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Inorganic and Applied Chemistry
As we have a basic solution we balance with 2 OH- on the right side of the reaction giving:
4Ag + 8CN- + O 2 4Ag(CN) 2 - + 4 OH-
Last step is to balance with water molecules such that the atoms match on both sides of the reaction arrow.
In the last case we balance with 2 H 2 O-molecules on the left side:
2H 2 O + 4Ag + 8CN- + O 2 4Ag(CN) 2 - + 4 OH-
As an extra control one may control that the charges match on both sides of the reaction arrow:
Left side: 8 · (-1) = -8
Right side: 4 · (-1) + 4 · (-1) = -8
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: 4 Ag, 8 C, 4 O, 8 N, 4 H
Right side: 4 Ag, 8 C, 4 O, 8 N, 4 H
Electrochemistry
what‘s missing in this equation?
maeRsK inteRnationaL teChnoLogY & sCienCe PRogRamme
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