Chemistry Metals

• Jasmine ngjiamin

  13 Sep 11, 22:09

  hi, how do u do this question?

  chromium Cr----------> insoluble in water---->reacts to give a green solution

  chromium(III) oxide--->insoluble in water---->reacts to give a green solution

  copper CU ------------->insoluble in water---->does not react

  copper (II) oxide------->insoluble in water---->reacts to give a blue solution

  a) construct an equation to show the reduction of copper (II) oxide by chromium.

  b) at the end of heating experiment, all 4 substances were present in the reaction mixture. suggest how pure dry copper can be obtained from this mixture.

  thank you.

• UltimaOnline

  13 Sep 11, 23:34

  Originally posted by Jasmine ngjiamin:

  hi, how do u do this question?

  chromium Cr----------> insoluble in water---->reacts to give a green solution

  chromium(III) oxide--->insoluble in water---->reacts to give a green solution

  copper CU ------------->insoluble in water---->does not react

  copper (II) oxide------->insoluble in water---->reacts to give a blue solution

  a) construct an equation to show the reduction of copper (II) oxide by chromium.

  b) at the end of heating experiment, all 4 substances were present in the reaction mixture. suggest how pure dry copper can be obtained from this mixture.

  thank you.

   

  OS = Oxidation State (also known as ON = Oxidation Number)

  a) Copper is reduced from CuO (OS of +2) to Cu (OS of 0), whilst Cr is oxidized from Cr (OS of 0) to Cr3+ (OS of +3). To ensure the overall redox equation is correctly balanced, you're advised to write reduction and oxidation half-equations (this is not usually taught at 'O' levels, but you can self-learn this for your own benefit).

  [O]   3H2O + 2Cr ---> Cr2O3 + 6H+ + 6e-

  [R]   2H+ + 2e- + CuO ---> Cu + H2O

  Multiply the reduction half-equation by 3 to equate the number of electrons with the oxidation half-equation

  [R]   6H+ + 6e- + 3CuO ---> 3Cu + 3H2O

  [Balanced Redox]   2Cr + 3CuO ---> Cr2O3 + 3Cu

   

  b) Use nitric(V) acid HNO3(aq), which will react with the basic oxides CuO and Cr2O3 (to form soluble copper(II) and chromium(III) nitrate(V) salts, and (HNO3) will also react with chromium metal (in a metal-acid redox reaction) to generate soluble chromium(III) nitrate(V) salts. But as long as dilute (rather than concentrated*) nitric(V) acid is used, copper will not react, and therefore can be separated as the residue of the filtration process.

  (* this is not usually taught at 'O' levels, but concentrated nitric(V) acid is a sufficiently powerful oxidizing agent that is able to oxidize copper and other usually non-reactive metals.)

• Efryterfsdrwef

  14 Sep 11, 10:04

  Carpe diem. Cest la vie. Quid pro quo. Veni, vidi, vici.