Redox reaction help!!!?
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The most reactive of the metals, which include the alkali metals, react with water to produce hydrogen gas. For example: 2Na(s) + 2H2O → 2Na+ + 2OH− + H2(g) Which of the following statements about this ReDox reaction are correct? Na is a reducing agent. Na is being oxidized. The oxidation number of Na decreases. O is being oxidized. Na is losing electrons.
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Answer:
Na --> Na^+ O^-2 ---> O^-2 H^+ ----> H^0 Na^+ is Oxidized so the reducing agent is Na Na is being oxidized Na is losing electrons
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Other answers
1) Na is a reducing agent since it is being oxidized (it reduces somebody else, the H in this case) 2) Na is being oxidized, its oxidation number changes from 0 (on Na(s)) to + 1 in Na+. 3) No, the oxidation number of Na(s) increases, from 0 to +1. 4) O does not change its oxidation number, starts as - 2 in H2O and ends at - 2 in OH-. Hydrogen is the element that is reduced, changing from + 1 in H2O to 0 in H2(g). 5) Na looses one electron per atom, changing from neutral (Na(s), oxidation number 0) to + 1 in Na+(aq).
Gladi ATorres
1) Na is a reducing agent since it is being oxidized (it reduces somebody else, the H in this case) 2) Na is being oxidized, its oxidation number changes from 0 (on Na(s)) to + 1 in Na+. 3) No, the oxidation number of Na(s) increases, from 0 to +1. 4) O does not change its oxidation number, starts as - 2 in H2O and ends at - 2 in OH-. Hydrogen is the element that is reduced, changing from + 1 in H2O to 0 in H2(g). 5) Na looses one electron per atom, changing from neutral (Na(s), oxidation number 0) to + 1 in Na+(aq).
Gladi ATorres
Na --> Na^+ O^-2 ---> O^-2 H^+ ----> H^0 Na^+ is Oxidized so the reducing agent is Na Na is being oxidized Na is losing electrons
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