Write the balanced half reaction for the reduction and oxidation reactions for the redox reaction:
2 Mg (s) + O2 (g) -> 2 MgO (s)

Notice: for this question you do not have to include a charge of zero and you do not have to include the states.

Below is an example of the form of the answer:
Oxidation: Ca ->Ca2+ + 2 e-
Reducation: Sn4+ + 2e- -> Sn2+

For this part of this question you do not have to include a charge of zero and you do not have to include the states.

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  1. Tell me what the problem is with this.

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  2. I wasn't sure how to balance the rest of this, I got this far:

    Oxidation: 2Mg -> ____

    Reduction: O2 + 4e- -> _____

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  3. 2 Mg (s) + O2 (g) -> 2 MgO (s)

    Mg on the left must go to Mg on the right but the Mg on the right is an ion. It has a charge of 2+; therefore,
    2Mg --> 2Mg^2+ + 4e
    O2 + 4e ==> 2O^2-

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  4. Thank you so much!!!

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  5. 2 Mg (s) + O2 (g) -> 2 MgO (s)

    You have to look up the half potentials for each, and determine which is being reduced and which one is being oxidized. The reduction potential for O2 is higher, so Mg loses electrons. This means you have to reverse the reaction that you see in the chart for Mg:

    Mg <-----> Mg2+ + 2 e-

    O2+4e- <-----> 2 O

    Balance electrons for Mg:

    2 x (Mg <-----> Mg2+ + 2 e- )


    2Mg <-----> 2 Mg2+ + 4 e-

    O2+4e- <-----> 2 O

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  6. I thought I posted a response for this a while ago, but when I looked back at my browser, I saw that I didn't. Also, I omitted ^2- charge on the O, which Dr.Bob222 didn't do.

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