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Topic: Redox Reaction questions  (Read 14671 times)

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Offline MakoEyes

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Redox Reaction questions
« on: October 08, 2009, 12:06:18 AM »
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.

Offline MrTeo

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Re: Redox Reaction questions
« Reply #1 on: October 08, 2009, 12:19:50 AM »
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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Offline MakoEyes

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Re: Redox Reaction questions
« Reply #2 on: October 08, 2009, 12:52:34 AM »
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.

Offline plankk

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Re: Redox Reaction questions
« Reply #3 on: October 08, 2009, 01:38:24 AM »
Li→Li++e- [x2]

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

Offline Borek

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Re: Redox Reaction questions
« Reply #4 on: October 08, 2009, 03:20:30 AM »
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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Offline MakoEyes

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Re: Redox Reaction questions
« Reply #5 on: October 08, 2009, 10:29:04 AM »
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?

Offline AWK

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Re: Redox Reaction questions
« Reply #6 on: October 08, 2009, 10:34:01 AM »
Li→Li++e-
then multiply twice
AWK

Offline MakoEyes

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Re: Redox Reaction questions
« Reply #7 on: October 08, 2009, 10:47:18 AM »
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?

Offline Borek

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Re: Redox Reaction questions
« Reply #8 on: October 08, 2009, 11:00:10 AM »
Li→Li++e-

That's a correct half reaction.

Quote
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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Offline MakoEyes

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Re: Redox Reaction questions
« Reply #9 on: October 08, 2009, 12:24:16 PM »
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)

Offline Borek

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Re: Redox Reaction questions
« Reply #10 on: October 08, 2009, 12:39:47 PM »
No, this is oxidation - metals charge goes up.
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