You need to know the rules for oxidation state. Here is a good source.
http://www.chemteam.info/Redox/Redox-Rules.html
I will do the first for you.
Al(s) + O2(g) -----> Al2O3(s)
a. The equation isn't balanced but that isn't necessary in order to determine what is oxidized and what is reduced.
Al on the left is zero oxidation state because all elements in the free state (uncombined in a compound) are zero. On the right it is +3. How do you know it's +3? Because oxygen is -2, there are three of them to make -6 total; therefore, Al must be +6 to balance the charge to make zero for the compound (All compounds are zero). There are two Al atoms so each must be +3. So Al moved from zero to +3, that is a loss of electrons; therefore, Al is oxidized.
O on the left is zero. On the right of -2 each. Therefore, O had to gain electrons to move from zero to -2. Thus reduction.
Remember the definitions.
Loss of electrons is oxidation.
Gain of electrons is reduction.
Here is a mnemonic to help.
Leo the lion goes grrrr. Focus on LEO. L for loss. E for electrons. O for oxidation.
I shall be happy to check your answers to #2.
Can someone please help me with the=is problem...
In each of the following, determine which reactant is being oxidized and which is being reduced:
A. Al(s) + O2(g) -----> Al2O3(s)
B. Ca(s) + F2(g) -----> CaF2(s)
Thanks so much for the help ;-)
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