0
$\begingroup$

For an assignment in my Chemistry class we have to build a voltaic cell using materials that are provided to us and that outputs a voltage as close to 0.8v as possible. The list of materials is as follows:

  • Glass plate
  • 4 filter paper strips

1 ml of the following 1.0M solutions

  • HCl
  • NaOH
  • NaCl
  • Water

Small metal strips:

  • Cu
  • Fe
  • Pb
  • Mg
  • Zn

I originally thought of doing something like this:

example

This is essentially a small scale version of a 2 beaker voltaic cell with a salt bridge that you see in textbooks. I would attach a multimeter between the Zn and Cu strips and read a voltage.

What I'm confused by however, is that I'm not provided with any solutions of aqueous metal ions, which are required for the half reactions powering this cell to go forward. I know that this has something to do with the HCl and NaOH that I'm provided with, but can't figure out exactly how they factor in.

I've done quite a bit of googling but I'm completely stuck. How would one make a voltaic cell out of the materials I've listed? Thanks for any help in advance.

$\endgroup$
  • $\begingroup$ Try something along the lines of a lead acid battery en.wikipedia.org/wiki/Lead%E2%80%93acid_battery. It doesn't need any metal ions. $\endgroup$ – popgalop Jun 8 '16 at 1:33
  • 2
    $\begingroup$ You can dissolve some of the metals to have aqueous metal ions.. $\endgroup$ – Greg Jun 8 '16 at 1:41
1
$\begingroup$

If you just put a copper strip and a zinc strip into some NaCl solution and try to write the Nernst equation for the whole system, you will notice that the electrochemical potential will be undefined. Since the electrochemical potentials involve the cation concentrations which are all zero by preparation.

What this means is that there is infinite driving force for some metal to get oxidized into the solution until there is some equilibrium. Therefore, as one of the comments to your question indicate, you will be dissolving some of the metals into the solution by the very act of putting them into such an arrangement.

$\endgroup$
0
$\begingroup$

Also think in terms of metal/air battery - just be careful about the expected potentials. Mg would not be good in this case (potential is somewhat high - about 3100 mV vs. SHE as stated in - see DOI: 10.1039/C3MH00059A).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.