| Contents for this page | Related topics |  |
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The ion-electron method Additional questions |
Electrolysis The Cu/Zn cell Standard electrode potentials |
 Data Glossary |
| Learning Outcomes | ||
| After studying this section, you will (a) understand the concept of half-reactions and (b) be able to balance redox equations using the ion-electron method. | ||
The following will demonstrate how the ion-electron method can be used to balance redox reactions. Let us take the reaction between potassium permanganate (KMnO4) and iron(II) sulphate (FeSO4), in which the permanganate ion (MnO4-) is reduced to manganese(II) cations (Mn2+), while the iron(II) cations (Fe2+) are oxidised to iron(III) cations (Fe3+). The reaction takes place in acid solution.
| Step 1: We first summarise the above chemical facts in a rough unbalanced reaction: Note that we ignore the presence of both the potassium and sulphate ions, which do not undergo any change. They are the spectator ions. |
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| Step 2: We now split this reaction into two half-reactions which separately describe the reduction and oxidation. In the next two steps, we will balance these half-reactions atomically and electronically. |
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| Step 3: Balancing the reduction half reaction: The oxygen atoms of the permanganate ion and the hydrogen ions must combine to form water. Since there are 4 oxygen atoms, 4 water molecules must appear on the right- hand side of the equation. To produce 4 water molecules, 8 hydrogen atoms are required. These must be derived from 8 H+ ions on the left-hand side. This gives us the reaction on the right |
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| The above does not balance electronically: on the left-hand side we have a net charge of 6+, while on the right hand-side we have 2+. This means that we must add 4 negative charges ( 4 electrons) on the left-hand side. This brings the total number of electrons on the left to 5, which gives us a the reaction on the right which is electronically balanced: |  | |
| Step 4: Balance the oxidation half-reaction: in this case, it is already balanced atomically and electronically: |  | |
| Step 5: The final reaction is the sum of the above two half-reactions, bearing in mind that the final reaction must not only balance atomically, but electronically as well. Just adding the two half-reactions above will
result in a reaction which will be balanced atomically, but there will be 5 e- on the left and only 1 e- on the right
All that needs to be done is to multiply the oxidation half-reaction by 5, and add it to the balanced reduction half-reaction. After cleaning it up by removing items which appear in equal amounts on both sides, we finally get:
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