10.1 Balancing oxidation-reduction reactions  (Page 4/12)

However, under basic conditions, ${\text{MnO}}_4{}^{\text{−}}$ normally reduces to MnO ₂ and iron will be present as either Fe(OH) ₂ or Fe(OH) ₃ . For these reasons, under basic conditions, this reaction will be

(Under very basic conditions ${\text{MnO}}_4{}^{\text{−}}$ will reduce to ${\text{MnO}}_4{}^{\mathrm{2-}},$ instead of MnO ₂ .)

It is still possible to balance any oxidation-reduction reaction as an acidic reaction and then, when necessary, convert the equation to a basic reaction. This will work if the acidic and basic reactants and products are the same or if the basic reactants and products are used before the conversion from acidic or basic. There are very few examples in which the acidic and basic reactions will involve the same reactants and products. However, balancing a basic reaction as acidic and then converting to basic will work. To convert to a basic reaction, it is necessary to add the same number of hydroxide ions to each side of the equation so that all the hydrogen ions (H ⁺ ) are removed and mass balance is maintained. Hydrogen ion combines with hydroxide ion (OH ⁻ ) to produce water.

Let us now try a basic equation. We will start with the following basic reaction:

Balancing this as acid gives

In this case, it is necessary to add two hydroxide ions to each side of the equation to convert the two hydrogen ions on the left into water:

Note that both sides of the equation show water. Simplifying should be done when necessary, and gives the desired equation. In this case, it is necessary to remove one H ₂ O from each side of the reaction arrows.

Again, check each side of the overall equation to make sure there are no errors:

Everything checks, so this is the overall equation in basic solution.

Balancing acidic 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 seven water molecules to the left side provides the necessary oxygen; the “left over” hydrogen appears as 14 H ⁺ on the right:

The left side of the equation has a total charge of [2 $\times$ (+3) = +6], and the right side a total charge of [−2 + 14 $\times$ (+1) = +12]. The difference is six; adding six electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):