Solve Ionic Equation: Cr + AgNO₃ → CrN₃ + Ag

[Jchem]

Ionic equation help

Hello, I was wondering if someone could help me work through an Ionic equation question step by step. (Its been a few years since I've done this stuff, and I think if I work through an example It'll come back)

This is a question from my textbook. I have the answer already...just not sure exactly how to get to it.

1) Chromium dipped into silver nitrate.

so thats:

Cr(s) + AgNO3(l) ---------->

Now i understand that a reaction WILL occur here because Cr is higher than Ag on the activity series.


Step 1 is to write a balanced equation...

...and here is where I start to need help..

Cr(s) + AgNO3(aq) ---------->

now I know that the NO3 will no be with Cr ... CrN03 and that Ag will be by itself.

Cr(s) + AgNO3(aq) ----------> CrN03 + Ag

.........ok Cr^+1 and Ag is ^+1, how do i figure out the electron arrangement of NO3 again?


thanks








BTW the final answer is

Cr(s) + 2Ag^+(aq) ---------------> Cr^2+(aq) + 2Ag (down)
Nitrate ions are the spectators

 

[Gokul43201]

The final answer tells you the correct oxidation state of Cr-ions in this case (+2), but you have used +1 instead.

Redo the original equation (balance it correctly) and we can take it from there...

 

[ravenprp]

, so they do not need to be included in the ionic equation. Now, let's balance the charges on each side.

On the left side, we have Cr with a charge of 0 and Ag with a charge of +1, so the total charge is +1.

On the right side, we have Cr with a charge of +2 and Ag with a charge of 0, so the total charge is also +1.

To balance the charges, we need to add one more Ag ion on the left side.

Cr(s) + 2Ag^+(aq) ---------------> Cr^2+(aq) + 2Ag (down)

This is the ionic equation for the reaction. The CrN03 and the extra Ag ion are the products of the reaction.

I hope this helps! Let me know if you have any other questions.