Balancing Redox Equations

Question

Cu + HNO₃ -> Cu(NO₃)₂ + No + H₂O

I can usually follow along when my professor does these, but I have trouble starting on my own.

I know I'm supposed to balance the equation and the charges, but I'm not sure how to do a more complex one with compounds instead of just elements.

Answer 1

Hopefully this will help you understand better.

You simply treat a compound like you would an element, because there is an element in that compound that is either being oxidized or reduced. However, we can eliminate spectator ions to make it easier.

For the problem you present, we have...

Cu(s) + HNO₃(aq) ==> Cu(NO₃)₂ + NO(g) + H₂O

A quick look tells us that Cu(s) is going to Cu(NO₃)₂ so Cu is changing from ox.no. zero to ox. no. 2+. This will be one half reaction, but NO₃^- is a spectator ion as it has the same ox.no. in HNO₃ as it is in Cu(NO₃)₂.

We also see that the NO₃ in HNO₃ is going to NO(g) so N is changing from ox.no. 5+ to 2+. This will be the other half reaction.

Now, to write the half reactions:

Cu(s) ==> Cu²⁺

NO₃^- ==> NO

Next, we will balance each half reaction for both mass and charge:

Cu(s) ==> Cu²⁺ ... balanced for mass

Cu(s) ==> Cu²⁺ + 2e- ... balanced for mass and charge

NO₃^- ==> NO

NO₃^- ==> NO + 2H2O ... balanced for oxygen

NO₃^- + 4H⁺ ==> NO + 2H2O ... balanced for oxygen and hydrogen so balanced for mass

NO₃^- + 4H⁺ + 3e-==> NO + 2H2O ... balanced for mass and charge

Finally, we need to equalize electrons for the oxidation and reduction half reactions, so multiply the oxidation half reaction by 3 and the reduction half reaction by 2 to get 6 e- in each.

3Cu(s) ==> 3Cu²⁺ + 6e- ... oxidation

2NO₃^- + 8H⁺ + 6e-==> 2NO + 4H2O ... reduction

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3Cu(s) + 2NO₃^- + 8H⁺ ==> 3Cu²⁺ + 2NO + 4H2O ... balanced net ionic equation

If you want the balanced molecular equation, we just add back the spectators to get...

3Cu(s) + 8HNO₃(aq) ==> 3Cu(NO₃)₂(aq) + 2NO(g) + 4H2O(l) ... balanced molecular equation