# How do I increase the amount of electrons transferred in a battery?

I made a simple battery with $\ce{Al}$ as anode electrode with $\pu{0.7 M}$ aluminum sulfate electrolyte. Platinum is my cathode electrode with $\pu{2 x 10^-4 M}$ $\ce{KOH}$ solution. I am generating $\ce{85 mA}$ (my cell produced $\pu{0.04 V}$ under a $\pu{47 \Omega}$ load). How can I increase amount of electrons to increase the current produced by my cell?

My half cell reactions are \begin{align} \ce{2Al &-> 2Al^3+ + 6e-}\\ \ce{3H2O + 6e- &-> 3H2 + 6OH-} \end{align}

• Can you provide a picture of your setup, and some more details of the shape/configuration/etc. of your electrodes? As the question sits, there's not enough information to help you debug your apparatus. – hBy2Py Mar 27 '17 at 4:48
• simply increasing your electrode areas by either physically using large electrodes, or by sandpapering your electrode surfaces to make rougher surfaces should help. I do agree with hby2py though, more details would help in debugging. – Burak Ulgut Mar 27 '17 at 6:10
• Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". – Martin - マーチン Mar 27 '17 at 11:37
• Have you acidified your aluminum sulfate electrolyte at all? Are you stirring either electrolyte chamber? – hBy2Py Mar 27 '17 at 22:43
• Also, why are you chilling your KOH? – hBy2Py Mar 27 '17 at 22:49