How Do I Balance This Redox Reaction Using Redox Number or Half Equation Method?

[Gorby]

Homework Statement

Can somebody help me balance this redox reaction using either the redox number method or the half equation method? Thank You

Na₂S + Ag + HNO₃ + Na₂CrO₄ ----> Ag₂S + Cr(NO₃)₃ + NaNO₃

Can you please explain your steps.

Thank You

Homework Equations

The Attempt at a Solution

I tried using both methods, but I find that I have to add AgNO₃ as a product to the right side. I've been told that I cannot do this, can someone explain why?

 

[Borek]

It can't be balanced as entered, as there is no hydrogen on the right, you need to add water to products.

There is no need for AgNO₃ on the RHS.

Have you recognized what is being oxidized and what is being reduced?

Chemically it doesn't make sense to me - sulfide is a reducing agent, nitric acid is an oxidizing agent, I don't see why they are just spectators.

 

[Gorby]

It can't be balanced as entered, as there is no hydrogen on the right, you need to add water to products.

There is no need for AgNO₃ on the RHS.

Have you recognized what is being oxidized and what is being reduced?

Chemically it doesn't make sense to me - sulfide is a reducing agent, nitric acid is an oxidizing agent, I don't see why they are just spectators.

Yes, our instructor taught us to add in water as a product while doing the solution.

This is what I get for my balanced net ionic equation, but I do not know what to do from there. Can someone help me put this back into the original equation:

8 H⁺ + CrO₄^2- + 3 Ag ---> 3 Ag⁺ + Cr³⁺ + 4 H₂O

I was told that I had to double the above, for the original equation to balance. I do not know why, and I am not sure how to do it either.

 

[Borek]

8 H⁺ + CrO₄^2- + 3 Ag ---> 3 Ag⁺ + Cr³⁺ + 4 H₂O

Thats OK.

I was told that I had to double the above, for the original equation to balance. I do not know why, and I am not sure how to do it either.

You need an even number of Ag⁺ for Ag₂S.

Simply multiply all coefficients by 2, add on the left enough S^2- to convert all Ag⁺ on RHS to Ag₂S, see how many Na⁺ you are missing and so on. One thing you can't do - you can't modify coefficients of the above redox equation separately, if you need to modify them, you have to modify all at the same time (this way redox skeleton stays always balanced).

 

[Gorby]

Thats OK.
You need an even number of Ag⁺ for Ag₂S.

Simply multiply all coefficients by 2, add on the left enough S^2- to convert all Ag⁺ on RHS to Ag₂S, see how many Na⁺ you are missing and so on. One thing you can't do - you can't modify coefficients of the above redox equation separately, if you need to modify them, you have to modify all at the same time (this way redox skeleton stays always balanced).

I see, that makes sense. I am wondering why it is not ok to have 3 Ag on the left side and then have Ag₂S along with an Ag⁺ ion by itself on the right side; so that there are 3 Ag on both sides.

 

[Borek]

It wouldn't be completely incorrect, just doesn't make much chemical sense.