Why couldn\'t you substitute 3M H_2 SO_4 for concentrated HNO_3 in the transform

Question

Why couldn't you substitute 3M H_2 SO_4 for concentrated HNO_3 in the transformation from Cu(s) to [Cu(H_2 O)_6]^2+? The necessary acid is the NO_3 ^- ion, which is not present in H_2 SO_4 solution The necessary oxidizing agent is the N03" ion, which is not present in H_2 SO_4 solution The necessary reducing agent is the NO_3 ^- ion, which is not present in H_2 SO_4 solution The necessary base is the NO_3 ^- ion, which is not present in H_2 SO_4 solution Why couldn't you substitute concentrated HNO_3 solution for 3M H_2 SO_4 in Parts IV and V? Neither acid should be used If HNO_3 solution were used in Parts IV and V, Cu(OH)_2 would not be reduced to Cu by Zn, because HNO_3 solution oxidizes Cu to Cu(OH)_2 Both acids could be used If HNO_3 solution were used in Parts IV and V, [Cu(H_2 O)_6]^2+ ion would not be reduced to Cu by Zn, because HNO_3 solution oxidizes Cu to [Cu(H_2 O)_6]^2+

Explanation / Answer

Copper does not reach with dilute sulphuric acid as its reduction potential is higher than that of hydrogen. Copper cannot displace hydrogen in sulphuric acid.

But in case of conc. Sulphuric acid, a redox reaction occurs wherein sulphuric acid is the oxidizing agent and Copper is the reducing agent.

Cu(s) + H2SO4 +2H+ --> Cu+2(aq) + SO2 +2H2O

Similarly, copper is oxidized by concentrated nitric acid to form Cu+2 ions.

Cu(s) + 4HNO3(aq) --> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Water displaces the nitrate ions to form {Cu(H2O)6}2+ and the colour changes from green to blue. This oxidized Copper complex with water can be reduced to Cu by zinc which has a lower reduction potential than Cu.

4. B

5. D

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