[MakoEyes]

For the complete redox reaction given here, (a) break down each reaction into its half reactions; (b)
identify the oxidizing agent; and (c) identify the reducing agent.
2 Li + S -> Li2S

Here is my work:

(a) Li -> Li+ + 2e-
and S + 2e- -> S


(b) The oxidizing agent is S (because its losing electrons).
(c) The reducing agent is Li  (because its gaining electrons).

Is this correct? I'm not really confident in my techniques here.

[MrTeo]

Li+S→Li₂S

Li→Li⁺+e^- [x2]
S+2e^-→S^2-

2Li+S→Li₂S

Your b and c statements are correct... I don't understand why you wrote the reduction semireaction of sulphur without specifing charges and the one regarding lithium using 2 electrons instead of one (we don't need them if we work with only one atom) otherwise they would have been correct.

Hope it's all clear now anyway 

[MakoEyes]

Li+S→Li₂S

Li→Li⁺+e^- [x2]
S+2e^-→S^2-

2Li+S→Li₂S

Your b and c statements are correct... I don't understand why you wrote the reduction semireaction of sulphur without specifing charges and the one regarding lithium using 2 electrons instead of one (we don't need them if we work with only one atom) otherwise they would have been correct.

Hope it's all clear now anyway 

Hey MrTeo,

thanks a lot. When you say "x2" though do you mean 2 electrons so...

Li→Li⁺+2e^-

or something different?

I didn't really understand this concept until now. I have to focus in on understanding it. Thanks again for your help.

[plankk]

Li→Li⁺+e^- [x2]

It concerns all reaction, so it means to double all reagents (products and substrats).

[Borek]

Li→Li⁺+2e^-

Note: for a reaction to be balanced not only atoms must be balanced, but also charge. You have 0 on the left and 1-2=-1 on the right.

[MakoEyes]

Li→Li⁺+2e^-

Note: for a reaction to be balanced not only atoms must be balanced, but also charge. You have 0 on the left and 1-2=-1 on the right.

Okay.

So back to the original question: when writing the half-reactions should I put in the balanced coefficients or leave it as MrTeo has written it?

[AWK]

Li→Li⁺+e^-
then multiply twice

[MakoEyes]

Li→Li⁺+e^-
then multiply twice

But how do you write this as an answer?

Just
Li→Li⁺+e^-

Or
2 Li→ 2Li⁺+2e^-

Or
Li→Li⁺+e^- (x2)


I'm not sure if that last one is even acceptable?

[Borek]

Li→Li⁺+e^-

That's a correct half reaction.

2 Li→ 2Li⁺+2e^-

Or
Li→Li⁺+e^- (x2)

These two are acceptable as a way of signalling how you are going to add half reactions - but they won't be accepted as correct half reactions, as balanced reaction equation should have smallest possible integer coefficients.

[MakoEyes]

Thanks a lot.

Li→Li++e-

This is the reduction half-reaction, correct? And the other one is the oxidation half reaction, right?

(Just want to be sure. )

[Borek]

No, this is oxidation - metals charge goes up.