# Does increasing one of the electrolytes in a galvanic cell increase the voltage?

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?

• Surface area does not affect the voltage of a cell. Temperature does affect the voltage but not much. The main effect is the concentration, according to the Nernst law. Apparently you do not know this law. Or do you ? In two words, in the zinc/copper cell, you may increase the cell voltage by increasing the copper concentration $\ce{[Cu^{2+}]}$ or decreasing the zinc concentration $\ce{[Zn^{2+}]}$ Mar 2 at 12:00
• It looks like you ask for explanation of electrochemistry basics. For fundaments, studying is better than asking. The latter is better for just filling the gaps. See electrochemistry basics on Hyperphysics, Libretexts and Wikipedia Mar 2 at 13:36

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)

$$\ce{[Cu(dmphen)_{2}]^2+}$$ 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.