I'm trying to find the electrochemical equivalent for zinc in an experiment. I am carrying out electrolysis of zinc sulfate solution. According to Faraday's law, m=zq. I made a 0.1 M solution and kept time for electrolysis constant (1 min). To get a value for z I plotted the mass over the charge (I*t). The current was varied. I'm getting a linear trend (mass deposited increases with increase in current) but a huge discrepancy of 100%. Also in the experiment, after the electrolysis I noticed that the zinc separated and got suspended in the solution when I disconnected the electrode to measure the new mass. This could have affected results. Any reasons for this? I might be able to redo the experiment if someone knows how I can solve this or get better results. The metal ions should undergo reduction at the cathode. Maybe that's not happening?

P.s I also did it for copper sulfate and it was worse. Almost no copper stayed on the electrode after electrolysis Also I used graphite electrodes for this. Before this, for another experiment I electroplated zinc with copper and all of it deposited. Maybe get better results with zinc/copper electrodes for this experiment instead of graphite?

Regards, Haseeb

  • $\begingroup$ Try reducing the current density, by increasing the electrode's size or by reducing current. $\endgroup$ – Gert Dec 5 '17 at 18:04
  • $\begingroup$ @Gert But I am already varying current from 1 to 5 A in steps of 1A. But do you think it would be possible to the same experiment and get the same relationship if I use copper electrodes instead of graphite. Because I didn't get an issue with the copper electrodes. $\endgroup$ – 0Darkvoid0 Dec 5 '17 at 18:24
  • $\begingroup$ $5\ \mathrm{A}$ is very high current for a 0.1 M solution. Graphite really isn't suitable. Better use the respective metals. $\endgroup$ – Gert Dec 5 '17 at 18:49
  • $\begingroup$ Ok, I will try that then. Do you know if the relationship still holds, m=zQ? (keeping time constant of course.) $\endgroup$ – 0Darkvoid0 Dec 5 '17 at 20:43
  • $\begingroup$ Yes, of course. Faraday's Law, basically. $\endgroup$ – Gert Dec 5 '17 at 20:48

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