Copper reacts with concentrated nitric acid according to the following equation:
Cu(s)+4HNO3(aq)→Cu(NO3)2(aq)+2NO2(g)+2H2O(l) \text{Cu(s)} + 4\text{HNO}_3\text{(aq)} \rightarrow \text{Cu(NO}_3)_2\text{(aq)} + 2\text{NO}_2\text{(g)} + 2\text{H}_2\text{O(l)} Cu(s)+4HNO3(aq)→Cu(NO3)2(aq)+2NO2(g)+2H2O(l)Which substance is reduced during this reaction?
Cu(s)\text{Cu(s)}Cu(s)
HNO3(aq)\text{HNO}_3\text{(aq)}HNO3(aq)
Cu(NO3)2(aq)\text{Cu(NO}_3)_2\text{(aq)}Cu(NO3)2(aq)
NO2(g)\text{NO}_2\text{(g)}NO2(g)