[candidateof]

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 25^o C

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

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

My answer was:
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]

Write reaction equation.

[candidateof]

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*(1.5^1/2))/(2*(2.3)^1/2))=8.9 mV

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

[candidateof]

But if we use E=E^o-RT/zF*lnK

then we will get -8.9 mV

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

I think eq. (1) is used to calculate E^o, right?

in this question E^o=0

so we get that.

[Borek]

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.

[candidateof]

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]

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).

[candidateof]

I see, and the sign is confusing indeed.

Thank you for your replies.