Writing half equation for reduction reaction

[Snorlax]

Are these two the same thing? Is one right, but the other not?

Cu²⁺ (aq) + 2e^- ---> Cu (s)           and,
Cu (aq) + 2e^- ---> Cu²⁺ (s)

A question has come up, and I get confused as to some questions the answers will have the first type, while others the latter.

Thanks :/

[brightsky]

the second one is wrong. compare charge on both sides. Cu can't really be reduced.

[Snorlax]

O₂ (g) +4e^- ---> 2O^2- (g)

Like, this is right?
But can you see the difference between the first (on first post) and this ^?
Like, how do you know which one to use?
I'm missing something really important :/

[brightsky]

both of them are reduction half-equations. Cu2+(aq) and O2(g) are getting electrons shoved into them, and becoming something different. an oxidation half-equation on the other hand is when a species loses electrons and becomes something different. don't get put off by the charges.

[lzxnl]

There are three parts to any half equation.
1. Whether it's reduction or oxidation. A reduction reaction will be written the same way as in the electrochemical series; species plus electrons giving you new species. Like, for instance, your reaction with oxygen gas and electrons. An oxidation reaction has electrons on the right hand side.

2. Balancing the charges. This is where your second reaction doesn't make sense. You're adding electrons to copper metal to get...a cation!? Oh my. Neutral plus negative unfortunately doesn't yield positive. The charges have to balance. Charge isn't created or destroyed. EVER. Prove that wrong and you'll win a Nobel prize.

3. Balancing coefficients. This isn't such a large problem.