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Derpoleohnekole
2 years ago

This is an O K2O (potassium oxide)

ChemLuis
2 years ago

You’ve done everything right up to the point where you brought together the oxidation partial equation and the reduction partial equation.

When you merge, you wrote K4O2 which was correct for the first time. However, potassium oxide is a salt.
For salts, no sum formula is given in the sense that the number of different ions (the lowest number behind the element symbol) is the actual number of ions in the salt.
Salts form an ion lattice which can be very large, for which reason only one ratio formula is given.
In water, the sum formula H2 O indicates that a water molecule consists exactly of 2 hydrogen and an oxygen atom.

Since the ion lattices of the salts cannot be determined exactly in size, it is also not possible to determine exactly how many ions are contained.

Thus, a ratio formula is given which describes the ratios of the positively charged ions (here potassium ion) and of the negatively charged ions (here oxide ion) in the salt. The ratios are chosen such that the positive and negative charges are exactly cancelled, with cation and anion always having the largest common divider in the numerical ratios to one another:

Although K4O2 is a neutral compound (four 1-fold positive charges of potassium ion and 2 times 2-fold negative charges of oxidion), it becomes clear at a glance of the ratios (calcium 4 to oxygen 2) that it can still be “short” because a 4:2 ratio is the same as a 2:1 ratio.

Thus, K4O2 describes the same salt as K2O, but the indication at K2O is the shortened, and thus the required. So you have to make sure that in such reactions, you always combine the resulting salts in the “smallest possible” relation to each other. So the smallest possible numbers of ions with which the connection becomes neutral.
In the case of a 2-fold negative oxidation and a 1-fold positive potassium ion, that is to say 2 potassium ions come to 1 oxidation, that is to say K2 O.

So that the number of ions on the left (4 potassium ions and 2 oxide ions) is correct with the number on the right, one writes after it has been found that K2O on the right has to stand a 2 ahead of the K2O so that: