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5.1.4Unit 5. Transition Metals and Organic Nitrogen Chemistry

Conventional Cell Representation — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 10 of 43 Unit quiz

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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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Part 1 of 5Rules of the Notation

The convention describes a cell completely in one line, including the states of every species.

Phase boundary
  • Phase boundary A single vertical line separates species in different phases, such as a solid electrode and its solution.
Salt bridge
  • Salt bridge A double vertical line represents the salt bridge linking the two half-cells.
Left electrode
  • Left electrode The half-cell on the left is written so that oxidation occurs there, and it is the negative terminal in a working cell.
Same phase
  • Same phase Species in the same solution are separated by a comma, for example Fe²⁺(aq), Fe³⁺(aq).
Inert electrode
  • Inert electrode Platinum is written next to the solution when no solid metal is present in the half-cell.

Part 2 of 5Examples

Examples
Metal cell
  • Metal cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) represents the zinc-copper cell with zinc oxidised on the left.
Ion cell
  • Ion cell Pt(s) | Fe²⁺(aq), Fe³⁺(aq) || Ag⁺(aq) | Ag(s) has an inert platinum electrode in the left-hand half-cell.
Hydrogen cell
  • Hydrogen cell Pt(s) | H₂(g) | H⁺(aq) || Cu²⁺(aq) | Cu(s) puts the hydrogen electrode on the left as a reference.

Part 3 of 5From Diagram to Voltage

1
Identify the left half-cell as the oxidation (the reversed half-equation) and the right half-cell as the reduction.
2
Look up the E⦵ values written as reductions for both half-cells.
3
Calculate E⦵cell = E⦵(right) - E⦵(left), which is positive when the cell as written produces a current spontaneously.
4
Write the electrons lost at the left and gained at the right, balance them, and add to give the overall equation.
5
If the answer is negative, the reaction goes in the reverse direction as written and the diagram should be reversed.

Part 4 of 5Worked Example

Question Calculate E⦵cell for Pt(s) | Fe²⁺(aq), Fe³⁺(aq) || Ag⁺(aq) | Ag(s), given E⦵(Fe³⁺/Fe²⁺) = +0.77 V and E⦵(Ag⁺/Ag) = +0.80 V.

  1. E⦵cell = E⦵(right) - E⦵(left) = +0.80 - (+0.77) = +0.03 V.
  2. Left (oxidation)
    • Fe²⁺ → Fe³⁺ + e⁻.
    • Right (reduction): Ag⁺ + e⁻ → Ag.

Answer

  • E⦵cell = +0.03 V;
    • Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)

Part 5 of 5Exam Checklist & Tips

5.1.4 – Key Points for the Exam

  • Cell diagram A shorthand showing electrodes, solutions and the salt bridge.
  • Anode Electrode where oxidation occurs, on the left of the diagram.
  • Cathode Electrode where reduction occurs, on the right.
  • Phase boundary The line between two different phases in the notation.

Common Mistakes

  • Placing the salt bridge with a single line instead of a double line.
  • Putting the reduction half-cell on the left.
  • Forgetting state symbols in a conventional representation.
  • Writing the platinum electrode in the middle of a half-cell.

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Cambridge 9700 AS Biology Core Formula

Magnification Triangle & Unit Conversion

Formula: M = I / A  ⇔  A = I / M  ⇔  I = M × A

Units: 1 mm = 1,000 μm = 1,000,000 nm.
Rule: Convert Image size (ruler mm) to μm by ×1,000 before dividing by Actual size.

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