How do we represent the oxidation of copper metal to #"cupric ion"#, with accompanying reduction of nitrate ion to #NO(g)#?

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Answer 1 · 2016-01-06T20:34:01

Copper metal is oxidized to cupric ion, #Cu^(2+)#. Nitrate (#N(V)#) is reduced to nitrous oxide (#N(II^+)#).

#"Oxidation half equation (i):"#

#Cu(s) rarr Cu^(2+) + 2e^-#

#"Reduction half equation (ii):"#

#NO_3^(-) + 4H^(+) + 3e^(-) rarr NO + 2H_2O# #(ii)#

#3xx(i) + 2xx(ii)#:

#3Cu(s) + 2NO_3^(-) + 8H^(+) rarr 3Cu^(2+) + 2NOuarr + 4H_2O#

Is this balanced with respect to both mass and charge? Should it be?