Ok, I've got a problem... The question is 'Solutions of silver nitrate and sodium chromate are mixed. Write the equation, balanced and in net ionic form. What is the oxidation number of chromium in the chromate ion?
I need help on if my equation is correct, and how to write it in net ionic form.
I think the equation is:
AgNO3+Na2CrO4=AgCrO4(s)+NaNO3
The oxidation number of Cr is +5? I am very confused, so any help would be greatly appreciated!
2007-05-10 14:56:54 · 2 answers · asked by J.S. 2 in Science & Mathematics ➔ Chemistry
The NET IONIC form means that you eliminate the spectator ions.
What you want to do is get rid of the nitrate and the sodium ion, because they are soluble and don't do anything in this reaction.
Here is the balanced equation in its net ionic form:
2 Ag(+)(aq) + CrO4(2-)(aq) ==> Ag2CrO4(s)
The chromate ion is CrO4 with a -2 charge.
Since each oxygen is -2, that would be a total of -8 for the four oxygen atoms. Now the overall charge on the ion is -2. So the Cr would have to be +6, so that the overall charge on the ion is equal to -2.
Make sense? If not, ask me another question about oxidation numbers, and I'd be happy to explain.
2007-05-10 15:03:32 · answer #1 · answered by mrfarabaugh 6 · 0⤊ 0⤋
2AgNO3 + Na2CrO4 ===> Ag2CrO4 + 2 NaNO3
The oxidation number of Cr is 6+. Count it up!
2007-05-10 22:02:37 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