So we're doing oxidation-reduction reactions and batteries and voltages.

We did a lab where we would take two metals, put it in a mixture (bleach/water/vinegar/salt) and used a voltmeter to see the DC voltage.

Now we're supposed to 'calculate the potential difference' for some of the combinations, given this info:

"STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT25°C" (google it for the link since jishka won't let me post it)

so I'll just ask for one example: zinc and aluminum. the DC voltage I got was 0.213 if it matters. how do I do it? thanks :)

also if someone knows how to find which copper charge i'm supposed to use since there are three different ones..

First, the usual one to use for Cu is

Cu ==> Cu^2+ + 2e

The potential for the Al/Zn couple, using standard reduction potential (but from my text which is several years old--it may not match your data).
Al + 3e = Al(aq) Eo = about -1.66 red pot
Zn + 2e ==> Zn(aq) Eo = -0.763
Reverse the more negative value and add to the other one; ie.,
Al ==> Al^3+ + 3e 1.66
Zn^2+ + 2e ==> Zn -0.763
Sum = 1.66 + (-0.763) = 0.897 or close to that is what you should have obtained in the lab IF you used the right measuring equipment.

And if the solutions were 1M.

To calculate the potential difference for the combination of zinc and aluminum, you can use the standard reduction potentials. Here are the steps to follow:

1. Look up the standard reduction potentials for zinc and aluminum in aqueous solution at 25°C from the link you mentioned.

2. Identify the half-reactions for the oxidation and reduction processes for each metal.

- Zinc: Zn(s) → Zn2+(aq) + 2e-
- Aluminum: Al(s) → Al3+(aq) + 3e-

3. Determine the standard reduction potential for each half-reaction.

- Zinc: E°(Zn2+/Zn) = -0.763 V
- Aluminum: E°(Al3+/Al) = -1.67 V

4. Remember that the potential difference (voltage) of a cell is given by the reduction potential difference between the two half-reactions. The cell potential (Ecell) can be calculated as:

Ecell = E°(reduction) - E°(oxidation)

Therefore,
Ecell = E°(Al3+/Al) - E°(Zn2+/Zn)
= -1.67 V - (-0.763 V)
= -0.907 V

5. Note that the potential difference you measured using a voltmeter (0.213 V) is the actual voltage difference of the cell you constructed. The sign is reversed since you measured it as Zn/Al and not Al/Zn. So, you need to consider the absolute value of the measured voltage. Therefore, the potential difference for the zinc and aluminum combination is approximately 0.213 V.

Regarding the copper charge, the three different charges for copper correspond to the oxidation states +1, +2, and +3. You mentioned that you are working with a mixture of bleach/water/vinegar/salt, but that information may not be sufficient to determine the specific copper charge. It depends on the context and the specific compounds present. If you provide more information about the experimental setup or the specific compounds involved, I can help you further determine which copper charge to consider.