I am using a $\pu{1.5 M}~\ce{KCl}$ solution as electrolyte and copper and aluminium electrodes but no salt bridge. So basically there is only one beaker with the two electrodes in it with the $\pu{1.5 M}~\ce{KCl}$ solution. I understand that its some variation of the standard Nernst Equation but most of the literature I saw applied the equation to cells with a salt bridge. I need help figuring out the reactions in the system and the potential generated by the cell.

  • $\begingroup$ Nernst equation does not care whether you have a single space or two half-cells with a bridge. $\endgroup$ – Ivan Neretin Feb 21 '19 at 7:39
  • $\begingroup$ It is a very good question. However, it is rather a non-trivial problem to calculate the potentials developed in your system. In order to apply the Nernst equation, the metal electrode must be in contact with its "own" ions. Currently, your Al and Cu in contact with K ions. You can certainly estimate the potential and the reactions from the standard half cells of aluminum and copper, however your actual potential can significantly lower than the predicted ones. As stated above the sole purpose of the salt bridge is to keep the metal ions separate. $\endgroup$ – M. Farooq Feb 21 '19 at 16:44
  • $\begingroup$ Let me see if I got that right, unless I use Cu or Al salts as electrolytes, the actual potential will be much less than the predicted one. That does make sense, we replaced Potassium Chloride with Calcium Chloride and then with Sodium Sulphite in varying concentrations only to get the same potential (roughly). I am not sure why that happened though. The way I understood it, Aluminium Chloride should be formed at the Al electrode and water should hydrolyse at the Cu one. $\endgroup$ – SVora Feb 22 '19 at 3:10

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