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Identify the oxidation-reduction reactions among the following:

1. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

2. 2Na(s) + Cl2(aq) → 2NaCl(s)

3. 2Mg(s) + O2(g) → 2MgO

1 and 2 only
2 and 3 only
1 and 3 only
All of 1, 2, and 3
None of 1, 2, and 3

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2 answers
  1. Oxidation is the loss of electrons. Reduction is the gain of e.electrons.
    For 1.
    Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

    Zn on left has oxidation state (OS) of 0. Zn on right has(OS) of +2. So that's a loss of 2e which makes Zn to Zn^2+ oxidation.
    Cu^2+ on the left has OS of 2+. On the right it is zero. That's a gain of 2e which makes Cu^2+ reduced.
    So check out 2 and 3 and you can answer the question.

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  2. Good Morning DrBob222, My answer is all of 1,2,and 3. Please let me know if I am on the right track.

    2) 2na=2na+2e (oxidation)
    Cl2+2e-=2Cl- (reduction)

    3) 2Mg=Mg2+ +4e (oxidation)
    O2+4e-=2O2 (oxidation)

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