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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: http://imgur.com/LysN7FZ.png

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!

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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 $$\ce{Cu + 8HNO3 -> \underbrace{3Cu^2+ + 6NO3- }_{3Cu(NO3)2} + 2NO + 4H2O}$$

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  • $\begingroup$ Thanks so much! I finally understood the mistakes I made and how to solve these problems. $\endgroup$ – Evan Apr 18 '16 at 7:33
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    $\begingroup$ I'd add that in the half-reactions the charge and the atoms have to balance. The utility is that you can use electrons which really don't just float around in aqueous solution. $\endgroup$ – MaxW Apr 18 '16 at 7:35

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