I would say number 3. This is because the primary concern in this half reaction is to balance the charges. The left hand side has a charge of 0 because there is only solid Zn. the other side as written is at 2+, so it needs a 2- somewhere to bring it to 0. This is done by adding 2e- to the right side of the equation making your half reaction Zn(s) --> Zn2+(aq) + 2e-
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