[jennielynn_1980]

I have to write ionic equations for the following and I am unsure of how to do so.  Here is the question and my attempt at each:

a) chromium dipped into silver nitrate
Cr³⁺ +Ag⁺(NO₃)^- --> Cr ³⁺(NO3)^- +Ag⁺

b) gold immersed in hydrocholoric acid
NO REACTION and therefore NO EQUATION

c) nickel pellets dropped into a bath of calcium acetate
Ni²⁺ + Ca²⁺ (C₂H₃O₂)^2- --> Ca²⁺ + Ni ²⁺ (C₂H₃O₂)₂^-

d) aluminium dropped into a bath of sulphuric acid
Al³⁺ + H₂²⁺(SO₄)^2- --> H₂²⁺ + Al³⁺(SO₄)^2-

e) zinc dipped inot lead(II) nitrate

Zn²⁺ + Pb²⁺ (NO₃^-)₂ --> Pb ²⁺ +Zn ²⁺ (NO₃)₂

I have the rules for writing the equations but the examples given are not helping me answer these questions.  I think my mistakes are:
1) I am only supposed to ionize (aq) substances and I ionized everything
2) when writing a compound I am supposed to just give the charge of the whole molecule instead of each part.

Other than that I am lost.

[Dan]

1) yes, this is the problem. "Chromium" is Chromium metal, Cr NOT the ion, likewise Nickel is Ni, etc.

2) You can give the charge for each part, that's fine, and even write them seperately as this will make it easier to spot the spectator ions.

eg. Lithium dropped into aqueous HBr

HBr_(aq) <-----> H⁺_(aq) + Br^-_(aq)

Li_(s) + H⁺_(aq) + Br^-_(aq) -----> LiBr_(aq) + ¹/₂H_2(g)

But,

LiBr_(aq) <-----> Li⁺_(aq) + Br^-_(aq)

So overall,

Li_(s) + H⁺_(aq) + Br^-_(aq) -----> Li⁺_(aq) + Br^-_(aq) + ¹/₂H_2(g)

So Bromide is a spectator ion, and the net ionic equation is

Li_(s) + H⁺_(aq) -----> Li⁺_(aq) + ¹/₂H_2(g)

[jennielynn_1980]

Okay so a) might look like this:

a) chromium dipped into silver nitrate
Cr(s) +Ag⁺(NO₃)^- (aq)--> Cr³⁺(NO₃)^- +Ag(s)

Then seperate the aqueous ions right? so:

Cr(s) + Ag⁺(aq) + NO₃^-(aq) --> Cr³⁺(aq) + NO₃^-(aq) + Ag(s)

Then we can see NO₃ is a spectator ion so we would rewrite it as:

Cr(s) + Ag⁺(aq) --> Cr³⁺(aq) + Ag(s)

But somehow I think this is still wrong    Also, I can't figure out how solubility plays in.  I am seriously confused and frustrated!

[Donaldson Tan]

Cr(s) +Ag⁺(NO₃)^- (aq)--> Cr³⁺(NO₃)^- +Ag(s

Your formula for Chromium(III) Nitrate is wrong and the charge is not balanced. It should be Cr(NO₃)₃

Cr(s) + Ag⁺(aq) --> Cr³⁺(aq) + Ag(s)

Your previous mistake carries forward itself here. It resulted in your charges are not balanced. It should be:

Cr(s) + 3 Ag⁺(aq) --> Cr³⁺(aq) + Ag(s)

[jennielynn_1980]

I figured out e) and c). 

I have a question about d.  If I end up with 6H (aq) on one side of the equation and 6H (g) on the other side, do I still cross it out as spectator ions?


For a) I am still trying to balance it out after correcting my mistake in the formula for chromium(III)nitrate.  Here is what I have with the revisions:

chromium + silver nitrate --> silver + chromium nitrate

3Cr(s) + 3Ag(NO₃)(aq) --> 3Ag(s) + Cr(NO₃)₃ (aq)

3Cr(s) + 3Ag⁺ (aq) + 3NO₃^-(aq) --> 3Ag(s) + 3Cr³⁺ (aq) + 3NO₃^- (aq)

eliminate the NO₃ as a spectator ion and we have:

3Cr(s) + 3Ag⁺ (aq) --> 3Ag(s) + 3Cr³⁺ (aq)

The equation is balanced but the charge is not.  What am I missing?

[Borek]

http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-redox

[Dan]

I have a question about d.  If I end up with 6H (aq) on one side of the equation and 6H (g) on the other side, do I still cross it out as spectator ions?

Do you mean 6H⁺_(aq) and 3H_2(g)?

If the species does not remain exactly the same, you do not cross it out.

chromium + silver nitrate --> silver + chromium nitrate

3Cr(s) + 3Ag(NO₃)(aq) --> 3Ag(s) + Cr(NO₃)₃ (aq)

Count your Cr, you have balanced it incorrectly, you have 3 on the left and one on the right.

[jennielynn_1980]

Thanks

I did realize that it was balanced incorrectly eventually.  I also had an incorrect formula for one of the other questions.  Now that is sorted, I think I got them all.