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Calculating E dot (1 Viewer)

H

housemouse

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Im not too sure if this is correct.
Do you change the negative or postive sign of the half cell that is written like this:
Zn --> Zn +2 + 2e- (i think this is called the oxidant or something like that. correct me if im wrong)

Then you just add the other half cell potenial value to that one???
 
P

pLuvia

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You have to use your standard potentials table, usually at the back of the periodic table

For that half cell it's 0.76 so yes
 

mitochondria

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There is a reason why the potentials in the table are also known as reduction potentials, and that is because the potentials quoted are the potentials for reduction reactions.

That might sound stupid (for stating the obvious) but many people have difficulties doing calculations involving potentials.

Now, whenever you turn the arrow of an equation on the table around (as you did for the reduction of Zn2+) you get an oxidation reaction and the potential of that equation is the same in magnitude to the reduction equation but has the opposite sign.

The equation that always works (for half cells under standard conditions) is:

Eo = E(reduction) - E(oxidation)

Where E(reduction) is the reduction potential for the reduction reaction and E(oxidation) is the reduction potential for the oxidation reaction. The " - " sign does all the magic for you :)


Hope that helps with the confusion ;)
 
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housemouse

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Can I just do this all the time:

Im given two half cell equations, I take out the standard potenials and look and see which is more negative and just change the sign of that and add them together?????
 

mitochondria

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Well... Yes and No.

Of course the one with a more negative potential means that it is more likely to be reduced than oxidise and you will naturally think that it is the oxidation reaction in your redox reaction. This is in fact true for questions like:

Given half cell A is such and such and half cell B is such and such, calculate Eo for this cell

However, if you recall electrolysis, this will be wrong because the signs will be the other way around.

I'm sure there are other more complicated senarios than electrolysis but I can't think of any :)

Bottom line: make sure you know which electrode is oxidation and which one is reduction. THEN you work the numbers out ;)

It's better to be safe than cleverly-lazy. ;)
 

Riviet

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housemouse said:
And I pronounce sine as SIN
Hehe, I have a friend that pronounces it like that. It always puts the class and teacher off when he mentions it in class. XD
 

funking_you

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I would like emphasise another point, which was breifly touched on above.

  • the table of standard reduction potentials gives in the HSC lists all the Eo values for reduction half reactions. These values are give in volts. HOWEVER, it is important to state that that these values are valid only under standard conditions, which in this case are 25oC, 1 atm AND 1 mol/L concentrations. Not many students are aware of the last condition, but it is important just to know. Eo values will change if the concentrations of the solutions is different to 1 mol/L. The good news is that in ALL HSC questions, it is assumed that the solution is 1 mol/L, even if the concentration of the solution is not given.


So the bottom line is, you should be able to STATE that the values in the table are valid for the standard conditions: 25oC, 1 atm AND 1 mol/L concentrations.
There was actually a question in a HSC Catholic Trail paper a few years ago testing whether student we aware of this and the importance of this. The question was worth 4 marks.


SIDE NOTE:
(in the shipwrecks option, you will encounter galvanic cells where the concentration of solution is not 1 mol/L, and you will see how this affected the reactions)


SIDE NOTE 2: If you are really interested in learning how concentrations can affect electrochemical cells, Google the 'Nernst Equation'.


Cheers,
George
 

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