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#### candidateof

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« on: March 02, 2012, 04:39:56 PM »
The question says:

For the cell:

Pt(s), H2(g,2.3 bar)|HCl (aq,3.5 m)||HCl (aq,2.0 m)|H2(g,1.5 bar),Pt(s)

Calculate the EMF at 25o C

a) -20 mV
b)  8.9 mV
c) -8.9 mV
d)  7.5 mv
d) -7.5 mV

-----------------

E=(RT/zF)ln(K)= 0.0257*ln(2*1.5/2.3*3.5)= -0.02537 V

And it is not one of the choices, what do you think is the mistake?

#### Borek

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##### Re: Question About Calculating EMF
« Reply #1 on: March 02, 2012, 04:56:15 PM »
Write reaction equation.
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#### candidateof

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##### Re: Question About Calculating EMF
« Reply #2 on: March 03, 2012, 04:41:40 AM »
Write reaction equation.

Oxd rxn:

1/2H2(g, 2.3 bar)---> H+(3.5 m)+e-

Red rxn:

H+(2 m)+e---> 1/2H2(1.5 bar)

Net rxn:

1/2H2(g, 2.3 bar)+H+(2 m)--> H+(3.5 m)+1/2H2(1.5 bar)

==> E=0.0257ln(((3.5*(1.51/2))/(2*(2.3)1/2))=8.9 mV

and the right answer is -8.9 mV, again there is a problem!

#### candidateof

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##### Re: Question About Calculating EMF
« Reply #3 on: March 03, 2012, 05:33:40 AM »
But if we use E=Eo-RT/zF*lnK

then we will get -8.9 mV

When we use E=RT/zF*lnK   (1)
and when we use E=Eo-RT/zF*lnK   (2) ?

I think eq. (1) is used to calculate Eo, right?

in this question Eo=0

so we get that.

#### Borek

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##### Re: Question About Calculating EMF
« Reply #4 on: March 03, 2012, 08:32:26 AM »
Sign is often a problem, and at the moment I don't have a time to dig in.

In general, the safest approach is to:

1. Write separate Nernst equations for each half cell:

$$E = E_0 + \frac{RT}{nF}ln\frac{[Ox]}{[ Red]}$$

(you can also flip Ox and Red, just remember to change the sign; whichever convention you use result will be the same)

2. Calculate EMF as a difference of both potentials.

This way you eliminate several possible sources of errors. This approach can be used to derive more complicated overall equation for a full cell, but IMHO it is not worth the effort - unless you use it very often.
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#### candidateof

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##### Re: Question About Calculating EMF
« Reply #5 on: March 03, 2012, 04:17:18 PM »
Thank you, I got it now. But the equation you wrote doesn't include the pressure when it is different than 1 bar, right?

#### Borek

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##### Re: Question About Calculating EMF
« Reply #6 on: March 03, 2012, 04:45:22 PM »
I used Ox and Red just as a way of marking what goes where, regardless of how complicated the reaction equation is. For example

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

yields

$$E = E_0 + \frac {RT}{5F} \ln \frac {[MnO_4^-][H^+]^8}{[Mn^{2+}]}$$

(side with the oxidized form goes into numerator, side with the reduced form into denominator).
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#### candidateof

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##### Re: Question About Calculating EMF
« Reply #7 on: March 03, 2012, 04:56:24 PM »
I see, and the sign is confusing indeed.