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anonymous

  • 4 years ago

In the reaction Cu+HNO3 →Cu2+ +NO2 +H2O Deduce the oxidation state of all the elements on both the reactant and the product side.

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  1. Xishem
    • 4 years ago
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    LHS: Cu: 0 H: +1 O: -2 N: +5 RHS: Cu: 0 N: +4 O: -2 H: +1 O: -2

  2. anonymous
    • 4 years ago
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    Rules for deciding oxidation number. Anything in elemental state is 0 (eg., Cu or Na or O2) All alkali metals(Na,K,Li,Rb) in any compunds are +1. eg, NaCl. Oxygen is always -2.(except O2 and O3). Hydrogen is always +1. All other elements areobtained by balancing oxidation number and charge If you apply these rules Cu is in elemental state on the left hence it is 0. H is always +1 and O is -2. In HNO3 net charge is 0. Hence sum of all oxidation numbers is0.If N has oxidation number x So 1-2*3+x=0 x=5. Hence N=5. Now Cu in Cu2+ is obviously +2. In NO2 O is -2 and net charge is 0. So x-2*2=0 x=4.

  3. anonymous
    • 4 years ago
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    which reactant was oxidised and which was reduced?

  4. Xishem
    • 4 years ago
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    When a reactant is oxidized, it means that it raises in oxidation number. By this... Cu was oxidized. N was reduced.

  5. anonymous
    • 4 years ago
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    thank you, ive never done chemistry before so it's taking awhile to all make sense

  6. Xishem
    • 4 years ago
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    Be sure you realize I wrote the oxidation for Cu(2+) wrong in my post. I didn't see the plus sign in your original equation. As shankvee pointed out, its oxidation number is actually +2.

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