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I read a statement given in the book and I did not understand it

"If any external battery is connected whose emf is more than the emf of galvanic cell ,then it will act as Electrolytic cell"

Also can you please explain what happens if we connect a external battery having emf less than that of galvanic cell?

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Think of a generic redox pair: $$\ce{Ox + ne- -> Red}$$ and use Nernst equation (and a few simplifications) to calculate its potential: $$ E = E^\circ + \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$

So, now you know what potential you would measure on such a half-cell. However, what would happen if you applied some arbitrary potential $E_{w}$ ?

Well, if $E_{w} > E$, then that also means that $$\frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]_w}{[Red]_w}\right) > \frac{\mathrm{RT}}{\mathrm{nF}}\mathrm{ln}\left(\frac{[Ox]}{[Red]}\right)$$

Finally, you get that this situation implies an increase of $\ce{Ox}$ at the expense of $\ce{Red}$, which is what you would expect from an electrolytic cell - the supply of electrical energy causes electrochemical reactions.

If you apply a potential that is below $E$, the only difference would be that $\ce{Ox}$ would decrease and $\ce{Red}$ would increase, but the overall treatment remains the same.

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  • $\begingroup$ That means if Ew<E then it still acts like galvanic cell, right ? $\endgroup$ – Dimenein Nov 17 '16 at 12:36
  • $\begingroup$ The thing that bothers me about sticking to names instead of the physical picture is that it may be very misleading. Galvanic cells are supposed to operate spontaneously, so, as long as nothing happens until you apply some potential, any value of $E_w$ should "turn" this cell into an electrolytic cell. $\endgroup$ – truffaut Nov 18 '16 at 1:40
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Truffaut's answer is comprehensive, I just thought I'd build further.

Given that the cell in question has a reversible setup, the result would be as specified in his answer. If the cell is not set up reversibly, (for instance most common galvanic cells (aka 'batteries') are set up for no charging) there is no chemical reaction to make use of the voltage difference, and the amps will increase until the voltage resistance drop into and through the electrolyte in the cell matches the external applied voltage.

$$ E=Eo + E\omega $$ This will generate heat, and will quite quickly burst the cell in question, and become a samsung note 7 type handgrenade.

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You didn't really specify how the batteries would be connected.

If the opposite poles were connected (both + to other -) then both batteries would be short circuited and a massive current flow would result.

If the same poles were connected (+ to + and - to -) of two batteries of different EMF values, then one battery would be discharging and one battery would be in the "charging" configuration.

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