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In the following unbalanced chemical equation the oxidation state of nitrogen ($+5$) is both decreased ($+5$ to $+2$) and unchanged ($+5$), so I am unable to balance it .

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

My attempt: I tried by making equal decrease in oxidation number and increase in oxidation number but I am not getting the given answer.

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There are two half reactions:

  • elemental copper to cupric ion
  • nitrate ion to nitric oxide

The copper half reaction should be easy to balance, but the reduction of nitrate to nitric oxide is more complicated. start with what you know:

  1. Nitrate goes to nitric oxide, one to one.
  2. Then balance oxygen with water since it is aqueous, presumably.
  3. Then balance hydrogen with protons since it is acidic from nitric acid.
  4. Finally, balance the half reaction with free electrons, which should be three electrons for a single nitrogen atom in the reactants to be consistent with nitrogen going from +5 to +2.

There are different numbers of of electrons in the copper oxidation and nitrogen reduction half reactions at this point, so you have to multiply each half reaction to get the numbers of electrons in each equal to cancel out. The free protons should be converted into nitric acid by adding the appropriate amount of nitrate ion to each side. your final balanced equation should have a ratio of elemental copper to nitric acid of 3 to 8, respectively.

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Consider those nitrogens two different elements for a while. $$\ce{Cu + \underbrace{HNO3}_{for\;redox} + \underbrace{HNO3}_{not\;for\;redox}->}\dots$$ This you supposedly know how to balance. The "first" nitrogen is an oxidant and should be balanced as such; as for the "second", you need 2 of it per each $\ce{Cu}$.

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