Writing net ionic equations ?

  • Thread starter mcandrewsr
  • Start date
  • #1

Homework Statement



For each of the following situations, identify those for which a reaction likely to occur. For those that do occur, white a net ionic equation.
a) Chromium dipped into silver nitrate.
b) Gold immersed in hydrochloric acid
c) Nickel pellets dropped into calcium acetate, Ca(C2H3O2)2.
d) Aluminum dropped into a bath of sulphuric acid.
e) Zinc dipped into a solution of lead(II) nitrate.


The Attempt at a Solution



This is what I got:
a)Cr(s) + 2AgNO3(aq) --> Cr(NO3)2 + 2Ag(s)
Separate the aqueous ions:
Cr(s) + 2Ag(aq) + 2NO3(aq) --> Cr(aq) + 2NO3(aq) + 2Ag(s)
Then I'm supposed to remove the ions common to both sides...but
then I end up with none at all b/c each one is on both sides...???
b) no reaction
c) no reaction
d) 2Al(s) + 3H2SO4(aq) --> Al2(SO4)3(aq) + 3H2(g)
Here I get the same problem as in (a) when I separate them.
e) Zn(s) + Pb(NO3)2(aq) --> Zn(NO3)2(aq) + Pb(s)
Again, same problem....if I separate the aqueous ions and get rid of the common ones I end up with nothing.

What am I doing wrong? If someone can show me how to do this that would be great, thanks!
 

Answers and Replies

  • #2
danago
Gold Member
1,122
4

Homework Statement



For each of the following situations, identify those for which a reaction likely to occur. For those that do occur, white a net ionic equation.
a) Chromium dipped into silver nitrate.
b) Gold immersed in hydrochloric acid
c) Nickel pellets dropped into calcium acetate, Ca(C2H3O2)2.
d) Aluminum dropped into a bath of sulphuric acid.
e) Zinc dipped into a solution of lead(II) nitrate.


The Attempt at a Solution



This is what I got:
a)Cr(s) + 2AgNO3(aq) --> Cr(NO3)2 + 2Ag(s)
Separate the aqueous ions:
Cr(s) + 2Ag(aq) + 2NO3(aq) --> Cr(aq) + 2NO3(aq) + 2Ag(s)
Then I'm supposed to remove the ions common to both sides...but
then I end up with none at all b/c each one is on both sides...???
b) no reaction
c) no reaction
d) 2Al(s) + 3H2SO4(aq) --> Al2(SO4)3(aq) + 3H2(g)
Here I get the same problem as in (a) when I separate them.
e) Zn(s) + Pb(NO3)2(aq) --> Zn(NO3)2(aq) + Pb(s)
Again, same problem....if I separate the aqueous ions and get rid of the common ones I end up with nothing.

What am I doing wrong? If someone can show me how to do this that would be great, thanks!
Be careful when you cancel off spectator species, make sure that they are indeed the same thing!

Firstly, note that the silver and nitrate exist as IONS in solution i.e.:

AgNO3(aq) --> Ag+(aq) + NO3-(aq)

So make sure you include the charges. This then indicates that the silver (and chromium) species on each side are NOT the same i.e. Ag+(aq) is completely different from Ag(s), so you cannot cancel them off, so overall, you are left with:

Cr(s) + 3Ag+(aq) --> Cr3+(aq) + 3Ag(s)
 
  • #3
How do I know the different charges of ions when they are in different states? Ex. How does Cr(s) have a different charge than the aqueous one? How do I figure this out for any others?
 
  • #4
1,100
0
There are a set of elements that have set oxidation numbers (for example oxygen normally has an oxidation number of 2-). By using these, you can determine what the other charges are.

The Bob
 
  • #5
Borek
Mentor
28,544
2,988
How do I know the different charges of ions when they are in different states? Ex. How does Cr(s) have a different charge than the aqueous one? How do I figure this out for any others?
Cr(s) is not ionic.

Otherwise - use periodic table. But some things you will just have to remember.
 

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