Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

8 H+ + 5 Fe2+ + MnO 4 -  Mn2+ + 5 Fe3+

The last step is to balance with H 2 O in order to make sure that there is the same number of atoms of both

side of the reaction arrow. In this case it is sufficient to place 4 H 2 O-molekules on the right side:

8 H+ + 5 Fe2+ + MnO 4 -  Mn2+ + 5 Fe3+ + 4 H 2 O

As en extra control one may check if the charges are the same of both sides of the reaction arrow:

Left side: 8 · (+1) + 5 · (+2) + (-1) = +17

Right side: (+2) + 5 · (+3) + 4 · 0 = +17

As an extra-extra control one may make sure that there are equally many of each type of atoms on both

sides of the reaction arrow:

Left side: 5 Fe, 1 Mn, 4 O, 8 H

Right side: 5 Fe, 1 Mn, 4 O, 8 H

Even though there is only a minor difference between the balance of redox reactions in acid and basic

solutions the principles differ in that the equation are balanced with OH- ions in stead of H+. this we will look

into in the following example:

Electrochemistry

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