Question #7eb16

1 Answer
Nov 30, 2015

Answer:

Copper metal is oxidized to #Cu^(2+)#. The nitrogen of nitric acid is reduced from #V+# to #II+#.

Explanation:

Oxidation half equation:

#Cu^0 rarr Cu^(2+) + 2e^-# #(i)#

#Hstackrel(+V)NO_3 +3e^(-) + 3H^(+) rarr stackrel(+II)NO +2H_2O# #(ii)#

Are both equation balanced with respect to mass and charge? I think they are (and they must be for a correct answer).

So now I cross multiply, #3xx(i)+2xx(ii)# (I do this to eliminate the electrons! which are included in the redox equations as conceptual particles):

#3Cu + 2HNO_3 + 6H^(+)rarr 3Cu^(2+) + 2NO + 4H_2O#

Of course, since I am producing #Cu(NO_3)_2(aq)# I could write:

#3Cu(s) + 8HNO_3(aq) rarr 3Cu(NO_3)_2(aq) + 2NO(g) + 4H_2O(l)#

The last hurdle: is this equation balanced with respect to mass and charge? I leave it to you to decide. I might have made a mistake.