Write the balanced half reaction for the reduction and oxidation reactions for the redox reaction:

Question

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.

Answer 1

Tell me what the problem is with this.

Answer 2

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

Oxidation: 2Mg -> ____

Reduction: O2 + 4e- -> _____

Answer 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-

Answer 4

Thank you so much!!!

Answer 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

Answer 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.

Answer 7

To determine the balanced half-reactions for the redox reaction, we need to identify the species that are being oxidized and reduced.

In the given reaction: 2 Mg (s) + O2 (g) -> 2 MgO (s)

Mg is being oxidized, as it goes from an oxidation state of 0 to +2 in MgO. This means it loses electrons.

Oxygen is being reduced, as it goes from an oxidation state of 0 to -2 in MgO. This means it gains electrons.

Now, we can write the balanced half-reactions:

Oxidation half-reaction:
Mg (s) -> Mg2+ + 2e-

Reduction half-reaction:
O2 (g) + 4e- -> 2O2-

To balance the number of electrons in the two half-reactions, we multiply the oxidation half-reaction by 2 and the reduction half-reaction by 4:

2Mg (s) -> 2Mg2+ + 4e-
O2 (g) + 4e- -> 2O2-

Now, both half-reactions have the same number of electrons, so we can add them together to get the balanced overall redox reaction:

2Mg (s) + O2 (g) -> 2Mg2+ + 2O2-

This overall reaction is now balanced in terms of both mass and charge.