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I can think of any other way to influence the value of $\Delta G$ than (1) change the temperature (2) providing an external source to overcome the negative cell potential

The relationship between $\Delta G$ and $E_{\mathrm{cell}}$ is $\Delta G =-nFE_{\mathrm{cell}}$

Between $\Delta H$, $\Delta S$ and $\Delta G$ is $\Delta G=\Delta H - T\Delta S$

Sometimes, temperature won't do anything to change the spontaneity. But if we change $\Delta G$ through $\Delta E_{\mathrm{cell}}$ by providing external voltage in excess of the negative voltage, and though $\Delta G$ we are also able to influence enthalpy and entropy(???).

This is the part I am not sure about. Besides the title question, a sub-question is: how in this way can we change the enthalpy and entropy?

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Natural variables of $\Delta G$ are $p$, $T$, and $N_i$.

$p$ = pressure

$T$ = Temperature

$N_i$ = number of particles (or number of moles)

The derivation showing this is listed on the Wikipedia.

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The Nernst Equation relates the reduction potential of a cell to the chemical reaction that is occurring in the cell, and is written:

$$E_{\mathrm{cell}}=E^\mathrm{o}-\frac{RT}{nF}\ln(Q)$$

$E_{\mathrm{cell}}$, and thereby $\Delta G$, can be affected by changes in the concentration of the ions in either half cell.

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  • $\begingroup$ But how will both enthalpy and entropy be affected in turn? $\endgroup$ – most venerable sir Apr 8 '15 at 1:15
  • $\begingroup$ Okay, so given: $E_{cell}=E^o−\frac{RT}{nF}ln(Q)$, $\Delta G=−nFE_{cell}$, and $\Delta G=\Delta H-\Delta S·T$, we can figure a few things out. We know that $\Delta H$ is a state function, so no matter the pathway we take for a given reaction, $\Delta H$ remains constant. We must, therefore, be affecting $\Delta S$ by changes in $Q$. An increase in the concentration of the products will increase $\Delta G$, meaning $\Delta S$ is decreasing. The opposite will be true for an increase in the concentration of the reactants. If you're still having trouble, try messing around with the equations. $\endgroup$ – ringo Apr 8 '15 at 1:35
  • $\begingroup$ You can't simply call only enthalpy a state function. A quick search shows a wiki list of state function. Both free energy and entropy are state functions. I think you need to adjust your terminology. $\endgroup$ – most venerable sir Apr 8 '15 at 13:08
  • $\begingroup$ I'll admit my initial reasoning may be faulty in saying that $\Delta H$ will not change because it is a state function, but I stand by the idea that $\Delta H$ won't be affected by a change in concentration. Since $\Delta H=\Delta U_{system}+P·\Delta V$ and $\Delta U_{system}=q-P·\Delta V$ by combining the two equations, we find that $\Delta H=q$, where $q$ is the heat flow into or out of the system. Since we are not adding or removing heat to the system by changing the concentration of solute, we can finally conclude that $\Delta H$ is not affected by changes in solute concentration. $\endgroup$ – ringo Apr 9 '15 at 3:38
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    $\begingroup$ $\Delta H^\circ$ is not the same as $\Delta H$. Both are state functions. But changing the concentration is changing the state. So $\Delta H$ will definitely change with solute concentration. $\endgroup$ – Curt F. May 7 '15 at 2:22

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