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We have from Nernst equation:

$\mathrm{E_{cell} = E^\circ - \dfrac{RT}{nF} \ln Q_c}$

But I was wondering that in a simple cell like Daniel cell if concentration of $\ce{Zn^{2+}}$ keeps increasing and that of $\ce{Cu^{2+}}$ keeps decreasing, won't the value of $\mathrm{Q_c}$ keep increasing? In that case, won't that lead to a decreasing emf? If it does, then what's the use of such a cell that yields decreasing emf?

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  • $\begingroup$ Batteries don't last forever. Aren't you just making that claim? $\endgroup$ – Zhe Oct 5 '18 at 13:36
  • $\begingroup$ Note that because you have a logarithm, the value of $\ln Q_{c}$ stays relatively constant for larger changes in the value of $Q_{c}$, implying that the battery can maintain a reasonable EMF as you use it up. $\endgroup$ – Zhe Oct 5 '18 at 13:36
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But I was wondering that in a simple cell like Daniel cell if concentration of $\ce{Zn^{2+}}$ keeps increasing and that of $\ce{Cu^{2+}}$ keeps decreasing, won't the value of $\mathrm{Q_c}$ keep increasing? In that case, won't that lead to a decreasing emf?

Have you ever wondered how battery meters on phones, laptops and other electronics can tell how charged your battery is? The question you just asked is the answer. As energy from the battery is consumed, the voltage will drop which can be correlated to a percent charge if known.

If it does, then what's the use of such a cell that yields decreasing emf?

These cell are still useful. Astute engineers know this and design around it using batteries that provide more voltage than needed when charged. If I have a $\pu{3.3 V}$ circuit, I can use a $\pu{3.7 V}$ Lithium battery (note $\pu{3.7V}$ is the charged voltage) to run the device without damaging it and will continue to work until the battery declines to below $\pu{3.3V}$ (actual/rated voltage). Similarly car batteries are typically rated for $\pu{12V}$ but if you measure it with a galvanometer (voltmeter) you will find the voltage is on the order of $\pu{13-14V}$.

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  • $\begingroup$ Doesn't decreasing emf cause any problems? Decreasing emf would lead to lower current... $\endgroup$ – Archer Oct 23 '18 at 5:46

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