Q in the Nernst Equation

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Niharika Reddy 1D
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Q in the Nernst Equation

Postby Niharika Reddy 1D » Tue Feb 03, 2015 5:01 pm

Why is it that partial pressure and concentration can be combined when calculating Q? For instance, for 13.41 a and c (I have attached a picture of the problem below), the solutions combine the given partial pressures of the gases with the concentrations of aqueous species to find Q, and multiply Q in the Nernst equation by 1.01325^2 for part c. I think this is to convert the units of pressure into bar but I'm not sure. If so, should we always convert partial pressures to bar when using them to calculate Q?

Kayla Denton 1A
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Re: Q in the Nernst Equation

Postby Kayla Denton 1A » Tue Feb 03, 2015 5:55 pm

My TA told us in class today that 1 bar is approximately equal to 1 mol/L, so that's probably why the solutions manual is converting partial pressures to bar (by multiplying by 1.01325^2) and then using this value in the Q expression alongside concentration values (as now bar is essentially the same as units of concentration).

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Re: Q in the Nernst Equation

Postby Chem_Mod » Wed Feb 04, 2015 2:19 am

The TA's explanation was not quite clear, but it is true that the purpose of multiplying 1.01325^2 is to convert two pressure values from atm to bar.

The exact reasons for this are quite technical and relate to the concept of standard states. In the end, if you want to use a "mixed" equilibrium expression, containing both concentrations and pressures, then the pressures must be in bars, and concentrations in molarity.

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