# Why does my voltaic pile produce a higher voltage then predicted?

In my chemistry class we are building our own voltaic cells and trying to produce as much current as possible with materials we are given using redox chemistry.

My lab partner made a battery without thinking about the actual chemistry and it produced about 1.7 V per cell and 5.5 V for 3 cells in series. We tried different voltmeters to confirm that our readings were not a hardware issue.

After consulting our textbook and our teacher we realized that the theoretical voltage produced from our battery should be .71 V per cell and about 2.1 V in series of 3 cells.

Our battery setup was as follows:

Mg(s)|Al(NO3)3(aq) || Mg(NO3)2(aq) | C(s)

with magnesium as an electrode and the strongest reducing agent and ionized aluminium as the strongest oxidizing agent with sodium nitrate in the salt bridge. I understand that the aluminium ion solution should be switched with the magnesium ion solution so that the aluminium doesn't cause the magnesium to oxidize.

My question is why and how does this battery produce such high voltage? We used distilled water to clean all of our beakers but even then I don't think a little bit of copper or other ions dissolved in our electrolyte would be enough to produce that much voltage and sustain it for the hour we spent trying to figure this out. So how did the battery produce a higher than expected voltage?

I have a diagram of what our battery looked like here https://docs.google.com/drawings/d/1u1ZjI9vHCNIeAACrRZmL3CqP_AgshX-HKg-00sS0rvc/edit?usp=sharing

• Did you use the nernst equation to account for using non-standard conditions? – Niels Kornerup Nov 7 '16 at 22:23
• No I had no idea it had existed as we had just started learning about this recently. I will look into the to figure out if the nernst equation provides the extra voltage that was recieved. – user37115 Nov 12 '16 at 6:25