

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Lesson summary

5.1 Redox Equilibria — electrode potentials, cells, feasibility, fuel cells and redox titrations — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Define and measure standard electrode potentials, write and use cell diagrams, predict feasibility and its limits, describe fuel cells and carr…

Standard electrode potential, E⦵
What the symbols mean
Zn / Cu cell
Standard hydrogen electrode
Salt bridge
Standard cell potential

Part 1 of 10Oxidation and reduction in the blocks

OIL RIG

OIL RIG oxidation is loss of electrons and reduction is gain of electrons.

Oxidation is also an increase in oxidation number; reduction is a decrease.

s-block metals are oxidised to 1+ and 2+ ions: Na → Na⁺ + e⁻, Ca → Ca²⁺ + 2e⁻.

p-block metals are oxidised to positive ions.

Al → Al³⁺ + 3e⁻ matches the group.
Sn → Sn²⁺ + 2e⁻ does not match the group.

p-block non-metals are reduced to negative ions with charge = group number − 8.

F + e⁻ → F⁻ and O + 2e⁻ → O²⁻.

d-block elements have variable oxidation states and are usually oxidised to positive ions, e.g. Cu²⁺, Cr³⁺, V⁵⁺.

In acidic conditions balance O with H₂O and H with H⁺; combine half-equations so electrons cancel.

Part 2 of 10Standard electrode potential

Standard electrode potential

The position of equilibrium, and so the electrode potential, depends on temperature, gas pressure and concentration.

Standard conditions

Ion concentration 1.00 mol dm⁻³.
Temperature 298 K.
Pressure 100 kPa.

Standard electrode potential, E⦵

Standard electrode potential, E⦵ the potential difference when a standard half-cell is connected to a standard hydrogen electrode under standard conditions.

A high-resistance voltmeter is used so that no current flows and the maximum potential difference is obtained.

Half-equations are written as reductions; a more positive E⦵ means the species on the left is more easily reduced (a stronger oxidising agent).

Half-equation	E⦵ / V
Na⁺ + e- ⇌ Na	-2.71
Zn²⁺ + 2e- ⇌ Zn	-0.76
2H+ + 2e- ⇌ H₂	0.00
Cu²⁺ + 2e- ⇌ Cu	+0.34
Fe³⁺ + e- ⇌ Fe²⁺	+0.77
Ag⁺ + e- ⇌ Ag	+0.80
Br₂ + 2e- ⇌ 2Br-	+1.09
MnO₄⁻ + 8H+ + 5e- ⇌ Mn²⁺ + 4H₂O	+1.52

Part 3 of 10The standard hydrogen electrode

The standard hydrogen electrode

It is the reference half-cell, given E⦵ = 0.00 V.

Components

H₂ gas at 100 kPa in equilibrium with H⁺ at 1.00 mol dm⁻³: 2H⁺ + 2e⁻ ⇌ H₂.
An inert platinum electrode in contact with both.
Platinum black (finely divided platinum) increases the surface area and the rate of attaining equilibrium.

Secondary standards such as a calomel electrode are easier to use than the SHE.

Part 4 of 10Measuring electrode potentials

Metal / metal ion half-cell (e.g. Ag⁺/Ag)

Connected to the SHE; E⦵cell = (+0.80) − (0.00) = +0.80 V.
The more positive half-cell is the positive pole (reduction); the SHE is the negative pole (oxidation).

Non-metal / non-metal ion half-cell (e.g. Br₂/Br⁻)

A platinum electrode makes contact; platinum is inert and the equilibrium is set up on its surface.

Ion / ion half-cell (e.g. MnO₄⁻/Mn²⁺)

A platinum electrode is used; H⁺ is also present because it is needed to convert MnO₄⁻ to Mn²⁺.

Keep the temperature at 298 K and gas pressure at 100 kPa, and measure with a high-resistance voltmeter.

Part 5 of 10The salt bridge

The salt bridge

A salt bridge completes the circuit by letting ions move between the half-cells, balancing charge as electrons flow through the wire.

A metal wire cannot be used because it carries electrons, not ions.

KNO₃ or KCl is used because the ions are very soluble, unreactive with the half-cell ions and unlikely to form precipitates.

Part 6 of 10Cell potential and conventional cell diagrams

What the symbols mean:

E⦵cell = E⦵right − E⦵left = E⦵reduction − E⦵oxidation, using the more positive E⦵ for the reduction.

What the symbols mean: E_cell = standard cell potential (V) · E_right = electrode potential of the positive (reduction) half-cell · E_left = potential of the negative half-cell (V)

Zn / Cu cell

(+0.34) − (−0.76) = +1.10 V; the Cu²⁺/Cu half-cell is the positive pole.

Conventional representation

Reduced form | oxidised form || oxidised form | reduced form.
The more negative half-cell (oxidation) is written on the left: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).
| is a phase boundary, || the salt bridge, and commas separate species in the same phase.
Use a platinum electrode when both species are aqueous or one is a gas: Pt | Fe²⁺(aq), Fe³⁺(aq).
Cl₂(g), 2Cl⁻(aq) | Pt.
Cell diagrams are not quantitative.

Left Is Oxidation; the order of species is Reduced/Oxidised || Oxidised/Reduced.

Part 7 of 10Feasibility, thermodynamics and limitations

A reaction is feasible when E⦵cell is positive; the reverse reaction has the opposite sign.

Cl₂/Cl⁻ (+1.36 V) and Cu²⁺/Cu (+0.34 V) give Cu + Cl₂ → Cu²⁺ + 2Cl⁻ with E⦵cell = +1.02 V.

