Question 1

How do you balance the redox reaction between permanganate and Fe2+?

Accepted Answer

First balance the reduction half-reaction as MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. The oxidation half-reaction is Fe2+ → Fe3+ + e-. Multiply it by 5, cancel the electrons, and obtain MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O.

Question 2

Why are five Fe2+ ions needed for one permanganate ion?

Accepted Answer

Each Fe2+ ion changes from oxidation number +2 to +3 and loses one electron, so five Fe2+ ions lose five electrons in total. Manganese changes from +7 to +2 and gains five electrons, so the electron losses and gains match one-to-one.

Question 3

How can you check that a redox equation is balanced?

Accepted Answer

Count every element on both sides and compare the net charges. For the displayed acidic-medium equation, both sides contain Mn 1, Fe 5, O 4 and H 8. The charge is also +17 on each side: -1 + 10 + 8 on the left and +2 + 15 on the right.

Question 4

What changes when this reaction is balanced in basic medium?

Accepted Answer

Hydrogen ions are replaced by hydroxide ions, and water is placed on the side that best balances oxygen and hydrogen. The corresponding equation is MnO4- + 5Fe2+ + 4H2O → Mn2+ + 5Fe3+ + 8OH-. Each side has charge +9 and contains Mn 1, Fe 5, O 8 and H 8.

Question 5

What is the difference between the ion-electron and oxidation-number methods?

Accepted Answer

The ion-electron method explicitly writes oxidation and reduction half-reactions, balances H, O and charge, and then matches electrons. The oxidation-number method first equalises the total positive and negative oxidation-number changes, then adjusts coefficients while balancing the remaining atoms and checking charge.

Balancing Redox in 3D — Split the Reaction, Count the Electrons

About this lab

Practice