bu puting your values in formula You need not to be concerned about the exact mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So which element was oxidized? Li does not have a charge while LiO is an ionic compound in which electrons are \"shared\" according to electronegativity. In this case oxygen is much more electronegative and it essentially obtains two electrons from Li through bonding. Now write out the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li-- ->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O (2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the original balanced equation. Solution bu puting your values in formula You need not to be concerned about the exact mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So which element was oxidized? Li does not have a charge while LiO is an ionic compound in which electrons are \"shared\" according to electronegativity. In this case oxygen is much more electronegative and it essentially obtains two electrons from Li through bonding. Now write out the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li-- ->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O (2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the original balanced equation..

1. bu puting your values in formula You need not to be concerned about the exact
mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So
which element was oxidized? Li does not have a charge while LiO is an ionic compound in
which electrons are "shared" according to electronegativity. In this case oxygen is much more
electronegative and it essentially obtains two electrons from Li through bonding. Now write out
the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li--
->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A
third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O
(2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the
original balanced equation.
Solution
bu puting your values in formula You need not to be concerned about the exact
mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So
which element was oxidized? Li does not have a charge while LiO is an ionic compound in
which electrons are "shared" according to electronegativity. In this case oxygen is much more
electronegative and it essentially obtains two electrons from Li through bonding. Now write out
the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li--
->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A
third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O
(2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the
original balanced equation.