This is an oxidation-reduction (redox) reaction:
OII + 2 e- → O0 (reduction)
O-II - 2 e- → O0 (oxidation)
F2O is an oxidizing agent, H2O is a reducing agent (synproportionation (comproportionation)).
Word equation
oxygen difluoride + water → hydrogen fluoride + oxygen
Input interpretation
F2O + H2O → HF + O2
Oxygen difluoride water hydrogen fluoride oxygen
Balanced equation
Balance the chemical equation algebraically:
F2O + H2O → HF + O2
Add stoichiometric coefficients, c,, to the reactants and products:
c1F2O + c2H2O → c3 HF + c4O2
Set the number of atoms in the reactants equal to the number of atoms in the products for F, O and H:
F: 2c1 = c3
O: c1 + c2 = 2c4
H: 2c2 = c3
Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c1 = 1 and solve the system of equations for the remaining coefficients:
c1 = 1 c2=1
c3 = 2 c4=1
Substitute the coefficients into the chemical reaction to obtain the balanced equation:
Answer:
F2O + H2O → HF + O2
