Oxidation Half-Reaction Paper

474 Words2 Pages
Step #1: Write the oxidation number of each element. Step #2: Write the half- reaction. In this case, there are three elements involved. The chlorine, Cl, did not change its oxidation number in the reactant to product. That’s why chlorine is called the spectator ion. Mg0  Mg+2 (Oxidation Half- Reaction) H+1 H20 (Reduction Half- Reaction) Step #3: Balance the atoms. Mg0  Mg+2 (Oxidation Half- Reaction) 2H+1 H20 (Reduction Half- Reaction) Step #4: Balance the charges. We can’t cancel out the number of electrons that we’ve added because it is not the same with the oxidation half- reaction and reduction half- reaction. Multiply the 1e to the common factor so that it will be equal to 2e. Mg0  Mg+2 + 2e (Oxidation Half- Reaction) 2[1e + 2H+1 H20] (Reduction…show more content…
a. agent b. Half- reaction c. oxidation number d. electrons 6.) What is the meaning of LEORA? a. Losing Electron is Oxidation, Reducing agent b. Lowering Electron is Oxidizing, Reduction agent c. Low Electron is Over, Reducing agent d. Low Electron is Over, Releasing agent 7.) What is GEROA? a. Go Electron is Releasing, Oxidizing agent b. Gaining Electrons is Reduction, Oxidizing agent c. Gaining Electron is Releasing, Over agent d. Go Electron is Reduction, Oxidizing agent 8.) Which of the following is an example of free element? a. NaCl b. MgCl c. Fe d. H2PO4 9.) A chemical that causes oxidation to occur. a. reducing agent b. oxidizing agent c. reactant d. product 10.) A chemical that causes reduction to occur a. reducing agent b. oxidizing agent c. reactant d. product 8 Answer Key Pre- Assessment 1. A 2. B 3. C 4. D 5. B 6. A 7.
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