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I would like to know how you could increase the voltage in zinc and copper galvanic cell:
$$\ce{Zn(s) + Cu^2+ → Zn^2+ + Cu(s)}$$
I know that temperature and surface area is a factor that affects the voltage, but I would like to specifically know about how concentration changes the voltage. If I want to increase the voltage, which electrolyte do I change? The anode or cathode? And what's the reason for that?
As suggested in the comments, the key is Nernst equation. The fundation is the Gibbs free energy given by: $\Delta G = \Delta G^0 + RT \ln Q_r$ where $\Delta G^0$ is the standard free energy for the process, while $Q_r$ is the reaction ratio. Remember that the reaction ratio is the analogous quantity of the equilibrium constant, but with the non equilibrium concentration. Remember also that the Gibbs free energy is proportional to the voltage. Then you need to increase the reaction ratio.
If I want to increase the voltage, which electrolyte do I change?
Hope you are familiar with the term electrolyte which is the solution where ions flow in the electrochemical cell
The electrolyte solutions used here are $\ce{ZnSO4(aq)}$ solution and $\ce{CuSO4(aq)}$ solution
The anode or cathode?
Yes you can chose both or one of them
Now lets just consider the anode
The half cell reaction is,
$$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$
But more correctly it is,
$$\ce{Zn(s) + 6H2O(l) -> [Zn(H_{2}O)_6]^{2+}(aq) + 2e^-}$$
This reaction happens in acidic medium or may be neutral medium but since Zn is a amphoteric metal it likes to get dissolved in basic medium rather than acidic medium.
The new reaction,
$$\ce{Zn(s) + 4OH^-(aq) -> [Zn(OH)_4]^2-(aq) + 2e^- }$$
Standard reduction potentials for acidic and basic status are $\ce{E^° -0.762 V, -1.190 V }$ receptivity
Thus for the anode try changing the medium to basic by adding $\ce{NaOH}$ in excess
But for the cathode there is not much solution easy as above but you can try adding $\ce{NH3}$ or $\ce{CH3COOH}$ maybe cream of tartar $\ce{KC4H5O6}$ in excess these will make complex ions in cathode side of $\ce{Cu^2+}$ ions which will either increase or decrease the half cell value for anode. But in order to have a higher voltage of the final cell you need a lower half cell reduction potential for cathode and for the anode it is higher cell potential value. (I have given only few most of them are house hold chemicals you can test)
Here Check this link it has different $\ce{Cu}$ complexes and reduction potentials!
$\ce{[Cu(dmphen)_{2}]^2+}$ looks good with $\ce{E 0.590 V}$ default is $\ce{0.339 V}$ but not verified.
I suggest you to look for nerst equation for understanding how to increase the voltage other than changing electrolyte (except concentration) it is defined for only for one reaction equation. Look for standard reduction potential table for changing the metal.
