How to balance a redox equation in molecular form?

Question

I've spent a while trying to balance this equation in molecular form with no success.

$$\ce{Cu + HNO3 -> Cu(NO3)2 + 2NO + H2O}$$

I need to find the coefficient of HNO3, which is 8. However, this was my attempt:

$$\ce{Cu^0 + H + (NO3)- -> Cu^2 + 2(NO3)- + 2NO-2 + H2O }$$

$$\ce{Cu^0 -> Cu^2 + 2e- }$$ $$\ce{2x (NO3- + 2H^1 + e--> 2NO^0 + H2O)}$$

$$\ce{Cu^0 + 4H+ +2NO3- -> Cu^2+ + 4NO^0 + 2H2O + NO3^-}$$

Image of process: https://i.sstatic.net/QrlST.jpg

NO3 is the spectator polyatomic ion I left out during my balancing. I have no idea whether I did it right and where to go from here. Thanks!

Answer 1

Judging from the picture you attached you have successfully identified the species that gets oxidised $$\ce{Cu -> Cu^2+ + 2e-}\tag1$$ and reduced, i.e. $\ce{NO3-}$. While balancing this reaction you have made a few mistakes. You cannot leave it out as a spectator ion, since some of it is involved in the reaction.
Consider the oxidation state in nitrate and nitrogen monoxide. How many electrons have to be transferred?

three, nitrogen goes from $+5$ to $+2$ (you have only written one)

How many water molecules will be produced? How many protons are needed?

Two water, four protons

Write down the full half reactions

$$\ce{\overset{+5}{N}O3- + 4H+ + 3e- -> \overset{+2}{N}O + 2H2O}\tag2$$

Now you need to find the factors to multiply $(1)$ and $(2)$ with to balance out the electrons.

$3\times(1)$ and $2\times(2)$

This results in the following equation

$$\require{cancel}\ce{Cu + 2NO3- + 8H+ + \cancel{6e- } -> 3Cu^2+ + \cancel{6e- } + 2NO + 4H2O}$$

How many more nitrate ions do you need to charge balance the protons? What is the final balanced equation?

Six more nitrate ions, giving a total of eight $$\ce{Cu + 8HNO3 -> \underbrace{3Cu^2+ + 6NO3- }_{3Cu(NO3)2} + 2NO + 4H2O}$$