# current of a galvanic cell

I am having a galvanic cell with the following half cell equations:

Anode--> 2Al(s) + 6OH --> 2Al(OH)3 + 6e-

Cathode--> 6H+ + 6e- --> 3H2(g)

So my cell is producing 6 electrons. Is there a way to calculate the current (ampere) generated using 6 electrons, theoretically?

P.S. sorry for the bad formatting but when I tried OH$^-$, it did not give me OH superscript - as shown.

• Current is charge per time, so without a timeframe, nope. – PlasmaHH Mar 6 '16 at 23:13
• @user510: Use back-slash, $to start and end the inline MathJAX on this site. Double-$ will put formula on separate line. – Transistor Mar 6 '16 at 23:23
• is that going to have to be through making a discharge curve? To figure out time – user510 Mar 6 '16 at 23:25

From Wikipedia, the ampere:

The SI unit of charge, the coulomb, "is the quantity of electricity carried in 1 second by a current of 1 ampere". Conversely, a current of one ampere is one coulomb of charge going past a given point per second:

$$1~ A=1~C/s$$

and the coulomb:

It is equivalent to the charge of approximately $6.242×10^{18} (1.036 ×10^{−5} mol)$ protons, and −1 C is equivalent to the charge of approximately $6.242×10^{18}$ electrons.

So, from the above:

$$1~A = -6.242 \cdot 10^{18} ~ electrons~ per ~ second.$$

To figure out the current you'll need to know the number of electrons produced per second.