E⦵cell is proportional to ΔS_total and to ln K.

ΔG = −nFE⦵cell and ΔG = −RT ln K.
A value above about +0.3 V often means the reaction goes almost to completion.

Limitations of predictions

E⦵ gives no information about rate;
- a high activation energy can make a feasible reaction kinetically stable, e.g. Cu²⁺ + H₂ → Cu + 2H⁺ (+0.34 V) does not occur.
Non-standard concentrations shift the equilibrium and change the electrode potential (Le Chatelier).
Many redox reactions are not aqueous.

For V³⁺ + e⁻ ⇌ V²⁺

More V³⁺ shifts the equilibrium right, removing electrons, so the potential becomes less negative.
More V²⁺ shifts it left, so the potential becomes more negative.

Part 8 of 10Fuel cells

Fuel cell

Fuel cell a cell in which a fuel donates electrons at one electrode and oxygen gains electrons at the other;
- fuel and oxidant are supplied continuously and the energy is not stored in the cell.

The anode is the negative electrode and the cathode is the positive electrode.

Hydrogen–oxygen cell

Porous platinum-coated electrodes separated by an electrolyte or membrane, with separate inlets for H₂ and O₂ and an outlet for water.
Alkaline
- anode H₂ + 2OH⁻ → 2H₂O + 2e⁻ (−0.83 V);
  - cathode O₂ + 2H₂O + 4e⁻ → 4OH⁻ (+0.40 V).
Acidic
- anode H₂ → 2H⁺ + 2e⁻ (0.00 V);
  - cathode O₂ + 4H⁺ + 4e⁻ → 2H₂O (+1.23 V).
Overall 2H₂ + O₂ → 2H₂O, E⦵cell = +1.23 V.

Advantages

Water is the only product.
More efficient than combustion because energy goes to electricity rather than heat and light, and no NO_x forms.
Used on spacecraft, where the water can be drunk.

Disadvantages

Hydrogen is highly flammable, hard to store and needs thick-walled cylinders.
Production often depends on fossil fuels, so widespread use is limited until it is cheap.
Hydrogen has high energy per gram but low energy per unit volume, so larger containers are needed.
Platinum catalysts are expensive.

Ethanol and methanol fuel cells also exist.

Part 9 of 10Redox titrations

Redox titrations

An oxidising agent is titrated against a reducing agent, with electrons transferred; transition metal ions often change colour so no indicator is needed.

Potassium manganate(VII)

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, mole ratio 1 : 5.
Acidify with dilute sulfuric acid, not HCl, which would be oxidised to Cl₂.
Self-indicatingpurple becomes colourless until the first permanent pale pink.
Used for the percentage purity of iron tablets and the formula of hydrated ethanedioic acid;
- MnO₄⁻ : C₂O₄²⁻ is 2 : 5 (warm to about 60 °C).

Iodine–thiosulfate: 2S₂O₃²⁻ + I₂ → 2I⁻ + S₄O₆²⁻

An oxidant (e.g. ClO⁻) liberates I₂ from excess KI, and the I₂ is titrated with thiosulfate.
Add starch when the solution is straw coloured; blue-black to colourless is the end point.

Titrant	Analyte	Mole ratio	End point
KMnO₄ (acidified)	Fe²⁺	1 : 5	First permanent pale pink
KMnO₄ (acidified)	C₂O₄₂-	2 : 5	Pale pink
Na₂S₂O₃	I₂	2 : 1	Blue-black to colourless (starch)
K₂Cr₂O₇	Fe²⁺	1 : 6	Redox indicator colour change

Part 10 of 10Redox titration calculations and uncertainty

Write the half-equations and the overall equation.

Calculate the moles of titrant (concentration × volume in dm³).

Use the mole ratio to find the moles of the reductant or oxidant in the aliquot.

Scale up to the whole solution, e.g. × 10 if 25.0 cm³ of 250 cm³ was used.

Find the concentration, mass or percentage purity.

Redox titration calculations and uncertainty

Percentage purity = mass of pure substance ÷ mass of impure sample × 100.

Percentage uncertainty = uncertainty ÷ measured value × 100; readings taken twice double the absolute uncertainty.

Reducing uncertainty in a titration: make the burette readings larger by diluting the solution in the burette.

Key terms

Term	Meaning
Standard electrode potential	The emf of a half-cell measured against the standard hydrogen electrode under standard conditions.
Standard hydrogen electrode	H₂ at 100 kPa over platinum in 1.00 mol dm⁻³ H⁺ at 298 K, with E⦵ = 0.00 V.
Salt bridge	A connection that allows ions to move between half-cells and completes the circuit.
Standard cell potential	E⦵cell = E⦵right − E⦵left.
Fuel cell	A cell that converts the chemical energy of a continuously supplied fuel directly into electrical energy.
Redox titration	A titration in which the reaction involves electron transfer.
Oxidising agent	A species that accepts electrons and is itself reduced.

Avoid these mistakes

Writing half-cells with the oxidation on the right — oxidation is on the left of the cell diagram.

Using HCl to acidify a manganate(VII) titration — it is oxidised to chlorine, so use dilute sulfuric acid.

Multiplying E⦵ by the number of electrons when balancing equations — E⦵ does not change with stoichiometry.

Saying a positive E⦵cell means the reaction will be fast — it only shows feasibility.

Forgetting platinum electrodes for ion/ion and gas half-cells.

Last updated: 18 October 2026