[guayguay]

Hi, my experiment is to create a galvanic cell using ZnCl2 and solid Zn, CuCl2-2H2O and solid Cu, and a NaCl salt bridge. According to the half reactions Cu2+ is the oxidizing agent, and Zn is the reducing agent. Zn is the anode, and Cu is the cathode.

Anyway, my goal was to see if decreasing the concentration of ZnCl2 would also decrease the voltage of the galvanic cell. My reasoning was Ohm's Law V=IR, if current (I) is decreased due to less electrons (less concentration) and resistance (R) is constant, then the voltage (V) must decrease. Unfortunately my hypothesis was incorrect. There was little to no change in voltage. 3.08V with 1M of ZnCl2, 3.08V with 0.6M, and 3.05V with 0.2M.

I can't figure out why this is. Is it because of human errors? Maybe I should have decreased concentration more? Did I decrease the concentration of the wrong ion? I did not learn about Nernst's law or anything yet, but if it is necessary for an explanation I can try to learn it.

Help is appreciated!

[Borek]

If all you are doing is just measuring the voltage, you use voltmeter which by definition has a very high internal resistance. That in turn means very small currents, and the voltage shouldn't change by much - so your hypothesis was difficult to prove form the start (if possible at all).

You should try change the concentration tenfold and hundredfold, that would give more interesting results. And yes, that's where the Nernst equation comes into play.

[guayguay]

If all you are doing is just measuring the voltage, you use voltmeter which by definition has a very high internal resistance. That in turn means very small currents, and the voltage shouldn't change by much - so your hypothesis was difficult to prove form the start (if possible at all).

You should try change the concentration tenfold and hundredfold, that would give more interesting results. And yes, that's where the Nernst equation comes into play.

Thank you for the reply. But wouldn't a very high R and a very small I still give me an accurate V? I mean a big number multiplied by a small number would still produce an accurate answer right?

I can't change my concentrations because the experiment is over, but I will talk about that in the analysis, assuming my reasoning is correct? Is it?

[Borek]

Let me put it differently - for the IR part to play any important role you need external R to be similar to the internal resistance of the cell. Using voltmeter you are basically measuring open circuit voltage.

[guayguay]

Let me put it differently - for the IR part to play any important role you need external R to be similar to the internal resistance of the cell. Using voltmeter you are basically measuring open circuit voltage.

What? I thought in a voltmeter you needed the external R to be a lot bigger than the internal R of the cell, because you only want to use a miniscule charge from the battery as to not affect its current. If the external R was similar to the internal R of the cell, wouldn't the charge be split/shared? Therefore giving you an inaccurate result?

[Borek]

Let me put it differently - for the IR part to play any important role you need external R to be similar to the internal resistance of the cell. Using voltmeter you are basically measuring open circuit voltage.

What? I thought in a voltmeter you needed the external R to be a lot bigger than the internal R of the cell

No contradiction here. You measured cell voltage with a voltmeter hence the external resistance (that of the voltmeter) was orders of magnitude higher than the internal resistance (that of the cell) - they were not similar, so the IR part doesn't play any substantial role in the result of your measurement.