---
title: "Balancing redox half-equations Template | MolDraw"
url: https://www.moldraw.com/templates/school-redox-half-equations/
description: "Balancing redox half-equations template: Step-by-step ion–electron (half-equation) method in acidic solution for GCSE, A-level and NCERT Class 11: MnO₄⁻/Fe²⁺…"
lang: en
publisher: Scidart Academy
generated: 2026-11-13
---
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# Balancing redox half-equations template

![Balancing redox half-equations template: Step-by-step ion–electron (half-equation) method in acidic solution for GCSE, A-level and NCERT Class 11: MnO₄⁻/Fe²⁺ (5 e⁻) and Cr₂O₇²⁻/I⁻ (6 e⁻) worked through each step, with charge checks, oxidation-state changes and colour changes.](https://www.moldraw.com/templates/school-redox-half-equations.webp)

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Step-by-step ion–electron (half-equation) method in acidic solution for GCSE, A-level and NCERT Class 11: MnO₄⁻/Fe²⁺ (5 e⁻) and Cr₂O₇²⁻/I⁻ (6 e⁻) worked through each step, with charge checks, oxidation-state changes and colour changes.

## What the template shows

- Balance atoms, then O with H₂O, H with H⁺ and charge with electrons; equalise the electrons and add
- Manganate(VII) oxidises iron(II) in acid
- Dichromate(VI) oxidises iodide in acid
- charge: −1 + 8 − 5 = +2 on both sides · Mn +7 → +2
- Other half
- × 5 so that both half-equations transfer 5 e⁻
- Purple MnO₄⁻ turns colourless: self-indicating titration (acidify with dilute H₂SO₄, not HCl).
- charge: −2 + 14 − 6 = +6 on both sides · each Cr +6 → +3
- × 3 so that both half-equations transfer 6 e⁻
- Orange Cr₂O₇²⁻ turns green Cr³⁺; the iodine released colours the solution brown.
- OIL RIG: oxidation is loss of electrons, reduction is gain. Electrons appear on the right of an oxidation half-equation and on the left of a reduction.
- In acid, balance O with H₂O and H with H⁺; in alkaline solution, then add the same number of OH⁻ to both sides to turn H⁺ into H₂O.
- The electrons gained must equal the electrons lost: multiply the half-equations so the electrons cancel. Check that atoms and total charge balance in the final equation.
- The change in oxidation state equals the electrons per atom: Mn +7 → +2 takes 5 e⁻; each Cr +6 → +3 takes 3 e⁻, 6 e⁻ for Cr₂O₇²⁻.
- Every part of this figure is editable · sources on the template card

Drawn with 5 shapes – every one editable in MolDraw.

## Sources

- [Housecroft & Sharpe, Inorganic Chemistry, 5th ed., Pearson 2018, ch. 8](https://openlibrary.org/isbn/9781292134147) Textbook
- [NCERT, Chemistry Part II, Class XI, ch. 7 Redox Reactions](https://ncert.nic.in/) Textbook

## How to use this template

1. Press **Edit in MolDraw**. The template opens on the canvas of the free MolDraw editor.
2. Change anything: move structures, edit atoms and labels, recolour highlights or delete what you do not need.
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## Frequently asked questions

### Is the Balancing redox half-equations template free?

You can preview the Balancing redox half-equations template here and open and edit it in MolDraw for free. Downloading or exporting it is part of MolDraw Pro, together with the other premium templates.

### Can I edit the Balancing redox half-equations template?

Yes. It opens as an ordinary MolDraw drawing, so every structure, arrow, label and colour can be changed, and you can add your own content around it.

### Is the chemistry checked?

Every structure is checked with RDKit and every curly arrow passes MolDraw’s electron-flow audit. The sources used are listed on this page.

### Which MO diagrams are included?

H₂ and He₂, N₂ and O₂ (showing s–p mixing and paramagnetic O₂), and carbon monoxide.

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