What are the equations that describes the reaction of Cu and HNO3?

1 Answer
Jacob T.
Aug 2, 2018

#3 color(white)(l) "Cu"(s) + 8 color(white)(l) "HNO"_3 (aq) to 3 color(white)(l) "Cu"("NO"_3)_2 (aq) + 2 color(white)(l) "NO" (g) + 4 color(white)(l) "H"_2"O"(l)#
#1 color(white)(l) "Cu"(s) + 4 color(white)(l) "HNO"_3 (aq, "conc.") to 1 color(white)(l) "Cu"("NO"_3)_2 (aq) + 2 color(white)(l) "NO"_2 (g) + 2 color(white)(l) "H"_2"O"(l)#

Explanation:

Nitric (V) acid demonstrates strong oxidizing properties even at low concentrations. In the reaction between copper #"Cu"(s)# and nitric (V) acid #"HNO"_3(aq)#, copper is oxidized from #0# to #+2# while depending on the concentration of the #("NO"_3)^(-)# ion, nitrogen is reduced from #+5# to #+4# (as #"NO"_2#, high concentrations) or #+2# (as #"NO"#, low concentrations).

The oxidation of one copper atom releases #2# electrons that produce either #1 color(white)(l) stackrel(+4)("N")"O"_2 # molecule or #2//3 color(white)(l) stackrel(+2)("N")"O"# molecule from one #"H"stackrel(+5)("N")"O"_3# molecule. The two equations can thus be balanced with reference to their changes in oxidation states:

  • Copper #"Cu"# and #"NO"_2# are consumed/produced at a #1:2# ratio;
  • Copper #"Cu"# and #"NO"# are consumed/produced at a #3:2# ratio.

Reference
"Nitric acid", https://en.wikipedia.org/wiki/Nitric_acid#Reactions_with_metals

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