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EMF of daniel cell


Arnav

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Consider a daniel cell.

according to nernst equation, the emf of the cell is given as E = E* - (RT.ln([Zn2+]/[Cu2+]))/nF

where E* = emf of the cell under standard conditions and the other symbols have their usual values.

Let the concentration of Cu2+ ions [Cu2+] be 1 M. As [Zn2+], the concentration of zinc ions in the oxidation half cell tends to zero, the equation predicts that the emf of the cell will approach infinity. How is this possible? What am i missing?

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Let's say i keep a zinc rod immersed in pure water as oxidation half cell, while a copper rod immersed in a 1 M solution of copper ions as reduction half cell. What then, will be the emf of the cell?

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16 hours ago, Arnav said:

Let's say i keep a zinc rod immersed in pure water as oxidation half cell, while a copper rod immersed in a 1 M solution of copper ions as reduction half cell. What then, will be the emf of the cell?

Yes, I'm very rusty on this but I think you raise quite a subtle point, actually.

Setting aside the electrochemistry angle, which gets in the way of the essential issue slightly, this is a general feature of the equation for Gibbs free energy, as a reaction proceeds: ΔG = ΔG⁰ + RT lnQ, Q being the reaction quotient, i.e. [products]/[reactants].

At the start, Q=0, so ln Q is undefined. Note however that this does not predict infinite Gibbs free energy or anything bonkers like that, as it is delta G i.e. dG/dξ, (using ξ to denote the reaction coordinate). So what this is saying is that the gradient of G, when graphed vs. extent of reaction, tends to infinity at the extremes. There is a discussion of this here: https://chemistry.stackexchange.com/questions/115634/calculating-Δg-at-the-extremes-of-reaction-extent/115701?noredirect=1#comment218308_115701

Returning to electrochemistry, since  ΔG = -zFE, you do indeed seem to get a theoretically infinite E, at the theoretical extremes in which the activities of products or reactants are mathematically zero. In practice there are a number of catches to that, however, since firstly such extremes are never quite realised in practice and secondly in practice kinetic effects are important, e.g. in your pure water example you have virtually no ionisation to allow the electrochemical reaction to get going.  

But I had to think quite hard about this and I may have got it a bit wrong after all these years, so I'd quite welcome comments from others. 

 

 

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