Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: MakoEyes on October 08, 2009, 12:06:18 AM
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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.
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Li+S→Li2S
Li→Li++e- [x2]
S+2e-→S2-
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 ;)
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Li+S→Li2S
Li→Li++e- [x2]
S+2e-→S2-
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.
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Li→Li++e- [x2]
It concerns all reaction, so it means to double all reagents (products and substrats).
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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.
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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?
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Li→Li++e-
then multiply twice
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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?
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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.
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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. :P)
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No, this is oxidation - metals charge goes up.