Let s say a galvanic cell based on the reaction of zinc and copper half-cells. How is it possible to get the voltage at 1V in stead of 1,1V?

My guess: maybe by modifying the salt bridge in some way as this influences the charges in some way, but I am really not sure.

  • $\begingroup$ Impurity. (6 more characters to go...) $\endgroup$ – Kenny Lau Aug 19 '16 at 17:33
  • $\begingroup$ @KennyLau that was what I first thought, but doping was not the expected answer. Other possibilities: electrode size, $Zn(NO_3)_2$ while $CuSO_4$ , different concentrations or volumes $\endgroup$ – LandonZeKepitelOfGreytBritn Aug 19 '16 at 17:43

You can get a smaller potential if you alter the concentration of one or both ion solutions away from ideal 1M concentration. This behavior is described by the Nernst equation:

$$E_{cell}=E_{cell}^{o} - \frac{RT}{zF}lnQ$$

where $E_{cell}$ is the actual potential in the cell, $E_{cell}^o$ is the potential of the cell under ideal conditions, R is the ideal gas constant, T is the temperature, F is Faraday's constant, z is the number of electrons in the reduction oxidation reaction and Q is the reaction quotient. By altering Q to be greater than 1, the actual voltage of the cell dips below the ideal value.

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