Posted by Peter on Sunday, April 19, 2009 at 5:21pm.
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.
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