What would the two half reactions be for 2Na+1MgBr2 arrow 2NaBr + Mg? I don't know how to determine this, other than that you separate the elements for the two half reactions. My other question is that when you're translating words and you have an element with Roman numerals such as Lead II, would it be Pb^2 or Pb2 (as a subscript?) Thanks for clearing this up.

Question

What would the two half reactions be for 2Na+1MgBr2 arrow 2NaBr + Mg? I don't know how to determine this, other than that you separate the elements for the two half reactions.  My other question is that when you're translating words and you have an element with Roman numerals such as Lead II, would it be Pb^2 or Pb2 (as a subscript?) Thanks for clearing this up.

DrBob222 · Answer

Separating the elements you see in the molecular equations into the two half reactions is EXACTLY what you do. 2Na + MgBr2 ==> 2NaBr + Mg The two half reactions are Na ==> Na^+ + e Mg^+2 + 2e ==> Mg I just wrote the ionic half reactions. Since Br is a spectator ion (notice it didn't change oxidation state; i.e., it is -1 on the left and -1 on the right.) To the second question, it depends. GENERALLY (that's most of the time), lead(II) means Pb^+2 and that goes for iron(II) or iron(III) or Cr(III) etc. However, in some cases, such as mercury(I), that is written as Hg₂²⁺ because Hg(I) is found as the dimer. [Interestingly, mercury(II) doesn't do that.] Mercury(I) chloride's formula is Hg₂Cl₂.