Use half-cells to balance the following redox reaction, and label the oxidation and reduction half-reactions.

Question

Use half-cells to balance the following redox reaction, and label the oxidation and reduction half-reactions.

Also, note that bromide is a spectator ion and can be eliminated.
FeBr(aq) +Mg(s) = Fe(s) + MgBr2(aq)
Fe^+2 -2e +Br^- e+ + Mg^+2 -2e =Fe^+2 -2e +MgBr2(aq) -2e e+

Idk what to from here?

I cancel out everything right?

Answer 1

You want to write half reactions but I think you made a typo with FeBr. Should it have been FeBr2

Fe^2+ 2e --> Fe
Mg --> Mg^2+ + 2e
-------------------
Then add the equations.

Answer 2

Use half-cells to balance the following redox reaction, and label the oxidation and reduction half-reactions.

Also, note that bromide is a spectator ion and can be eliminated.

FeBr3(aq) + Mg(s) à Fe(s) + MgBr2(aq)

Answer 3

To balance the redox reaction using half-cells, you need to separate the reaction into two half-reactions: one for the oxidation process (loss of electrons) and one for the reduction process (gain of electrons).

First, let's identify the oxidation half-reaction:
FeBr(aq) → Fe(s) + Br^-(aq)

In this half-reaction, the iron (Fe) is going from a higher oxidation state (+2) to a lower oxidation state (0) and losing two electrons. Therefore, the oxidation half-reaction is:
FeBr(aq) → Fe(s) + 2e^-

Next, let's identify the reduction half-reaction:
Mg(s) → MgBr2(aq)

In this half-reaction, the magnesium (Mg) is going from its elemental form (0 oxidation state) to a higher oxidation state (+2) by gaining two electrons. Therefore, the reduction half-reaction is:
Mg(s) + 2e^- → MgBr2(aq)

Now that you have separated the redox reaction into its half-reactions, you can balance them individually. To balance the number of electrons transferred in both half-reactions, you can multiply the half-reactions by appropriate coefficients.

To balance the oxidation half-reaction, we need to multiply it by 2:
2FeBr(aq) → 2Fe(s) + 4e^-

To balance the reduction half-reaction, we need to multiply it by 2:
2Mg(s) + 4e^- → 2MgBr2(aq)

Now, add the two balanced half-reactions together to get the overall balanced redox reaction:
2FeBr(aq) + 2Mg(s) + 4e^- → 2Fe(s) + 2MgBr2(aq) + 4e^-

Finally, cancel out the electrons on both sides of the equation:
2FeBr(aq) + 2Mg(s) → 2Fe(s) + 2MgBr2(aq)

Note that bromide (Br^-) is a spectator ion and can be eliminated because it appears on both sides of the equation.