17.1 Balancing oxidation-reduction reactions  (Page 5/12)

Checking the half-reaction:

Now work on the reduction. It is necessary to convert the four oxygen atoms in the permanganate into four water molecules. To do this, add eight H ⁺ to convert the oxygen into four water molecules:

Then add five electrons to the left side to balance the charge:

Make sure to check the half-reaction:

Collecting what we have so far:

The least common multiple for the electrons is 30, so multiply the oxidation half-reaction by five, the reduction half-reaction by six, combine, and simplify:

Checking each side of the equation:

This is the balanced equation in acidic solution.

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Balancing basic oxidation-reduction reactions

Solution

Starting with the oxidation half-reaction, we can balance the chromium

In acidic solution, we can use or generate hydrogen ions (H ⁺ ). Adding one water molecule to the left side provides the necessary oxygen; the “left over” hydrogen appears as five H ⁺ on the right side:

The left side of the equation has a total charge of [0], and the right side a total charge of [−2 + 5 $\times$ (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):

Checking the half-reaction:

Now work on the reduction. It is necessary to convert the four O atoms in the MnO ₄ ⁻ minus the two O atoms in MnO ₂ into two water molecules. To do this, add four H ⁺ to convert the oxygen into two water molecules: