Thank you very much!
It's too late to ask the teacher, so I'll go with the last option.
I have a few more questions: for the experiment I used a aluminium electrode in magnesium sulphate in the half-cell that was used to change the concentration (cause there was no magnesium), is this correct, what are the effects of not using a magnesium electrode (in the MgSO4), and how would I talk about this in the report?
Plus, I had trouble with getting the voltage to show up on the voltage meter, i suspect it's due to the use of aluminium instead of magnesium, is this correct?
Happy to help!
Yes, you are correct about the voltage. Your voltmeter won't show any voltage because your redox reaction in that half-cell simply cannot, spontaneously, occur. This is because the two species at play in that half cell here are Al solid metal and Mg2+ (which is the MgSO4 as the electrolyte). As seen in the electrochemical series, Mg is more reactive than Al (in other words, more likely to oxidise or lose electrons). So in order for your galvanic cell to work, the Mg would have to take in electrons (reduce) but it won't allow that because Mg is stronger than Al to do what it wants (so Mg2+ likes its ion state). You can also see this in the standard electrode potential table where the value for Mg reduction is much lower than Al suggesting that Mg does not like to be reduced.
To answer your first question, to make a galvanic cell work, the electrode is supposed to be either an inert metal or the metal of the solution. So, in your case, your aluminium electrode is neither inert nor (obviously) Mg (as MgSO4 is your electrolyte solution). If you place a metal (Mg) into its own metal solution (MgSO4) and keep your Cu electrode in your CuSO4 solution, then you will certainly get a voltage. This is because, now the Mg electrode will freely break down into its ions (as it likes to be oxidised) and the Cu2+ ions will deposit as copper solid onto the copper electrode (as it likes to be reduced). This movement of electrons (and ions) will produce a voltage equal to the difference of the electrode potentials in the table.
As for what you would do in the report, I recommend you talk about this theory I talked about above (with more elaboration) and thus, explain why you didn't get a voltage with your respective electrodes. The issue is that, assuming you didn't get any voltage values at all for your experiment, you wouldn't have any graphs/tables to do any analysis in which you may lose marks on the two criteria below:
- "appropriate application of algorithms, visual and graphical representations of data about chemical equilibrium systems or oxidation and reduction demonstrated by correct and relevant processing of data"
- "thorough identification of relevant trends, patterns or relationships"
In the end, you want to discuss all of these issues in your evaluation area where you talk about your errors, however, I really don't have any advice to help you in the two criteria above as you don't have any graphs/tables to process and analyse (really sorry about that). Remember, you will still have raw data and it will be all 0 as you didn't get any voltages.
I hope this helped but try the best you can to analyse and relate it to your theory/rationale. Again, feel free to ask more questions.